{
  "index": "2004-B-6",
  "type": "NT",
  "tag": [
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "Let $\\mathcal{A}$\nbe a non-empty set of positive integers, and let $N(x)$ denote\nthe number of elements of $\\mathcal{A}$ not exceeding $x$.\nLet $\\mathcal{B}$ denote the set\nof positive integers $b$ that can be written in the form $b = a - a'$ with\n$a \\in \\mathcal{A}$  and $a' \\in  \\mathcal{A}$. Let $b_1 < b_2 < \\cdots$\nbe the members of $\\mathcal{B}$,\nlisted in increasing order. Show that if the sequence $b_{i+1} - b_i$ is\nunbounded, then\n\\[\n\\lim_{x \\to\\infty}  N(x)/x  = 0.\n\\]\n\n\\end{itemize}\n\\end{document}",
  "solution": "\\textbf{First solution:}\n(based on a solution of Dan Bernstein)\nNote that for any $b$, the condition that $b \\notin \\calB$ already\nforces $\\limsup N(x)/x$ to be at most 1/2: pair off $2mb + n$ with\n$(2m+1)b+n$ for $n=1,\\dots, b$, and note that at most one member of each\npair may belong to $\\calA$.\nThe idea of the proof is to do something\nsimilar with pairs replaced by larger clumps, using long runs of\nexcluded elements of $B$.\n\nSuppose we have positive integers\n$b_0 = 1, b_1, \\dots, b_n$ with\nthe following properties:\n\\begin{enumerate}\n\\item[(a)] For $i=1, \\dots, n$,\n$c_i = b_i/(2b_{i-1})$ is an integer.\n\\item[(b)] For $e_i \\in \\{-1,0,1\\}$,\n$|e_1b_1 + \\cdots + e_n b_n| \\notin \\calB$.\n\\end{enumerate}\nEach nonnegative integer $a$ has a unique ``base expansion''\n\\[\na = a_0 b_0 + \\cdots + a_{n-1} b_{n-1} + m b_n \\qquad (0 \\leq a_i < 2c_i);\n\\]\nif two integers have expansions with the same value of $m$, and values\nof $a_i$ differing by at most 1 for $i=0, \\dots, n-1$, then their\ndifference is not in $\\calB$, so at most one of them lies in $\\calA$.\nIn particular,\nfor any $d_i \\in \\{0, \\dots, c_i - 1\\}$, any $m_0 \\in \\{0, 2c_0 - 1\\}$\nand any $m_n$, the set\n\\begin{multline*}\n\\{m_0 b_0 + (2d_1 +e_1)b_0 + \\cdots \\\\\n+ (2d_{n-1} + e_{n-1})b_{n-1} + (2m_n + e_n)b_n\\},\n\\end{multline*}\nwhere each $e_i$ runs over $\\{0,1\\}$,\ncontains at most one element of $\\calA$; consequently,\n$\\limsup N(x)/x \\leq 1/2^{n}$.\n\nWe now produce such $b_i$ recursively, starting with $b_0 = 1$ (and both\n(a) and (b) holding vacuously).  Given $b_0, \\dots, b_n$ satisfying (a)\nand (b), note that $b_0 + \\cdots + b_{n-1} < b_n$ by induction on $n$.\nBy the hypotheses of the problem, we can find a set $S_n$ of $6b_n$\nconsecutive integers, none of which belongs to $\\calB$. Let $b_{n+1}$\nbe the second-smallest multiple of $2b_n$ in $S_n$; then $b_{n+1} + x \\in\nS_n$ for $-2b_n \\leq x \\leq 0$ clearly, and also for $0 \\leq x \\leq 2b_n$\nbecause there are most $4b_n-1$ elements of $S_n$ preceding $b_{n+1}$.\nIn particular, the analogue of (b) with $n$ replaced by $n+1$ holds for\n$e_{n+1} \\neq 0$; of course it holds for $e_{n+1} = 0$ because (b) was\nalready known. Since the analogue of (a) holds by construction, we have\ncompleted this step of the construction and the recursion may continue.\n\nSince we can construct $b_0, \\dots, b_n$ satisfying (a) and (b) for any $n$,\nwe have $\\limsup N(x)/x \\leq 1/2^n$ for any $n$, yielding $\\lim N(x)/x =\n0$ as desired.\n\n\\textbf{Second solution:} (by Paul Pollack)\nLet $S$ be the set of possible values of $\\limsup N(x)/x$;\nsince $S \\subseteq [0,1]$ is bounded, it has a least upper bound $L$.\nSuppose by way of contradiction that $L>0$;\nwe can then choose $\\calA, \\calB$ satisfying the conditions of the problem such\nthat $\\limsup N(x)/x > 3L/4$.\n\nTo begin with, we can\ncertainly find some positive integer $m \\notin \\calB$, so that\n$\\calA$ is disjoint from $\\calA + m = \\{a +m: a \\in \\calA\\}$.\nPut $\\calA' = \\calA \\cup (\\calA + m)$ and let\n$N'(x)$ be the size of $\\calA' \\cap \\{1, \\dots, x\\}$; then\n$\\limsup N'(x)/x = 3L/2 > L$, so $\\calA'$ cannot obey the conditions\nof the problem statement. That is, if we let $\\calB'$ be the set of positive\nintegers that occur as differences between elements of $\\calA'$,\nthen there exists an integer $n$ such that among any $n$ consecutive\nintegers, at least one lies in $\\calB'$.\nBut\n\\[\n\\calB' \\subseteq \\{b + em: b \\in \\calB, e \\in \\{-1,0,1\\}\\},\n\\]\nso among any $n+2m$ consecutive integers, at least one lies in $\\calB$.\nThis contradicts the condition of the problem statement.\n\nWe conclude that it is impossible to have $L>0$, so $L = 0$\nand $\\lim N(x)/x = 0$ as desired.\n\n\\textbf{Remark:} A hybrid between these two arguments is to\nnote that if we can produce $c_1, \\dots, c_n$ such that\n$|c_i - c_j| \\notin \\calB$ for $i,j = 1, \\dots, n$, then\nthe translates $\\calA + c_1, \\dots, \\calA + c_n$ are disjoint\nand so $\\limsup N(x)/x \\leq 1/n$. Given $c_1 \\leq \\dots \\leq c_n$\nas above, we can then choose $c_{n+1}$ to be the largest element of a\nrun of $c_n+1$ consecutive integers, none of which lie in $\\calB$.\n\n\\end{itemize}\n\n\n\\end{document}",
  "vars": [
    "x",
    "a",
    "a_0",
    "a_i",
    "a_n-1",
    "b",
    "b_0",
    "b_1",
    "b_2",
    "b_i",
    "b_i-1",
    "b_i+1",
    "b_n",
    "b_n+1",
    "c_i",
    "c_1",
    "c_n+1",
    "d_i",
    "e_i",
    "i",
    "j",
    "m",
    "m_0",
    "m_n",
    "n",
    "N",
    "S_n",
    "A",
    "B",
    "L"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "coordxval",
        "a": "elementa",
        "a_0": "elementazero",
        "a_i": "elementaindex",
        "a_n-1": "elementaprev",
        "b": "elementb",
        "b_0": "elementbzero",
        "b_1": "elementbone",
        "b_2": "elementbtwo",
        "b_i": "elementbindex",
        "b_i-1": "elementbprev",
        "b_i+1": "elementbnext",
        "b_n": "elementblast",
        "b_n+1": "elementbafter",
        "c_i": "coefficientci",
        "c_1": "coefficientone",
        "c_n+1": "coefficientafter",
        "d_i": "digitdindex",
        "e_i": "epsilonindex",
        "i": "indexi",
        "j": "indexj",
        "m": "indexm",
        "m_0": "indexmzero",
        "m_n": "indexmlast",
        "n": "indexn",
        "N": "countn",
        "S_n": "setsequence",
        "A": "setalpha",
        "B": "setbeta",
        "L": "limitl"
      },
