{
  "index": "1984-B-5",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Problem B-5\nFor each nonnegative integer \\( k \\), let \\( d(k) \\) denote the number of 1 's in the binary expansion of \\( k \\) (for example. \\( d(0)=0 \\) and \\( d(5)=2 \\) ). Let \\( m \\) be a positive integer. Express\n\\[\n\\sum_{k=0}^{2^{m}-1}(-1)^{d(k)} k^{m}\n\\]\nin the form \\( (-1)^{m} a^{f(m)}(g(m))! \\), where \\( a \\) is an integer and \\( f \\) and \\( g \\) are polynomials.",
  "solution": "B-5.\nDefine\n\\[\nD(x)=(1-x)\\left(1-x^{2}\\right)\\left(1-x^{4}\\right) \\cdots\\left(1-x^{2^{n-1}}\\right)\n\\]\n\nSince binary expansions are unique, each monomial \\( x^{k}\\left(0 \\leqslant k \\leqslant 2^{n}-1\\right) \\) appears exactly once in the expansion of \\( D(x) \\), with coefficient \\( (-1)^{d(k)} \\). That is,\n\\[\nD(x)=\\sum_{k=0}^{2^{n}-1}(-1)^{d(k)} x^{k}\n\\]\n\nApplying the operator \\( \\left(x \\frac{d}{d x}\\right) \\) to \\( D(x) m \\) times, we obtain\n\\[\n\\left(x \\frac{d}{d x}\\right)^{m} D(x)=\\sum_{k=0}^{2^{n}-1}(-1)^{d(k)} k^{m} x^{k}\n\\]\nso that\n\\[\n\\left.\\left(x \\frac{d}{d x}\\right)^{m} D(x)\\right]_{x=1}=\\sum_{k=0}^{2^{n}-1}(-1)^{d(k)} k^{m}\n\\]\n\nDefine \\( F(x)=D(x+1) \\), so that\n\\[\n\\left.\\left.\\left(x \\frac{d}{d x}\\right)^{m} D(x)\\right]_{x=1}=\\left[(x+1) \\frac{d}{d x}\\right]^{m} F(x)\\right]_{x=0}\n\\]\n\nBut\n\\[\n\\begin{aligned}\nF(x) & =\\prod_{\\alpha=1}^{m}\\left[1-(x+1)^{2^{\\alpha-1}}\\right]=\\prod_{\\alpha=1}^{m}\\left[-2^{\\alpha-1} x+O\\left(x^{2}\\right)\\right], \\quad(x \\rightarrow 0) \\\\\n& =(-1)^{m} 2^{m(m-1) / 2} x^{m}+O\\left(x^{m+1}\\right)\n\\end{aligned}\n\\]\nand by observing that \\( [(x+1) d / d x] x^{n}=n x^{n}+n x^{n-1} \\), we see that\n\\[\n\\left[(x+1) \\frac{d}{d x}\\right]^{m}\\left(A x^{m}+O\\left(x^{m+1}\\right)\\right)=m!A+O(x)\n\\]\n\nSo\n\\[\n\\left.\\left.\\left[(x+1) \\frac{d}{d x}\\right]^{m} F(x)\\right]_{x=0}=(-1)^{m} 2^{m(m-1) / 2} m!+O(x)\\right]_{x=0}=(-1)^{m} 2^{m(m-1) / 2} m!\n\\]",
  "vars": [
    "k",
    "d",
    "m",
    "x",
    "n",
    "\\\\alpha"
  ],
  "params": [
    "a",
    "f",
    "g",
    "D",
    "F",
    "A",
    "O"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "k": "binidx",
        "d": "bitcount",
        "m": "posint",
        "x": "inputx",
        "n": "powern",
        "\\alpha": "alphaidx",
        "a": "intconst",
        "f": "polyfunf",
        "g": "polyfung",
        "D": "polydfunc",
        "F": "polyffunc",
        "A": "constbig",
        "O": "bigoterm"
      },
      "question": "Problem B-5\nFor each nonnegative integer \\( binidx \\), let \\( bitcount(binidx) \\) denote the number of 1 's in the binary expansion of \\( binidx \\) (for example. \\( bitcount(0)=0 \\) and \\( bitcount(5)=2 \\) ). Let \\( posint \\) be a positive integer. Express\n\\[\n\\sum_{binidx=0}^{2^{posint}-1}(-1)^{bitcount(binidx)} binidx^{posint}\n\\]\nin the form \\( (-1)^{posint} intconst^{polyfunf(posint)}(polyfung(posint))! \\), where \\( intconst \\) is an integer and \\( polyfunf \\) and \\( polyfung \\) are polynomials.",