      "question": "Let $\\mathcal{setalpha}$\nbe a non-empty set of positive integers, and let $\\countn(\\coordxval)$ denote\nthe number of elements of $\\mathcal{setalpha}$ not exceeding $\\coordxval$.\nLet $\\mathcal{setbeta}$ denote the set\nof positive integers $\\elementb$ that can be written in the form $\\elementb = \\elementa - \\elementa'$ with\n$\\elementa \\in \\mathcal{setalpha}$  and $\\elementa' \\in  \\mathcal{setalpha}$. Let $\\elementbone < \\elementbtwo < \\cdots$\nbe the members of $\\mathcal{setbeta}$,\nlisted in increasing order. Show that if the sequence $\\elementbnext - \\elementbindex$ is\nunbounded, then\n\\[\n\\lim_{\\coordxval \\to\\infty}  \\countn(\\coordxval)/\\coordxval  = 0.\n\\]\n",
      "solution": "\\textbf{First solution:}\n(based on a solution of Dan Bernstein)\nNote that for any $\\elementb$, the condition that $\\elementb \\notin \\calsetbeta$ already\nforces $\\limsup \\countn(\\coordxval)/\\coordxval$ to be at most 1/2: pair off $2\\indexm\\elementb + \\indexn$ with\n$(2\\indexm+1)\\elementb+\\indexn$ for $\\indexn=1,\\dots, \\elementb$, and note that at most one member of each\npair may belong to $\\calsetalpha$.\nThe idea of the proof is to do something\nsimilar with pairs replaced by larger clumps, using long runs of\nexcluded elements of $\\setbeta$.\n\nSuppose we have positive integers\n$\\elementbzero = 1, \\elementbone, \\dots, \\elementblast$ with\nthe following properties:\n\\begin{enumerate}\n\\item[(a)] For $\\indexi=1, \\dots, \\indexn$,\n$\\coefficientci = \\elementbindex/(2\\elementbprev)$ is an integer.\n\\item[(b)] For $\\epsilonindex \\in \\{-1,0,1\\}$,\n$|\\epsilonindex_1\\elementbone + \\cdots + \\epsilonindex \\elementblast| \\notin \\calsetbeta$.\n\\end{enumerate}\nEach nonnegative integer $\\elementa$ has a unique ``base expansion''\n\\[\n\\elementa = \\elementazero \\elementbzero + \\cdots + \\elementaprev b_{n-1} + \\indexm \\elementblast \\qquad (0 \\leq \\elementaindex < 2\\coefficientci);\n\\]\nif two integers have expansions with the same value of $\\indexm$, and values\nof $\\elementaindex$ differing by at most 1 for $\\indexi=0, \\dots, \\indexn-1$, then their\ndifference is not in $\\calsetbeta$, so at most one of them lies in $\\calsetalpha$.\nIn particular,\nfor any $\\digitdindex \\in \\{0, \\dots, \\coefficientci - 1\\}$, any $\\indexmzero \\in \\{0, 2c_0 - 1\\}$\nand any $\\indexmlast$, the set\n\\begin{multline*}\n\\{\\indexmzero \\elementbzero + (2d_1 +e_1)\\elementbzero + \\cdots \\\\\n+ (2d_{n-1} + e_{n-1})b_{n-1} + (2\\indexmlast + e_n)\\elementblast\\},\n\\end{multline*}\nwhere each $\\epsilonindex$ runs over $\\{0,1\\}$,\ncontains at most one element of $\\calsetalpha$; consequently,\n$\\limsup \\countn(\\coordxval)/\\coordxval \\leq 1/2^{\\indexn}$.\n\nWe now produce such $\\elementbindex$ recursively, starting with $\\elementbzero = 1$ (and both\n(a) and (b) holding vacuously).  Given $\\elementbzero, \\dots, \\elementblast$ satisfying (a)\nand (b), note that $\\elementbzero + \\cdots + b_{n-1} < \\elementblast$ by induction on $\\indexn$.\nBy the hypotheses of the problem, we can find a set $\\setsequence$ of $6\\elementblast$\nconsecutive integers, none of which belongs to $\\calsetbeta$. Let $\\elementbafter$\nbe the second-smallest multiple of $2\\elementblast$ in $\\setsequence$; then $\\elementbafter + \\coordxval \\in\n\\setsequence$ for $-2\\elementblast \\leq \\coordxval \\leq 0$ clearly, and also for $0 \\leq \\coordxval \\leq 2\\elementblast$\nbecause there are most $4\\elementblast-1$ elements of $\\setsequence$ preceding $\\elementbafter$.\nIn particular, the analogue of (b) with $\\indexn$ replaced by $\\indexn+1$ holds for\n$e_{\\indexn+1} \\neq 0$; of course it holds for $e_{\\indexn+1} = 0$ because (b) was\nalready known. Since the analogue of (a) holds by construction, we have\ncompleted this step of the construction and the recursion may continue.\n\nSince we can construct $\\elementbzero, \\dots, \\elementblast$ satisfying (a) and (b) for any $\\indexn$,\nwe have $\\limsup \\countn(\\coordxval)/\\coordxval \\leq 1/2^{\\indexn}$ for any $\\indexn$, yielding $\\lim \\countn(\\coordxval)/\\coordxval =\n0$ as desired.\n\n\\textbf{Second solution:} (by Paul Pollack)\nLet $S$ be the set of possible values of $\\limsup \\countn(\\coordxval)/\\coordxval$;\nsince $S \\subseteq [0,1]$ is bounded, it has a least upper bound $\\limitl$.\nSuppose by way of contradiction that $\\limitl>0$;\nwe can then choose $\\calsetalpha, \\calsetbeta$ satisfying the conditions of the problem such\nthat $\\limsup \\countn(\\coordxval)/\\coordxval > 3\\limitl/4$.\n\nTo begin with, we can\ncertainly find some positive integer $\\indexm \\notin \\calsetbeta$, so that\n$\\calsetalpha$ is disjoint from $\\calsetalpha + \\indexm = \\{\\elementa +\\indexm: \\elementa \\in \\calsetalpha\\}$.\nPut $\\calsetalpha' = \\calsetalpha \\cup (\\calsetalpha + \\indexm)$ and let\n$\\countn'(\\coordxval)$ be the size of $\\calsetalpha' \\cap \\{1, \\dots, \\coordxval\\}$; then\n$\\limsup \\countn'(\\coordxval)/\\coordxval = 3\\limitl/2 > \\limitl$, so $\\calsetalpha'$ cannot obey the conditions\nof the problem statement. That is, if we let $\\calsetbeta'$ be the set of positive\nintegers that occur as differences between elements of $\\calsetalpha'$,\nthen there exists an integer $\\indexn$ such that among any $\\indexn$ consecutive\nintegers, at least one lies in $\\calsetbeta'$.\nBut\n\\[\n\\calsetbeta' \\subseteq \\{\\elementb + e\\indexm: \\elementb \\in \\calsetbeta, e \\in \\{-1,0,1\\}\\},\n\\]\nso among any $\\indexn+2\\indexm$ consecutive integers, at least one lies in $\\calsetbeta$.\nThis contradicts the condition of the problem statement.\n\nWe conclude that it is impossible to have $\\limitl>0$, so $\\limitl = 0$\nand $\\lim \\countn(\\coordxval)/\\coordxval = 0$ as desired.\n\n\\textbf{Remark:} A hybrid between these two arguments is to\nnote that if we can produce $\\coefficientone, \\dots, c_{n}$ such that\n$|\\coefficientci - c_{j}| \\notin \\calsetbeta$ for $\\indexi,j = 1, \\dots, \\indexn$, then\nthe translates $\\calsetalpha + \\coefficientone, \\dots, \\calsetalpha + c_{n}$ are disjoint\nand so $\\limsup \\countn(\\coordxval)/\\coordxval \\leq 1/\\indexn$. Given $\\coefficientone \\leq \\dots \\leq c_{n}$\nas above, we can then choose $\\coefficientafter$ to be the largest element of a\nrun of $c_{n}+1$ consecutive integers, none of which lie in $\\calsetbeta$.\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandstone",