      "solution": "B-5.\nDefine\n\\[\npolydfunc(inputx)=(1-inputx)\\left(1-inputx^{2}\\right)\\left(1-inputx^{4}\\right)\\cdots\\left(1-inputx^{2^{powern-1}}\\right)\n\\]\n\nSince binary expansions are unique, each monomial \\( inputx^{binidx}\\left(0 \\leqslant binidx \\leqslant 2^{powern}-1\\right) \\) appears exactly once in the expansion of \\( polydfunc(inputx) \\), with coefficient \\( (-1)^{bitcount(binidx)} \\). That is,\n\\[\npolydfunc(inputx)=\\sum_{binidx=0}^{2^{powern}-1}(-1)^{bitcount(binidx)} inputx^{binidx}\n\\]\n\nApplying the operator \\( \\left(inputx \\frac{bitcount}{bitcount inputx}\\right)^{posint} \\) to \\( polydfunc(inputx) \\) posint times, we obtain\n\\[\n\\left(inputx \\frac{bitcount}{bitcount inputx}\\right)^{posint} polydfunc(inputx)=\\sum_{binidx=0}^{2^{powern}-1}(-1)^{bitcount(binidx)} binidx^{posint} inputx^{binidx}\n\\]\nso that\n\\[\n\\left.\\left(inputx \\frac{bitcount}{bitcount inputx}\\right)^{posint} polydfunc(inputx)\\right]_{inputx=1}=\\sum_{binidx=0}^{2^{powern}-1}(-1)^{bitcount(binidx)} binidx^{posint}\n\\]\n\nDefine \\( polyffunc(inputx)=polydfunc(inputx+1) \\), so that\n\\[\n\\left.\\left.\\left(inputx \\frac{bitcount}{bitcount inputx}\\right)^{posint} polydfunc(inputx)\\right]_{inputx=1}=\\left[(inputx+1) \\frac{bitcount}{bitcount inputx}\\right]^{posint} polyffunc(inputx)\\right]_{inputx=0}\n\\]\n\nBut\n\\[\n\\begin{aligned}\npolyffunc(inputx) & =\\prod_{alphaidx=1}^{posint}\\left[1-(inputx+1)^{2^{alphaidx-1}}\\right]=\\prod_{alphaidx=1}^{posint}\\left[-2^{alphaidx-1} inputx+bigoterm\\!\\left(inputx^{2}\\right)\\right], \\quad(inputx \\rightarrow 0) \\\\\n& =(-1)^{posint} 2^{posint(posint-1) / 2} inputx^{posint}+bigoterm\\!\\left(inputx^{posint+1}\\right)\n\\end{aligned}\n\\]\nand by observing that \\( [(inputx+1) bitcount / bitcount inputx] inputx^{powern}=powern inputx^{powern}+powern inputx^{powern-1} \\), we see that\n\\[\n\\left[(inputx+1) \\frac{bitcount}{bitcount inputx}\\right]^{posint}\\left(constbig\\, inputx^{posint}+bigoterm\\!\\left(inputx^{posint+1}\\right)\\right)=posint!\\, constbig+bigoterm(inputx)\n\\]\n\nSo\n\\[\n\\left.\\left.\\left[(inputx+1) \\frac{bitcount}{bitcount inputx}\\right]^{posint} polyffunc(inputx)\\right]_{inputx=0}=(-1)^{posint} 2^{posint(posint-1) / 2} posint!+bigoterm(inputx)\\right]_{inputx=0}=(-1)^{posint} 2^{posint(posint-1) / 2} posint!\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "k": "sandpaper",
        "d": "teapotlid",
        "m": "raincloud",
        "x": "buttercup",
        "n": "jellybean",
        "\\\\alpha": "marigolds",
        "a": "paperboat",
        "f": "lemonade",
        "g": "toothfairy",