        "a": "tangerine",
        "a_0": "porchlamp",
        "a_i": "blueguitar",
        "a_n-1": "swiftcabin",
        "b": "moonlight",
        "b_0": "greenfield",
        "b_1": "starflower",
        "b_2": "honeycomb",
        "b_i": "riverstone",
        "b_i-1": "quietbrook",
        "b_i+1": "mistyvalley",
        "b_n": "cloudberry",
        "b_n+1": "frostgarden",
        "c_i": "copperleaf",
        "c_1": "silverpine",
        "c_n+1": "amberhollow",
        "d_i": "sunlitpath",
        "e_i": "nightowl",
        "i": "driftwood",
        "j": "hillcrest",
        "m": "glimmering",
        "m_0": "whispering",
        "m_n": "lanternlit",
        "n": "brookside",
        "N": "meadowlark",
        "S_n": "starlitbay",
        "A": "frostwhale",
        "B": "thornapple",
        "L": "marigolds"
      },
      "question": "\nLet $\\mathcal{frostwhale}$\nbe a non-empty set of positive integers, and let $meadowlark(sandstone)$ denote\nthe number of elements of $\\mathcal{frostwhale}$ not exceeding $sandstone$.\nLet $\\mathcal{thornapple}$ denote the set\nof positive integers moonlight that can be written in the form moonlight = tangerine - tangerine' with\ntangerine \\in \\mathcal{frostwhale}  and tangerine' \\in  \\mathcal{frostwhale}. Let moonlight_1 < moonlight_2 < \\cdots\nbe the members of $\\mathcal{thornapple}$,\nlisted in increasing order. Show that if the sequence moonlight_{driftwood+1} - moonlight_{driftwood} is\nunbounded, then\n\\[\n\\lim_{sandstone \\to\\infty}  meadowlark(sandstone)/sandstone  = 0.\n\\]\n\n\\end{itemize}\n\\end{document}",
      "solution": "\\textbf{First solution:}\n(based on a solution of Dan Bernstein)\nNote that for any moonlight, the condition that moonlight \\notin \\calthornapple already\nforces \\limsup meadowlark(sandstone)/sandstone to be at most 1/2: pair off $2m moonlight + nightowl$ with\n$(2m+1)moonlight+nightowl$ for $nightowl=1,\\dots, moonlight$, and note that at most one member of each\npair may belong to \\calfrostwhale.\nThe idea of the proof is to do something\nsimilar with pairs replaced by larger clumps, using long runs of\nexcluded elements of thornapple.\n\nSuppose we have positive integers\ncloudberry_0 = 1, cloudberry_1, \\dots, cloudberry_brookside with\nthe following properties:\n\\begin{enumerate}\n\\item[(a)] For driftwood=1, \\dots, brookside,\n copperleaf_{driftwood} = cloudberry_{driftwood}/(2cloudberry_{driftwood-1}) is an integer.\n\\item[(b)] For nightowl_{driftwood} \\in \\{-1,0,1\\},\n$|nightowl_1 cloudberry_1 + \\cdots + nightowl_brookside cloudberry_brookside| \\notin \\calthornapple$.\n\\end{enumerate}\nEach nonnegative integer tangerine has a unique ``base expansion''\n\\[\ntangerine = porchlamp greenfield + \\cdots + blueguitar_{brookside-1} cloudberry_{brookside-1} + glimmering cloudberry_{brookside} \\qquad (0 \\leq porchlamp < 2copperleaf_{1});\n\\]\nif two integers have expansions with the same value of glimmering, and values\nof blueguitar_{driftwood} differing by at most 1 for driftwood=0, \\dots, brookside-1, then their\ndifference is not in \\calthornapple, so at most one of them lies in \\calfrostwhale.\nIn particular,\nfor any sunlitpath_{driftwood} \\in \\{0, \\dots, copperleaf_{driftwood} - 1\\}$, any whispering \\in \\{0, 2copperleaf_{0} - 1\\}$\nand any lanternlit, the set\n\\begin{multline*}\n\\{whispering greenfield + (2sunlitpath_1 +nightowl_1)greenfield + \\cdots \\\\\n+ (2sunlitpath_{brookside-1} + nightowl_{brookside-1})cloudberry_{brookside-1} + (2lanternlit + nightowl_{brookside})cloudberry_{brookside}\\},\n\\end{multline*}\nwhere each nightowl_{driftwood} runs over \\{0,1\\},\ncontains at most one element of \\calfrostwhale; consequently,\n$\\limsup meadowlark(sandstone)/sandstone \\leq 1/2^{brookside}$.\n\nWe now produce such cloudberry_{driftwood} recursively, starting with greenfield = 1 (and both\n(a) and (b) holding vacuously).  Given greenfield, \\dots, cloudberry_{brookside} satisfying (a)\nand (b), note that greenfield + \\cdots + cloudberry_{brookside-1} < cloudberry_{brookside} by induction on brookside.\nBy the hypotheses of the problem, we can find a set starlitbay_{brookside} of $6cloudberry_{brookside}$\nconsecutive integers, none of which belongs to thornapple. Let cloudberry_{brookside+1}\nbe the second-smallest multiple of $2cloudberry_{brookside}$ in starlitbay_{brookside}; then cloudberry_{brookside+1} + sandstone \\in\nstarlitbay_{brookside} for $-2cloudberry_{brookside} \\leq sandstone \\leq 0$ clearly, and also for $0 \\leq sandstone \\leq 2cloudberry_{brookside}$\nbecause there are most $4cloudberry_{brookside}-1$ elements of starlitbay_{brookside} preceding cloudberry_{brookside+1}$.\nIn particular, the analogue of (b) with brookside replaced by brookside+1 holds for\nnightowl_{brookside+1} \\neq 0$; of course it holds for nightowl_{brookside+1} = 0$ because (b) was\nalready known. Since the analogue of (a) holds by construction, we have\ncompleted this step of the construction and the recursion may continue.\n\nSince we can construct greenfield, \\dots, cloudberry_{brookside} satisfying (a) and (b) for any brookside,\nwe have $\\limsup meadowlark(sandstone)/sandstone \\leq 1/2^{brookside}$ for any brookside, yielding $\\lim meadowlark(sandstone)/sandstone =\n0$ as desired.\n\n\\textbf{Second solution:} (by Paul Pollack)\nLet $S$ be the set of possible values of $\\limsup meadowlark(sandstone)/sandstone$;\nsince $S \\subseteq [0,1]$ is bounded, it has a least upper bound marigolds.\nSuppose by way of contradiction that marigolds>0$;\nwe can then choose \\calfrostwhale, \\calthornapple satisfying the conditions of the problem such\nthat $\\limsup meadowlark(sandstone)/sandstone > 3marigolds/4$.\n\nTo begin with, we can\ncertainly find some positive integer glimmering \\notin \\calthornapple$, so that\n\\calfrostwhale$ is disjoint from \\calfrostwhale + glimmering = \\{tangerine +glimmering: tangerine \\in \\calfrostwhale\\}$.\nPut \\calfrostwhale' = \\calfrostwhale \\cup (\\calfrostwhale + glimmering)$ and let\nmeadowlark'(sandstone)$ be the size of \\calfrostwhale' \\cap \\{1, \\dots, sandstone\\}$; then\n$\\limsup meadowlark'(sandstone)/sandstone = 3marigolds/2 > marigolds$, so \\calfrostwhale'$ cannot obey the conditions\nof the problem statement. That is, if we let \\calthornapple'$ be the set of positive\nintegers that occur as differences between elements of \\calfrostwhale'$, then there exists an integer brookside$ such that among any brookside$ consecutive\nintegers, at least one lies in \\calthornapple'.$\nBut\n\\[\n\\calthornapple' \\subseteq \\{moonlight + nightowl glimmering: moonlight \\in \\calthornapple, nightowl \\in \\{-1,0,1\\}\\},\n\\]\nso among any brookside+2glimmering$ consecutive integers, at least one lies in \\calthornapple.$\nThis contradicts the condition of the problem statement.\n\nWe conclude that it is impossible to have marigolds>0$, so marigolds = 0$\nand $\\lim meadowlark(sandstone)/sandstone = 0$ as desired.\n\n\\textbf{Remark:} A hybrid between these two arguments is to\nnote that if we can produce copperleaf_1, \\dots, copperleaf_brookside$ such that\n$|copperleaf_{driftwood} - copperleaf_{hillcrest}| \\notin \\calthornapple$ for driftwood,hillcrest = 1, \\dots, brookside$, then\nthe translates \\calfrostwhale + copperleaf_1, \\dots, \\calfrostwhale + copperleaf_brookside$ are disjoint\nand so $\\limsup meadowlark(sandstone)/sandstone \\leq 1/brookside$. Given copperleaf_1 \\leq \\dots \\leq copperleaf_brookside$\nas above, we can then choose copperleaf_{brookside+1}$ to be the largest element of a\nrun of copperleaf_brookside+1$ consecutive integers, none of which lie in \\calthornapple$.\n\n\\end{itemize}\n\n\n\\end{document}"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "limitless",
        "a": "omegaelem",
        "a_0": "omegazero",
        "a_i": "omegacoef",
        "a_n-1": "omegaprev",
        "b": "alphavar",
        "b_0": "alphabase",
        "b_1": "alphafirst",
        "b_2": "alphasecnd",
        "b_i": "alphagen",
        "b_i-1": "alphaprev",
        "b_i+1": "alphanext",
        "b_n": "alphatop",
        "b_n+1": "alphasucc",
        "c_i": "gammadigi",
        "c_1": "gammafirst",
        "c_n+1": "gammansucc",
        "d_i": "deltagen",
        "e_i": "epsilonvar",
        "i": "zerovario",
        "j": "onevario",
        "m": "staticelem",
        "m_0": "staticzero",
        "m_n": "statictop",
        "n": "tinyindex",
        "N": "emptymap",
        "S_n": "voidblock",
        "A": "emptyset",
        "B": "additions",
        "L": "lowerlim"
      },
      "question": "Let $\\mathcal{emptyset}$\nbe a non-empty set of positive integers, and let $emptymap(limitless)$ denote\nthe number of elements of $\\mathcal{emptyset}$ not exceeding $limitless$.\nLet $\\mathcal{additions}$ denote the set\nof positive integers $alphavar$ that can be written in the form $alphavar = omegaelem - omegaelem'$ with\n$omegaelem \\in \\mathcal{emptyset}$  and $omegaelem' \\in  \\mathcal{emptyset}$. Let $alphafirst < alphasecnd < \\cdots$\nbe the members of $\\mathcal{additions}$,\nlisted in increasing order. Show that if the sequence $alphanext - alphagen$ is\nunbounded, then\n\\[\n\\lim_{limitless \\to\\infty}  emptymap(limitless)/limitless  = 0.\n\\]",
      "solution": "\\textbf{First solution:}\n(based on a solution of Dan Bernstein)\nNote that for any $alphavar$, the condition that $alphavar \\notin \\caladditions$ already\nforces $\\limsup emptymap(limitless)/limitless$ to be at most 1/2: pair off $2staticelem alphavar + tinyindex$ with\n$(2staticelem+1)alphavar+tinyindex$ for $tinyindex=1,\\dots, alphavar$, and note that at most one member of each\npair may belong to $\\calemptyset$.\nThe idea of the proof is to do something\nsimilar with pairs replaced by larger clumps, using long runs of\nexcluded elements of $additions$.\n\nSuppose we have positive integers\n$alphabase = 1, alphafirst, \\dots, alphatop$ with\nthe following properties:\n\\begin{enumerate}\n\\item[(a)] For $zerovario=1, \\dots, tinyindex$,\n$gammadigi = alphagen/(2alphaprev)$ is an integer.\n\\item[(b)] For $epsilonvar \\in \\{-1,0,1\\}$,\n$|epsilonvar_1 alphafirst + \\cdots + epsilonvar_{tinyindex} alphatop| \\notin \\caladditions$.\n\\end{enumerate}\nEach nonnegative integer $omegaelem$ has a unique ``base expansion''\n\\[\nomegaelem = omegazero alphabase + \\cdots + omegaprev alphaprev + staticelem alphatop \\qquad (0 \\leq omegacoef < 2gammadigi);\n\\]\nif two integers have expansions with the same value of $staticelem$, and values\nof $omegacoef$ differing by at most 1 for $zerovario=0, \\dots, tinyindex-1$, then their\ndifference is not in $\\caladditions$, so at most one of them lies in $\\calemptyset$.\nIn particular,\nfor any $deltagen \\in \\{0, \\dots, gammadigi - 1\\}$, any $staticzero \\in \\{0, 2gammafirst - 1\\}$\nand any $statictop$, the set\n\\begin{multline*}\n\\{staticzero alphabase + (2deltagen_1 +epsilonvar_1)alphabase + \\cdots \\\\\n+ (2deltagen_{tinyindex-1} + epsilonvar_{tinyindex-1})alphaprev + (2statictop + epsilonvar_{tinyindex})alphatop\\},\n\\end{multline*}\nwhere each $epsilonvar$ runs over $\\{0,1\\}$,\ncontains at most one element of $\\calemptyset$; consequently,\n$\\limsup emptymap(limitless)/limitless \\leq 1/2^{tinyindex}$.\n\nWe now produce such $alphagen$ recursively, starting with $alphabase = 1$ (and both\n(a) and (b) holding vacuously).  Given $alphabase, \\dots, alphatop$ satisfying (a)\nand (b), note that $alphabase + \\cdots + alphaprev < alphatop$ by induction on $tinyindex$.\nBy the hypotheses of the problem, we can find a set $voidblock$ of $6alphatop$\nconsecutive integers, none of which belongs to $\\caladditions$. Let $alphasucc$\nbe the second-smallest multiple of $2alphatop$ in $voidblock$; then $alphasucc + limitless \\in\nvoidblock$ for $-2alphatop \\leq limitless \\leq 0$ clearly, and also for $0 \\leq limitless \\leq 2alphatop$\nbecause there are most $4alphatop-1$ elements of $voidblock$ preceding $alphasucc$.\nIn particular, the analogue of (b) with $tinyindex$ replaced by $tinyindex+1$ holds for\n$epsilonvar_{tinyindex+1} \\neq 0$; of course it holds for $epsilonvar_{tinyindex+1} = 0$ because (b) was\nalready known. Since the analogue of (a) holds by construction, we have\ncompleted this step of the construction and the recursion may continue.