        "D": "candlewax",
        "F": "dragonfly",
        "A": "moonlight",
        "O": "evergreen"
      },
      "question": "Problem B-5\nFor each nonnegative integer \\( sandpaper \\), let \\( teapotlid(sandpaper) \\) denote the number of 1 's in the binary expansion of \\( sandpaper \\) (for example. \\( teapotlid(0)=0 \\) and \\( teapotlid(5)=2 \\) ). Let \\( raincloud \\) be a positive integer. Express\n\\[\n\\sum_{sandpaper=0}^{2^{raincloud}-1}(-1)^{teapotlid(sandpaper)} sandpaper^{raincloud}\n\\]\nin the form \\( (-1)^{raincloud} paperboat^{lemonade(raincloud)}(toothfairy(raincloud))! \\), where \\( paperboat \\) is an integer and \\( lemonade \\) and \\( toothfairy \\) are polynomials.",
      "solution": "B-5.\nDefine\n\\[\ncandlewax(buttercup)=(1-buttercup)\\left(1-buttercup^{2}\\right)\\left(1-buttercup^{4}\\right) \\cdots\\left(1-buttercup^{2^{jellybean-1}}\\right)\n\\]\n\nSince binary expansions are unique, each monomial \\( buttercup^{sandpaper}\\left(0 \\leqslant sandpaper \\leqslant 2^{jellybean}-1\\right) \\) appears exactly once in the expansion of \\( candlewax(buttercup) \\), with coefficient \\( (-1)^{teapotlid(sandpaper)} \\). That is,\n\\[\ncandlewax(buttercup)=\\sum_{sandpaper=0}^{2^{jellybean}-1}(-1)^{teapotlid(sandpaper)} buttercup^{sandpaper}\n\\]\n\nApplying the operator \\( \\left(buttercup \\frac{d}{d buttercup}\\right) \\) to \\( candlewax(buttercup)^{raincloud} \\) times, we obtain\n\\[\n\\left(buttercup \\frac{d}{d buttercup}\\right)^{raincloud} candlewax(buttercup)=\\sum_{sandpaper=0}^{2^{jellybean}-1}(-1)^{teapotlid(sandpaper)} sandpaper^{raincloud} buttercup^{sandpaper}\n\\]\nso that\n\\[\n\\left.\\left(buttercup \\frac{d}{d buttercup}\\right)^{raincloud} candlewax(buttercup)\\right]_{buttercup=1}=\\sum_{sandpaper=0}^{2^{jellybean}-1}(-1)^{teapotlid(sandpaper)} sandpaper^{raincloud}\n\\]\n\nDefine \\( dragonfly(buttercup)=candlewax(buttercup+1) \\), so that\n\\[\n\\left.\\left.\\left(buttercup \\frac{d}{d buttercup}\\right)^{raincloud} candlewax(buttercup)\\right]_{buttercup=1}=\\left[(buttercup+1) \\frac{d}{d buttercup}\\right]^{raincloud} dragonfly(buttercup)\\right]_{buttercup=0}\n\\]\n\nBut\n\\[\n\\begin{aligned}\ndragonfly(buttercup) & =\\prod_{marigolds=1}^{raincloud}\\left[1-(buttercup+1)^{2^{marigolds-1}}\\right]=\\prod_{marigolds=1}^{raincloud}\\left[-2^{marigolds-1} buttercup+evergreen\\left(buttercup^{2}\\right)\\right], \\quad(buttercup \\rightarrow 0) \\\\\n& =(-1)^{raincloud} 2^{raincloud(raincloud-1) / 2} buttercup^{raincloud}+evergreen\\left(buttercup^{raincloud+1}\\right)\n\\end{aligned}\n\\]\nand by observing that \\( [(buttercup+1) d / d buttercup] buttercup^{jellybean}=jellybean buttercup^{jellybean}+jellybean buttercup^{jellybean-1} \\), we see that\n\\[\n\\left[(buttercup+1) \\frac{d}{d buttercup}\\right]^{raincloud}\\left(moonlight buttercup^{raincloud}+evergreen\\left(buttercup^{raincloud+1}\\right)\\right)=raincloud! \\, moonlight+evergreen(buttercup)\n\\]\n\nSo\n\\[\n\\left.