\n\nSince we can construct $alphabase, \\dots, alphatop$ satisfying (a) and (b) for any $tinyindex$,\nwe have $\\limsup emptymap(limitless)/limitless \\leq 1/2^{tinyindex}$ for any $tinyindex$, yielding $\\lim emptymap(limitless)/limitless =\n0$ as desired.\n\n\\textbf{Second solution:} (by Paul Pollack)\nLet $S$ be the set of possible values of $\\limsup emptymap(limitless)/limitless$;\nsince $S \\subseteq [0,1]$ is bounded, it has a least upper bound $lowerlim$.\nSuppose by way of contradiction that $lowerlim>0$;\nwe can then choose $\\calemptyset, \\caladditions$ satisfying the conditions of the problem such\nthat $\\limsup emptymap(limitless)/limitless > 3lowerlim/4$.\n\nTo begin with, we can\ncertainly find some positive integer $staticelem \\notin \\caladditions$, so that\n$\\calemptyset$ is disjoint from $\\calemptyset + staticelem = \\{omegaelem +staticelem: omegaelem \\in \\calemptyset\\}$.\nPut $\\calemptyset' = \\calemptyset \\cup (\\calemptyset + staticelem)$ and let\n$emptymap'(limitless)$ be the size of $\\calemptyset' \\cap \\{1, \\dots, limitless\\}$; then\n$\\limsup emptymap'(limitless)/limitless = 3lowerlim/2 > lowerlim$, so $\\calemptyset'$ cannot obey the conditions\nof the problem statement. That is, if we let $\\caladditions'$ be the set of positive\nintegers that occur as differences between elements of $\\calemptyset'$,\nthen there exists an integer $tinyindex$ such that among any $tinyindex$ consecutive\nintegers, at least one lies in $\\caladditions'$.\nBut\n\\[\n\\caladditions' \\subseteq \\{alphavar + epsilonvar staticelem: alphavar \\in \\caladditions, e \\in \\{-1,0,1\\}\\},\n\\]\nso among any $tinyindex+2staticelem$ consecutive integers, at least one lies in $\\caladditions$.\nThis contradicts the condition of the problem statement.\n\nWe conclude that it is impossible to have $lowerlim>0$, so $lowerlim = 0$\nand $\\lim emptymap(limitless)/limitless = 0$ as desired.\n\n\\textbf{Remark:} A hybrid between these two arguments is to\nnote that if we can produce $gammafirst, \\dots, gammadigi$ such that\n$|gammadigi - gammadigi| \\notin \\caladditions$ for $i,j = 1, \\dots, tinyindex$, then\nthe translates $\\calemptyset + gammafirst, \\dots, \\calemptyset + gammadigi$ are disjoint\nand so $\\limsup emptymap(limitless)/limitless \\leq 1/tinyindex$. Given $gammafirst \\leq \\dots \\leq gammadigi$\nas above, we can then choose $gammansucc$ to be the largest element of a\nrun of $gammadigi+1$ consecutive integers, none of which lie in $\\caladditions$. \n\n"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "a": "hjgrksla",
        "a_0": "pcvldmfr",
        "a_i": "sngchtlb",
        "a_n-1": "mkqdrsve",
        "b": "lfhqznco",
        "b_0": "rmpxagfu",
        "b_1": "uzvtgqke",
        "b_2": "nyhslvwo",
        "b_i": "tjrqmavb",
        "b_i-1": "xysdcvle",
        "b_i+1": "wavjkpto",
        "b_n": "dgmehutb",
        "b_n+1": "ftrdkezs",
        "c_i": "voseklqn",
        "c_1": "gyqnwklp",
        "c_n+1": "jqheflzt",
        "d_i": "kmrtbqsa",
        "e_i": "praqsmle",
        "i": "bczdfqgv",
        "j": "hlmxgrta",
        "m": "wvnrasjo",
        "m_0": "ysjntpke",
        "m_n": "qudpbfza",
        "n": "kejsdavn",
        "N": "fmgxavul",
        "S_n": "dpscnqwv",
        "A": "oprytieg",
        "B": "xvalerdp",
        "L": "gchmsdkw"
      },
      "question": "Let $\\mathcal{A}$\nbe a non-empty set of positive integers, and let $fmgxavul(qzxwvtnp)$ denote\nthe number of elements of $\\mathcal{A}$ not exceeding $qzxwvtnp$.\nLet $\\mathcal{B}$ denote the set\nof positive integers $lfhqznco$ that can be written in the form $lfhqznco = hjgrksla - hjgrksla'$ with\n$hjgrksla \\in \\mathcal{A}$  and $hjgrksla' \\in  \\mathcal{A}$. Let $uzvtgqke < nyhslvwo < \\cdots$\nbe the members of $\\mathcal{B}$,\nlisted in increasing order. Show that if the sequence $wavjkpto - tjrqmavb$ is\nunbounded, then\n\\[\n\\lim_{qzxwvtnp \\to\\infty}  fmgxavul(qzxwvtnp)/qzxwvtnp  = 0.\n\\]",
      "solution": "\\textbf{First solution:}\n(based on a solution of Dan Bernstein)\nNote that for any $lfhqznco$, the condition that $lfhqznco \\notin \\calB$ already\nforces $\\limsup fmgxavul(qzxwvtnp)/qzxwvtnp$ to be at most 1/2: pair off $2wvnrasjo lfhqznco + kejsdavn$ with\n$(2wvnrasjo+1)lfhqznco+kejsdavn$ for $kejsdavn=1,\\dots, lfhqznco$, and note that at most one member of each\npair may belong to $\\calA$.\nThe idea of the proof is to do something\nsimilar with pairs replaced by larger clumps, using long runs of\nexcluded elements of $xvalerdp$.\n\nSuppose we have positive integers\n$rmpxagfu = 1, uzvtgqke, \\dots, dgmehutb$ with\nthe following properties:\n\\begin{enumerate}\n\\item[(a)] For $bczdfqgv=1, \\dots, kejsdavn$,\n$voseklqn = tjrqmavb/(2xysdcvle)$ is an integer.\n\\item[(b)] For $praqsmle_{bczdfqgv} \\in \\{-1,0,1\\}$,\n$|praqsmle_1 uzvtgqke + \\cdots + praqsmle_{kejsdavn} dgmehutb| \\notin \\calB$.\n\\end{enumerate}\nEach nonnegative integer $hjgrksla$ has a unique ``base expansion''\n\\[\nhjgrksla = pcvldmfr rmpxagfu + \\cdots + mkqdrsve xysdcvle + wvnrasjo dgmehutb \\qquad (0 \\leq sngchtlb < 2voseklqn);\n\\]\nif two integers have expansions with the same value of $wvnrasjo$, and values\nof $sngchtlb$ differing by at most 1 for $bczdfqgv=0, \\dots, kejsdavn-1$, then their\ndifference is not in $\\calB$, so at most one of them lies in $\\calA$.\nIn particular,\nfor any $kmrtbqsa \\in \\{0, \\dots, voseklqn - 1\\}$, any $ysjntpke \\in \\{0, 2c_0 - 1\\}$\nand any $qudpbfza$, the set\n\\begin{multline*}\n\\{ysjntpke rmpxagfu + (2kmrtbqsa +praqsmle_1)rmpxagfu + \\cdots \\\\\n+ (2kmrtbqsa_{kejsdavn-1} + praqsmle_{kejsdavn-1})b_{kejsdavn-1} + (2qudpbfza + praqsmle_{kejsdavn})dgmehutb\\},\n\\end{multline*}\nwhere each $praqsmle_{bczdfqgv}$ runs over $\\{0,1\\}$,\ncontains at most one element of $\\calA$; consequently,\n$\\limsup fmgxavul(qzxwvtnp)/qzxwvtnp \\leq 1/2^{kejsdavn}$.\n\nWe now produce such $tjrqmavb$ recursively, starting with $rmpxagfu = 1$ (and both\n(a) and (b) holding vacuously).  Given $rmpxagfu, \\dots, dgmehutb$ satisfying (a)\nand (b), note that $rmpxagfu + \\cdots + b_{kejsdavn-1} < dgmehutb$ by induction on $kejsdavn$.\nBy the hypotheses of the problem, we can find a set $dpscnqwv$ of $6dgmehutb$\nconsecutive integers, none of which belongs to $\\calB$. Let $ftrdkezs$\nbe the second-smallest multiple of $2dgmehutb$ in $dpscnqwv$; then $ftrdkezs + qzxwvtnp \\in\ndpscnqwv$ for $-2dgmehutb \\leq qzxwvtnp \\leq 0$ clearly, and also for $0 \\leq qzxwvtnp \\leq 2dgmehutb$\nbecause there are most $4dgmehutb-1$ elements of $dpscnqwv$ preceding $ftrdkezs$.\nIn particular, the analogue of (b) with $kejsdavn$ replaced by $kejsdavn+1$ holds for\n$praqsmle_{kejsdavn+1} \\neq 0$; of course it holds for $praqsmle_{kejsdavn+1} = 0$ because (b) was\nalready known. Since the analogue of (a) holds by construction, we have\ncompleted this step of the construction and the recursion may continue.\n\nSince we can construct $rmpxagfu, \\dots, dgmehutb$ satisfying (a) and (b) for any $kejsdavn$,\nwe have $\\limsup fmgxavul(qzxwvtnp)/qzxwvtnp \\leq 1/2^{kejsdavn}$ for any $kejsdavn$, yielding $\\lim fmgxavul(qzxwvtnp)/qzxwvtnp =\n0$ as desired.\n\n\\textbf{Second solution:} (by Paul Pollack)\nLet $S$ be the set of possible values of $\\limsup fmgxavul(qzxwvtnp)/qzxwvtnp$;\nsince $S \\subseteq [0,1]$ is bounded, it has a least upper bound $gchmsdkw$.\nSuppose by way of contradiction that $gchmsdkw>0$;\nwe can then choose $\\calA, \\calB$ satisfying the conditions of the problem such\nthat $\\limsup fmgxavul(qzxwvtnp)/qzxwvtnp > 3gchmsdkw/4$.\n\nTo begin with, we can\ncertainly find some positive integer $wvnrasjo \\notin \\calB$, so that\n$\\calA$ is disjoint from $\\calA + wvnrasjo = \\{hjgrksla +wvnrasjo: hjgrksla \\in \\calA\\}$.\nPut $\\calA' = \\calA \\cup (\\calA + wvnrasjo)$ and let\nfmgxavul'(qzxwvtnp) be the size of $\\calA' \\cap \\{1, \\dots, qzxwvtnp\\}$; then\n$\\limsup fmgxavul'(qzxwvtnp)/qzxwvtnp = 3gchmsdkw/2 > gchmsdkw$, so $\\calA'$ cannot obey the conditions\nof the problem statement. That is, if we let $\\calB'$ be the set of positive\nintegers that occur as differences between elements of $\\calA'$, then there exists an integer $kejsdavn$ such that among any $kejsdavn$ consecutive\nintegers, at least one lies in $\\calB'$.\nBut\n\\[\n\\calB' \\subseteq \\{lfhqznco + e wvnrasjo: lfhqznco \\in \\calB, e \\in \\{-1,0,1\\}\\},\n\\]\nso among any $kejsdavn+2wvnrasjo$ consecutive integers, at least one lies in $\\calB$.\nThis contradicts the condition of the problem statement.\n\nWe conclude that it is impossible to have $gchmsdkw>0$, so $gchmsdkw = 0$\nand $\\lim fmgxavul(qzxwvtnp)/qzxwvtnp = 0$ as desired.\n\n\\textbf{Remark:} A hybrid between these two arguments is to\nnote that if we can produce $gyqnwklp, \\dots, c_{kejsdavn}$ such that\n$|voseklqn - c_{hlmxgrta}| \\notin \\calB$ for $bczdfqgv,hlmxgrta = 1, \\dots, kejsdavn$, then\nthe translates $\\calA + gyqnwklp, \\dots, \\calA + c_{kejsdavn}$ are disjoint\nand so $\\limsup fmgxavul(qzxwvtnp)/qzxwvtnp \\leq 1/kejsdavn$. Given $gyqnwklp \\leq \\dots \\leq c_{kejsdavn}$\nas above, we can then choose jqheflzt to be the largest element of a\nrun of $c_{kejsdavn}+1$ consecutive integers, none of which lie in $\\calB$.  "
    },
    "kernel_variant": {
      "question": "Fix an integer $d\\ge 2$ and write $\\mathbb N=\\{1,2,3,\\dots\\}$.\nFor $n=(n_1,\\dots ,n_d)\\in\\mathbb Z^{d}$ put\n\\[\n|n|_{\\infty}:=\\max_{1\\le i\\le d}|n_i|.\n\\]\n\nFor a non–empty subset\n\\[\n\\mathcal A\\subset\\mathbb N^{d}:=\\bigl\\{(n_1,\\dots ,n_d):n_i\\in\\mathbb N\\bigr\\}\n\\]\ndefine, for $x\\ge 1$,\n\\[\nN(x):=\\bigl|\\mathcal A\\cap[1,x]^{d}\\bigr|.\n\\]\n\nGlobal difference set\n\\[\n\\Delta(\\mathcal A):=\\{a-a':a,a'\\in\\mathcal A,\\;a\\ne a'\\}\\subset\\mathbb Z^{d}.\n\\]\n\nFor every coordinate $j\\in\\{1,\\dots ,d\\}$ put\n\\[\n\\mathcal B_{j}:=\\bigl\\{|a_j-a'_j|:a,a'\\in\\mathcal A,\\;a_j\\ne a'_j\\bigr\\}\n              =(b_{j,1}<b_{j,2}<\\dots)\\subset\\mathbb N .\n\\]\n\nSimultaneous coordinate lacunarity.\nAssume that there is a sequence of positive integers\n\\[\n1=L_{1},\\qquad L_{n+1}=4L_{n}\\qquad(n\\ge 1) \\tag{$\\Lambda$}\n\\]\nsuch that, for every $n\\ge 1$,\n\\[\n(L_{n},2L_{n}]\\;\\cap\\;\\mathcal B_{j}=\\varnothing\\qquad(j=1,\\dots ,d).\\tag{$\\diamondsuit_{n}$}\n\\]\nIn words:  for each $n$ the interval $(L_{n},2L_{n}]$ is simultaneously free of\none–dimensional differences along every coordinate direction.\n\nProblem. \nShow that the counting function of $\\mathcal A$ satisfies the power–saving\ndensity bound\n\\[\nN(x)\\;\\ll_{d}\\;x^{\\,d-\\tfrac12}\\qquad(x\\ge 4),\\tag{1}\n\\]\nwhere the implied constant depends only on $d$.  In particular\n\\[\n\\lim_{x\\to\\infty}\\frac{N(x)}{x^{\\,d}}=0.\\tag{2}\n\\]\n\nThe task is to deduce the exponent $\\tfrac12$ in (1) from the genuinely\n$d$–dimensional lacunarity \\((\\diamondsuit_{n})\\) together with the uniform\nradix–$4$ growth \\((\\Lambda)\\).\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout $c,c_{1},c_{2},\\dots$ denote positive constants depending only\non the ambient dimension $d$.\n\nStep 1.  Choosing “digit’’ vectors strictly inside the forbidden gaps.  \nFix once and for all the constant\n\\[\n\\alpha:=\\frac54\\quad\\Bigl(1<\\alpha<\\frac32\\Bigr).\n\\]\nFor every integer $n\\ge 2$ and every coordinate $j\\in\\{1,\\dots ,d\\}$ put\n\\[\nv_{n,j}:=\\Bigl\\lceil\\alpha\\,L_{n}\\Bigr\\rceil\\in (L_{n},2L_{n}],\\qquad\nv_{n}:=(v_{n,1},\\dots ,v_{n,d})\\in\\mathbb N^{d}. \\tag{3}\n\\]\nBecause $\\alpha<\\tfrac32$ and $L_{n}\\ge 4$ for $n\\ge 2$,  \n\\[\n\\alpha L_{n}<v_{n,j}\\le\\alpha L_{n}+1\\le\\frac32L_{n}. \\tag{4}\n\\]\n\nStep 2.  Mixed signed sums of the digit vectors avoid $\\Delta(\\mathcal A)$.  \nFor $n\\ge 2$ set\n\\[\nS_{n}:=\\Bigl\\{\\sum_{k=2}^{n}\\varepsilon_{k}v_{k}:\n              \\varepsilon_{k}\\in\\{-1,0,1\\}\\Bigr\\}\\setminus\\{0\\}. \\tag{5}\n\\]\nFix $\\delta=\\sum_{k=2}^{n}\\varepsilon_{k}v_{k}\\in S_{n}$ and let\n$m$ be the largest index $\\ge 2$ with $\\varepsilon_{m}\\ne 0$.\nWrite\n\\[\n\\sigma_{j}:=\\sum_{k=2}^{m-1}|\\varepsilon_{k}|\\,v_{k,j}\\qquad(1\\le j\\le d).\n\\]\nUsing (4) and $L_{k}=4^{\\,k-1}$,\n\\[\n\\begin{aligned}\n\\sigma_{j}\n &\\le \\alpha\\sum_{r=1}^{m-2}L_{m-r}\n     =\\alpha L_{m-1}\\sum_{r=0}^{m-3}4^{-r}\n     \\le\\alpha L_{m-1}\\cdot\\tfrac43\n     =\\frac{4\\alpha}{3}L_{m-1}.