\\left.\\left[(buttercup+1) \\frac{d}{d buttercup}\\right]^{raincloud} dragonfly(buttercup)\\right]_{buttercup=0}=(-1)^{raincloud} 2^{raincloud(raincloud-1) / 2} raincloud!+evergreen(buttercup)\\right]_{buttercup=0}=(-1)^{raincloud} 2^{raincloud(raincloud-1) / 2} raincloud!\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "k": "fixedvalue",
        "d": "zerocounter",
        "m": "negativeinteger",
        "x": "constantinput",
        "n": "tinyvalue",
        "\\alpha": "latinletter",
        "a": "irrationalnum",
        "f": "exponentialfunc",
        "g": "transcendentalfunc",
        "D": "separatorpoly",
        "F": "staticfunction",
        "A": "variableconst",
        "O": "smallorder"
      },
      "question": "Problem B-5\nFor each nonnegative integer \\( fixedvalue \\), let \\( zerocounter(fixedvalue) \\) denote the number of 1 's in the binary expansion of \\( fixedvalue \\) (for example. \\( zerocounter(0)=0 \\) and \\( zerocounter(5)=2 \\) ). Let \\( negativeinteger \\) be a positive integer. Express\n\\[\n\\sum_{fixedvalue=0}^{2^{negativeinteger}-1}(-1)^{zerocounter(fixedvalue)} fixedvalue^{negativeinteger}\n\\]\nin the form \\( (-1)^{negativeinteger} irrationalnum^{exponentialfunc(negativeinteger)}(transcendentalfunc(negativeinteger))! \\), where \\( irrationalnum \\) is an integer and \\( exponentialfunc \\) and \\( transcendentalfunc \\) are polynomials.",
      "solution": "B-5.\nDefine\n\\[\nseparatorpoly(constantinput)=(1-constantinput)\\left(1-constantinput^{2}\\right)\\left(1-constantinput^{4}\\right) \\cdots\\left(1-constantinput^{2^{tinyvalue-1}}\\right)\n\\]\n\nSince binary expansions are unique, each monomial \\( constantinput^{fixedvalue}\\left(0 \\leqslant fixedvalue \\leqslant 2^{tinyvalue}-1\\right) \\) appears exactly once in the expansion of \\( separatorpoly(constantinput) \\), with coefficient \\( (-1)^{zerocounter(fixedvalue)} \\). That is,\n\\[\nseparatorpoly(constantinput)=\\sum_{fixedvalue=0}^{2^{tinyvalue}-1}(-1)^{zerocounter(fixedvalue)} constantinput^{fixedvalue}\n\\]\n\nApplying the operator \\( \\left(constantinput \\frac{zerocounter}{zerocounter constantinput}\\right) \\) to \\( separatorpoly(constantinput)\\, negativeinteger \\) times, we obtain\n\\[\n\\left(constantinput \\frac{zerocounter}{zerocounter constantinput}\\right)^{negativeinteger} separatorpoly(constantinput)=\\sum_{fixedvalue=0}^{2^{tinyvalue}-1}(-1)^{zerocounter(fixedvalue)} fixedvalue^{negativeinteger} constantinput^{fixedvalue}\n\\]\nso that\n\\[\n\\left.\\left(constantinput \\frac{zerocounter}{zerocounter constantinput}\\right)^{negativeinteger} separatorpoly(constantinput)\\right]_{constantinput=1}=\\sum_{fixedvalue=0}^{2^{tinyvalue}-1}(-1)^{zerocounter(fixedvalue)} fixedvalue^{negativeinteger}\n\\]\n\nDefine \\( staticfunction(constantinput)=separatorpoly(constantinput+1) \\), so that\n\\[\n\\left.