\n\\end{aligned}\n\\]\nSince $L_{m-1}=L_{m}/4$,\n\\[\n\\sigma_{j}\\le\\frac{\\alpha}{3}L_{m}. \\tag{6}\n\\]\nWith $\\alpha=\\tfrac54$ this gives $\\sigma_{j}\\le\\tfrac{5}{12}L_{m}$.\nConsequently\n\\[\n\\sigma_{j}\\le \\frac12L_{m}-1\\qquad(m\\ge 3),\\tag{7}\n\\]\nbecause $L_{m}\\ge 16$ for $m\\ge 3$.  \nIf $m=2$ then the sum for $\\sigma_{j}$ is empty, hence $\\sigma_{j}=0$ and\n(7) is trivially satisfied as well.\n\nNext, by (4) and (7),\n\\[\n\\begin{aligned}\n|\\delta_{j}|\n&=|\\,\\varepsilon_{m}v_{m,j}\\pm\\sigma_{j}\\,|\n\\;\\ge\\;v_{m,j}-\\sigma_{j}\\\\\n&\\ge\\;\\alpha L_{m}-\\Bigl(\\tfrac12L_{m}-1\\Bigr)\n      \\;=\\;\\Bigl(\\alpha-\\tfrac12\\Bigr)L_{m}+1\n      \\;>\\;L_{m},\\tag{8}\n\\end{aligned}\n\\]\nwhile\n\\[\n|\\delta_{j}|\n\\;\\le\\;v_{m,j}+\\sigma_{j}\n\\;\\le\\;\\tfrac32L_{m}+\\Bigl(\\tfrac12L_{m}-1\\Bigr)\n      \\;=\\;2L_{m}-1\n      \\;<\\;2L_{m}.\\tag{9}\n\\]\nThus for every coordinate\n\\[\nL_{m}<|\\delta_{j}|\\le 2L_{m}\\qquad(1\\le j\\le d).\n\\]\nEach coordinate of $\\delta$ therefore lies inside the lacunary interval\n$(L_{m},2L_{m}]$ and,\nthanks to $(\\diamondsuit_{m})$, no such vector can belong\nto $\\Delta(\\mathcal A)$.  Hence\n\\[\nS_{n}\\cap\\Delta(\\mathcal A)=\\varnothing.\\tag{Lemma}\n\\]\n\nStep 3.  A large family of pairwise disjoint translates.  \nFor $\\varepsilon=(\\varepsilon_{2},\\dots ,\\varepsilon_{n})\\in\\{0,1\\}^{\\,n-1}$\nput\n\\[\nT(\\varepsilon):=\\sum_{k=2}^{n}\\varepsilon_{k}v_{k}\\in\\mathbb N^{d}. \\tag{10}\n\\]\nIf $\\varepsilon\\ne\\eta$ then\n$T(\\varepsilon)-T(\\eta)\\in S_{n}\\setminus\\{0\\}$; by the Lemma\n\\[\n\\bigl(\\mathcal A+T(\\varepsilon)\\bigr)\\cap\n\\bigl(\\mathcal A+T(\\eta)\\bigr)=\\varnothing\n\\qquad(\\varepsilon\\ne\\eta).\\tag{11}\n\\]\nThus we have $2^{\\,n-1}$ pairwise disjoint translates of $\\mathcal A$.\n\nStep 4.  Localising the translates.  \nUsing (4) and $L_{k}=4^{\\,k-1}$,\n\\[\n\\|T(\\varepsilon)\\|_{\\infty}\n \\le\\alpha\\sum_{k=2}^{n}L_{k}\n <\\alpha L_{n}\\sum_{r=0}^{n-2}4^{-r}\n <\\tfrac43\\alpha\\,L_{n}\n <2L_{n}\n =\\tfrac12L_{n+1}. \\tag{12}\n\\]\nPut $X:=L_{n+1}=4L_{n}$.  Then each translate\n\\[\n\\bigl(\\mathcal A\\cap[1,X]^{d}\\bigr)+T(\\varepsilon)\n\\;\\subset\\;[1,2X]^{d},\\tag{13}\n\\]\nand the $2^{\\,n-1}$ such sets are pairwise disjoint.  Therefore\n\\[\n2^{\\,n-1}\\,N(X)\\;\\le\\;|[1,2X]^{d}|\n       \\;=\\;(2X)^{d}\n       \\;=\\;2^{\\,d}\\,X^{d}.\\tag{14}\n\\]\n\nStep 5.  Relating $n$ and $X$.  \nBecause $(\\Lambda)$ is an equality,\n\\[\nX=L_{n+1}=4^{\\,n},\\qquad\\text{hence}\\qquad\n2^{\\,n}=X^{1/2}.\\tag{15}\n\\]\nConsequently\n\\[\n2^{\\,n-1}=2^{-1}X^{1/2}.\\tag{16}\n\\]\nInserting (16) into (14) gives\n\\[\nN(X)\\;\\le\\;2^{\\,d+1}\\,X^{\\,d-\\tfrac12}.\\tag{17}\n\\]\n\nStep 6.  From special abscissae to all $x$.  \nThe right–hand side of (17) is monotone increasing, and every\n$x\\ge 4$ lies in some interval $[L_{n+1},L_{n+2})$.\nUsing $N(x)\\le N(L_{n+2})$ and (17) with $X=L_{n+2}$ yields\n\\[\nN(x)\\;\\ll_{d}\\;x^{\\,d-\\tfrac12}\\qquad(x\\ge 4),\\tag{18}\n\\]\nwhich is exactly the asserted bound (1).\n\nStep 7.  Vanishing upper density.  \nDivide (18) by $x^{d}$ and let $x\\to\\infty$ to obtain (2).\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.790270",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension.  \n   • The problem is lifted from subsets of \\(\\mathbb N\\) to subsets of\n     \\(\\mathbb N^{d}\\) with \\(d\\ge 2\\).  \n   • Density is measured in \\(d\\)-dimensional boxes\n     \\([1,x]^{d}\\), so counting arguments must handle an\n     extra power of \\(x\\).\n\n2. Several interacting difference sets.  \n   • One must control \\emph{all} coordinate-wise difference sets\n     \\(\\mathcal B_{1},\\dots ,\\mathcal B_{d}\\) simultaneously.  \n   • The hypothesis is weaker (“there exist unbounded gaps distributed\n     among the coordinates’’) yet must still force sparsity.\n\n3. Multidimensional packing.  \n   • Translating the set along a \\emph{vector} uses vector arithmetic\n     and multidimensional volume estimates rather than simple intervals.\n   • The argument must make sure that the union of many disjoint\n     translates still fits into a controlled \\(d\\)-dimensional box.\n\n4. Quantitative contradiction.  \n   • The final inequality needs an additional power of \\(x\\) and an\n     intricate balance between the number of translates and the volume\n     of the enlarged box.  \n   • Careful parameter choices (the constant \\(L\\), bounds on \\(x\\),\n     etc.) are indispensable; a naïve one-dimensional adaptation fails.\n\nAll of these features force the solver to blend combinatorial packing,\ncoordinate projections, and asymptotic analysis in higher dimensions,\nmaking the enhanced kernel variant markedly more technical and\nsubstantially harder than the original one-dimensional problem."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer $d\\ge 2$ and write $\\mathbb N=\\{1,2,3,\\dots\\}$.\nFor $n=(n_{1},\\dots ,n_{d})\\in\\mathbb Z^{d}$ put\n\\[\n|n|_{\\infty}:=\\max_{1\\le i\\le d}|n_{i}|.\n\\]\nFor a non–empty subset\n\\[\n\\mathcal A\\subset\\mathbb N^{d}:=\\bigl\\{(n_{1},\\dots ,n_{d}):n_{i}\\in\\mathbb N\\bigr\\}\n\\]\ndefine, for $x\\ge 1$,\n\\[\nN(x):=\\bigl|\\mathcal A\\cap[1,x]^{d}\\bigr|.\n\\]\n\nGlobal difference set\n\\[\n\\Delta(\\mathcal A):=\\{a-a':a,a'\\in\\mathcal A,\\;a\\ne a'\\}\\subset\\mathbb Z^{d}.\n\\]\n\nFor every coordinate $j\\in\\{1,\\dots ,d\\}$ let\n\\[\n\\mathcal B_{j}:=\\bigl\\{|a_{j}-a'_{j}|:a,a'\\in\\mathcal A,\\;a_{j}\\ne a'_{j}\\bigr\\}\n              =\\{b_{j,1}<b_{j,2}<\\dots \\}\\subset\\mathbb N .\n\\]\n\nSimultaneous coordinate lacunarity.\nAssume that there is a sequence of positive integers\n\\[\n1=L_{1},\\qquad L_{n+1}=4L_{n}\\qquad(n\\ge 1) \\tag{$\\Lambda$}\n\\]\nsuch that, for every $n\\ge 1$,\n\\[\n(L_{n},2L_{n}]\\,\\cap\\,\\mathcal B_{j}=\\varnothing\\qquad(j=1,\\dots ,d).\\tag{$\\diamondsuit_{n}$}\n\\]\nIn words: for each $n$ the interval $(L_{n},2L_{n}]$ is simultaneously free of\none–dimensional differences along every coordinate direction.\n\nProblem.\nProve that the counting function of $\\mathcal A$ satisfies the power–saving\ndensity bound\n\\[\nN(x)\\;\\ll_{d}\\;x^{\\,d-\\tfrac12}\\qquad (x\\ge 4),\\tag{1}\n\\]\nwhere the implied constant may depend on $d$ (on no other feature of $\\mathcal A$).\nIn particular\n\\[\n\\lim_{x\\to\\infty}\\frac{N(x)}{x^{\\,d}}=0.\\tag{2}\n\\]\n\nThe task is to deduce the exponent $\\tfrac12$ in (1) from the genuinely\n$d$–dimensional lacunarity $(\\diamondsuit_{n})$ together with the uniform\n“radix–$4$’’ growth $(\\Lambda)$.",