\\left.\\left(constantinput \\frac{zerocounter}{zerocounter constantinput}\\right)^{negativeinteger} separatorpoly(constantinput)\\right]_{constantinput=1}=\\left[(constantinput+1) \\frac{zerocounter}{zerocounter constantinput}\\right]^{negativeinteger} staticfunction(constantinput)\\right]_{constantinput=0}\n\\]\n\nBut\n\\[\n\\begin{aligned}\nstaticfunction(constantinput) & =\\prod_{latinletter=1}^{negativeinteger}\\left[1-(constantinput+1)^{2^{latinletter-1}}\\right]=\\prod_{latinletter=1}^{negativeinteger}\\left[-2^{latinletter-1} constantinput+smallorder\\left(constantinput^{2}\\right)\\right], \\quad(constantinput \\rightarrow 0) \\\\\n& =(-1)^{negativeinteger} 2^{negativeinteger(negativeinteger-1) / 2} constantinput^{negativeinteger}+smallorder\\left(constantinput^{negativeinteger+1}\\right)\n\\end{aligned}\n\\]\nand by observing that \\( [(constantinput+1) zerocounter / zerocounter constantinput] constantinput^{tinyvalue}=tinyvalue constantinput^{tinyvalue}+tinyvalue constantinput^{tinyvalue-1} \\), we see that\n\\[\n\\left[(constantinput+1) \\frac{zerocounter}{zerocounter constantinput}\\right]^{negativeinteger}\\left(variableconst constantinput^{negativeinteger}+smallorder\\left(constantinput^{negativeinteger+1}\\right)\\right)=negativeinteger!variableconst+smallorder(constantinput)\n\\]\n\nSo\n\\[\n\\left.\\left.\\left[(constantinput+1) \\frac{zerocounter}{zerocounter constantinput}\\right]^{negativeinteger} staticfunction(constantinput)\\right]_{constantinput=0}=(-1)^{negativeinteger} 2^{negativeinteger(negativeinteger-1) / 2} negativeinteger!+smallorder(constantinput)\\right]_{constantinput=0}=(-1)^{negativeinteger} 2^{negativeinteger(negativeinteger-1) / 2} negativeinteger!\n\\]"
    },
    "garbled_string": {
      "map": {
        "k": "zvdqmmtr",
        "d": "xlfkqzja",
        "m": "ubnztrse",
        "x": "pdhycwle",
        "n": "qbrmvgsa",
        "\\\\alpha": "swtnhfrc",
        "a": "nhktpbez",
        "f": "cgyxrlom",
        "g": "rmavzqit",
        "D": "yzoxqvha",
        "F": "lqejgkfs",
        "A": "skvrdmha",
        "O": "qbhyfzwe"
      },
      "question": "Problem B-5\nFor each nonnegative integer \\( zvdqmmtr \\), let \\( xlfkqzja(zvdqmmtr) \\) denote the number of 1 's in the binary expansion of \\( zvdqmmtr \\) (for example. \\( xlfkqzja(0)=0 \\) and \\( xlfkqzja(5)=2 \\) ). Let \\( ubnztrse \\) be a positive integer. Express\n\\[\n\\sum_{zvdqmmtr=0}^{2^{ubnztrse}-1}(-1)^{xlfkqzja(zvdqmmtr)} zvdqmmtr^{ubnztrse}\n\\]\nin the form \\( (-1)^{ubnztrse} nhktpbez^{cgyxrlom(ubnztrse)}(rmavzqit(ubnztrse))! \\), where \\( nhktpbez \\) is an integer and \\( cgyxrlom \\) and \\( rmavzqit \\) are polynomials.",
      "solution": "B-5.\nDefine\n\\[\nyzoxqvha(pdhycwle)=(1-pdhycwle)\\left(1-pdhycwle^{2}\\right)\\left(1-pdhycwle^{4}\\right) \\cdots\\left(1-pdhycwle^{2^{qbrmvgsa-1}}\\right)\n\\]\n\nSince binary expansions are unique, each monomial \\( pdhycwle^{zvdqmmtr}\\left(0 \\leqslant zvdqmmtr \\leqslant 2^{qbrmvgsa}-1\\right) \\) appears exactly once in the expansion of \\( yzoxqvha(pdhycwle) \\), with coefficient \\( (-1)^{xlfkqzja(zvdqmmtr)} \\). That is,\n\\[\nyzoxqvha(pdhycwle)=\\sum_{zvdqmmtr=0}^{2^{qbrmvgsa}-1}(-1)^{xlfkqzja(zvdqmmtr)} pdhycwle^{zvdqmmtr}\n\\]\n\nApplying the operator \\( \\left(pdhycwle \\frac{d}{d pdhycwle}\\right) \\) to \\( yzoxqvha(pdhycwle) ubnztrse \\) times, we obtain\n\\[\n\\left(pdhycwle \\frac{d}{d pdhycwle}\\right)^{ubnztrse} yzoxqvha(pdhycwle)=\\sum_{zvdqmmtr=0}^{2^{qbrmvgsa}-1}(-1)^{xlfkqzja(zvdqmmtr)} zvdqmmtr^{ubnztrse} pdhycwle^{zvdqmmtr}\n\\]\nso that\n\\[\n\\left.\\left(pdhycwle \\frac{d}{d pdhycwle}\\right)^{ubnztrse} yzoxqvha(pdhycwle)\\right]_{pdhycwle=1}=\\sum_{zvdqmmtr=0}^{2^{qbrmvgsa}-1}(-1)^{xlfkqzja(zvdqmmtr)} zvdqmmtr^{ubnztrse}\n\\]\n\nDefine \\( lqejgkfs(pdhycwle)=yzoxqvha(pdhycwle+1) \\), so that\n\\[\n\\left.\\left.\\left(pdhycwle \\frac{d}{d pdhycwle}\\right)^{ubnztrse} yzoxqvha(pdhycwle)\\right]_{pdhycwle=1}=\\left[(pdhycwle+1) \\frac{d}{d pdhycwle}\\right]^{ubnztrse} lqejgkfs(pdhycwle)\\right]_{pdhycwle=0}\n\\]\n\nBut\n\\[\n\\begin{aligned}\nlqejgkfs(pdhycwle) & =\\prod_{swtnhfrc=1}^{ubnztrse}\\left[1-(pdhycwle+1)^{2^{swtnhfrc-1}}\\right]=\\prod_{swtnhfrc=1}^{ubnztrse}\\left[-2^{swtnhfrc-1} pdhycwle+qbhyfzwe\\left(pdhycwle^{2}\\right)\\right], \\quad(pdhycwle \\rightarrow 0) \\\\\n& =(-1)^{ubnztrse} 2^{ubnztrse(ubnztrse-1) / 2} pdhycwle^{ubnztrse}+qbhyfzwe\\left(pdhycwle^{ubnztrse+1}\\right)\n\\end{aligned}\n\\]\nand by observing that \\( [(pdhycwle+1) d / d pdhycwle] pdhycwle^{qbrmvgsa}=qbrmvgsa pdhycwle^{qbrmvgsa}+qbrmvgsa pdhycwle^{qbrmvgsa-1} \\), we see that\n\\[\n\\left[(pdhycwle+1) \\frac{d}{d pdhycwle}\\right]^{ubnztrse}\\left(skvrdmha pdhycwle^{ubnztrse}+qbhyfzwe\\left(pdhycwle^{ubnztrse+1}\\right)\\right)=ubnztrse!skvrdmha+qbhyfzwe(pdhycwle)\n\\]\n\nSo\n\\[\n\\left.\\left.\\left[(pdhycwle+1) \\frac{d}{d pdhycwle}\\right]^{ubnztrse} lqejgkfs(pdhycwle)\\right]_{pdhycwle=0}=(-1)^{ubnztrse} 2^{ubnztrse(ubnztrse-1) / 2} ubnztrse!+qbhyfzwe(pdhycwle)\\right]_{pdhycwle=0}=(-1)^{ubnztrse} 2^{ubnztrse(ubnztrse-1) / 2} ubnztrse!\n\\]"
    },
    "kernel_variant": {
      "question": "Let n and s be positive integers.  \nFor an integer k ≥ 0 write k in binary and let  \nd(k) = (number of 1’s in the binary expansion of k).  \nDefine  \n\n  T(n,s)=  ∑_{k_1=0}^{2^{\\,n}-1}  …  ∑_{k_s=0}^{2^{\\,n}-1} (-1)^{\\,d(k_1)+\\dots +d(k_s)}\\,\n           (k_1+k_2+\\dots +k_s)^{\\,n\\,s} .\n\nShow that T(n,s) can be written in the form  \n\n  T(n,s)=(-1)^{\\,n s}\\,2^{\\,\\frac12 s\\,n\\,(n-1)}\\,(n s)! .\n\n(That is, b = 2 and q(n,s)=s n(n−1)/2 in the stipulated notation.)\n\n--------------------------------------------------------------------",