      "solution": "Throughout $c,c_{1},c_{2},\\dots$ denote positive constants depending only\non the ambient dimension $d$.\n\nStep 1.  “Digit’’ vectors inside the forbidden gaps.  \nBecause of $(\\diamondsuit_{n})$ we may, for every $n\\ge 1$ and every\n$j\\in\\{1,\\dots ,d\\}$, choose an integer\n\\[\nv_{n,j}:=\\Bigl\\lceil\\tfrac32L_{n}\\Bigr\\rceil\\in(L_{n},2L_{n}].\\tag{3}\n\\]\n(The ceiling is introduced only to secure integrality.)\nPut\n\\[\nv_{n}:=(v_{n,1},\\dots ,v_{n,d})\\in\\mathbb N^{d}.\\tag{4}\n\\]\nThen for each coordinate\n\\[\n\\tfrac32L_{n}\\;\\le\\;v_{n,j}\\;<\\;\\tfrac32L_{n}+1\\;\\le\\;\\tfrac74L_{n}.\\tag{5}\n\\]\n\nStep 2.  Mixed signed sums of the digits never hit $\\Delta(\\mathcal A)$.  \nFor $n\\ge 1$ set\n\\[\nS_{n}:=\\Bigl\\{\\sum_{k\\le n}\\varepsilon_{k}v_{k}:\n              \\varepsilon_{k}\\in\\{-1,0,1\\}\\Bigr\\}\\setminus\\{0\\}.\\tag{6}\n\\]\nFix $\\delta=\\sum_{k\\le n}\\varepsilon_{k}v_{k}\\in S_{n}$ and let\n$m$ be the largest index with $\\varepsilon_{m}\\ne 0$.\nWrite\n\\[\n\\sigma_{j}:=\\sum_{k<m}|\\varepsilon_{k}|\\,v_{k,j}\\qquad(1\\le j\\le d).\n\\]\nSince $L_{k}=4^{\\,k-1}$,  \n\\[\n\\begin{aligned}\n\\sigma_{j}\n &\\le \\tfrac32\\sum_{r=1}^{m-1}L_{m-r}\n   =\\tfrac32L_{m-1}\\sum_{r=0}^{m-2}4^{-r}\n   =\\tfrac32L_{m-1}\\cdot\\tfrac43\\bigl(1-4^{-(m-1)}\\bigr)  \\\\\n &<2L_{m-1}= \\tfrac12L_{m}.\\tag{7}\n\\end{aligned}\n\\]\nObserve moreover that every $v_{k,j}$ is a multiple of $1$, hence\n$\\sigma_{j}$ is an integer.  Combining the integrality with (7) yields\n\\[\n\\sigma_{j}\\;\\le\\;\\tfrac12L_{m}-1\\qquad(1\\le j\\le d).\\tag{8}\n\\]\n\nNow, by (5) and (8),\n\\[\n|\\delta_{j}|\n\\;=\\;|\\,\\varepsilon_{m}v_{m,j}\\pm\\sigma_{j}\\,|\n\\;\\ge\\;v_{m,j}-\\sigma_{j}\n\\;\\ge\\;\\tfrac32L_{m}-(\\tfrac12L_{m}-1)\n\\;=\\;L_{m}+1\n\\;>\\;L_{m},\\tag{9}\n\\]\nwhile\n\\[\n|\\delta_{j}|\n\\;\\le\\;v_{m,j}+\\sigma_{j}\n\\;<\\;\\bigl(\\tfrac32L_{m}+1\\bigr)+\\tfrac12L_{m}\n\\;=\\;2L_{m}+1\n\\;\\le\\;2L_{m}\\qquad(m\\ge 2).\\tag{10}\n\\]\n(For $m=1$ one has $L_{1}=1,\\;v_{1,j}=2$, so\n$|\\delta_{j}|\\in\\{2\\}\\subset(L_{1},2L_{1}]$ as well.)\nThus each coordinate of $\\delta$ lies in the lacunary interval\n$(L_{m},2L_{m}]$; by $(\\diamondsuit_{m})$ no such vector can belong\nto $\\Delta(\\mathcal A)$.  Hence\n\\[\nS_{n}\\cap\\Delta(\\mathcal A)=\\varnothing.\\tag{Lemma 1}\n\\]\n\nStep 3.  A large family of pairwise disjoint translates.  \nFor $\\varepsilon=(\\varepsilon_{1},\\dots ,\\varepsilon_{n})\\in\\{0,1\\}^{n}$\nput\n\\[\nT(\\varepsilon):=\\sum_{k=1}^{n}\\varepsilon_{k}v_{k}\\in\\mathbb N^{d}.\\tag{11}\n\\]\nIf $\\varepsilon\\ne\\eta$, then\n$T(\\varepsilon)-T(\\eta)\\in S_{n}\\setminus\\{0\\}$; by Lemma 1\n\\[\n\\bigl(\\mathcal A+T(\\varepsilon)\\bigr)\\cap\n\\bigl(\\mathcal A+T(\\eta)\\bigr)=\\varnothing\n\\qquad(\\varepsilon\\ne\\eta).\\tag{12}\n\\]\nThus we have $2^{n}$ pairwise disjoint translates of $\\mathcal A$.\n\nStep 4.  Localising the translates.  \nUsing (5) and the exact radix–$4$ rule $L_{k}=4^{\\,k-1}$,\n\\[\n\\|T(\\varepsilon)\\|_{\\infty}\n \\le \\tfrac32\\sum_{k=1}^{n}L_{k}\n =\\tfrac32L_{n}\\sum_{r=0}^{n-1}4^{-r}\n <2L_{n}=\\tfrac12L_{n+1}.\\tag{13}\n\\]\nPut $X:=L_{n+1}$.  Then each translate\n\\[\n\\bigl(\\mathcal A\\cap[1,X]^{d}\\bigr)+T(\\varepsilon)\n\\;\\subset\\;[1,2X]^{d},\\tag{14}\n\\]\nand the $2^{n}$ such sets are pairwise disjoint.\nTherefore\n\\[\n2^{n}\\,N(X)\\;\\le\\;|[1,2X]^{d}|\n       \\;=\\;(2X)^{d}\n       \\;=\\;2^{\\,d}\\,X^{d}.\\tag{15}\n\\]\n\nStep 5.  Relating $n$ and $X$.  \nBecause $(\\Lambda)$ is an equality,\n\\[\nX=L_{n+1}=4L_{n}=4^{\\,n},\\qquad\\text{hence}\\qquad\n2^{n}=X^{1/2}.\\tag{16}\n\\]\nInsert (16) into (15):\n\\[\nN(X)\\;\\le\\;2^{\\,d}\\,X^{d-\\tfrac12}.\\tag{17}\n\\]\n\nStep 6.  From special abscissae to all $x$.  \nThe right–hand side of (17) is monotone increasing, and every\n$x\\ge 4$ lies in some interval $[L_{n+1},L_{n+2})$.\nUsing $N(x)\\le N(L_{n+2})$ and (17) with $X=L_{n+2}$ gives\n\\[\nN(x)\\;\\ll_{d}\\;x^{\\,d-\\tfrac12}\\qquad(x\\ge 4),\\tag{18}\n\\]\nexactly the asserted bound (1).\n\nStep 7.  Vanishing upper density.  \nDivide (18) by $x^{d}$ and let $x\\to\\infty$ to obtain (2).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.603938",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension.  \n   • The problem is lifted from subsets of \\(\\mathbb N\\) to subsets of\n     \\(\\mathbb N^{d}\\) with \\(d\\ge 2\\).  \n   • Density is measured in \\(d\\)-dimensional boxes\n     \\([1,x]^{d}\\), so counting arguments must handle an\n     extra power of \\(x\\).\n\n2. Several interacting difference sets.  \n   • One must control \\emph{all} coordinate-wise difference sets\n     \\(\\mathcal B_{1},\\dots ,\\mathcal B_{d}\\) simultaneously.  \n   • The hypothesis is weaker (“there exist unbounded gaps distributed\n     among the coordinates’’) yet must still force sparsity.\n\n3. Multidimensional packing.  \n   • Translating the set along a \\emph{vector} uses vector arithmetic\n     and multidimensional volume estimates rather than simple intervals.\n   • The argument must make sure that the union of many disjoint\n     translates still fits into a controlled \\(d\\)-dimensional box.\n\n4. Quantitative contradiction.  \n   • The final inequality needs an additional power of \\(x\\) and an\n     intricate balance between the number of translates and the volume\n     of the enlarged box.  \n   • Careful parameter choices (the constant \\(L\\), bounds on \\(x\\),\n     etc.) are indispensable; a naïve one-dimensional adaptation fails.\n\nAll of these features force the solver to blend combinatorial packing,\ncoordinate projections, and asymptotic analysis in higher dimensions,\nmaking the enhanced kernel variant markedly more technical and\nsubstantially harder than the original one-dimensional problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}