      "solution": "Step 1.  A multivariate generating function.  \nFor one variable set  \n\n  D(x)=∑_{k=0}^{2^{\\,n}-1}(-1)^{d(k)}x^{\\,k}\n       =∏_{j=0}^{\\,n-1}(1-x^{2^{\\,j}})                         (1)\n\n(the product representation is classical: every 0–1 string of length n occurs exactly once).\n\nFor s variables let  \n\n  G_s(x)=D(x)^{\\,s}=∑_{m≥0}c_m\\,x^{\\,m},  c_m:=∑_{k_1+\\dots+k_s=m}(-1)^{d(k_1)+\\dots+d(k_s)}.   (2)\n\nStep 2.  Converting the required sum to a derivative.  \nIntroduce the Euler operator L:=x(d/dx).  Because\n L^{\\,r}x^{\\,m}=m^{\\,r}x^{\\,m}, we have\n\n  L^{\\,n s} G_s(x)=∑_{m≥0}c_m\\,m^{\\,n s}x^{\\,m}.                 (3)\n\nEvaluating at x=1 therefore yields exactly the desired sum:\n\n  T(n,s)= (L^{\\,n s} G_s)(1).                                (4)\n\nStep 3.  The local behaviour of D(x) at x=1.  \nEach factor in (1) vanishes simply at x=1, so x=1 is a zero of order n of D(x).  \nSet y=x−1; then, for small y,\n\n  D(1+y)=(-1)^{\\,n}\\,2^{\\,n(n-1)/2}\\,y^{\\,n}+O\\!\\bigl(y^{\\,n+1}\\bigr).   (5)\n\n(The coefficient is obtained exactly as in the original B-5 solution, using\n 1−(1+y)^{2^{j}}=−2^{j}y+O(y^{2}).)\n\nRaising (5) to the s-th power gives\n\n  G_s(1+y)=(-1)^{\\,n s}\\,2^{\\,\\frac12 s\\,n(n-1)}\\,y^{\\,n s}+O\\!\\bigl(y^{\\,n s+1}\\bigr).   (6)\n\nStep 4.  Action of L^{ns} at x=1.  \nWrite x=1+y, so that L=(1+y)d/dy.  \nIf F(y)=A\\,y^{\\,n s}+O\\!\\bigl(y^{\\,n s+1}\\bigr), the elementary identity  \n\n  [(1+y)d/dy]^{\\,n s} \\bigl( A\\,y^{\\,n s}\\bigr)\\bigl|_{y=0}= (n s)! \\,A              (7)\n\nfollows immediately from [(1+y)d/dy]\\,y^{r}=r\\,y^{r}+r\\,y^{r-1}.  \nApplying (7) with A from (6) gives  \n\n  T(n,s)=(-1)^{\\,n s}\\,2^{\\,\\frac12 s\\,n(n-1)}\\,(n s)!.              (8)\n\nThis matches the required form with  \n\n  b=2 and q(n,s)=½ s n (n−1). □\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.680296",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional structure:  the original single sum has been replaced by an s-fold\n   sum over an s-dimensional discrete cube; the exponent also scales to n s, coupling the\n   two parameters.\n\n2. Additional parameters and interactions:  the answer now depends simultaneously and\n   non-trivially on both n and s, and one must keep track of how the zero of order n in\n   D(x) propagates to a zero of order n s in G_s(x)=D(x)^{s}.\n\n3. More elaborate generating-function manipulations:  one has to raise D(x) to the s-th\n   power, expand near x=1, and control the leading term, all while\n   handling the multivariate sum implicitly encoded in (2).\n\n4. Deeper operator calculus:  the derivative operator L^{n s} must be applied to a\n   function with a high-order zero; proving (7) in this context, and recognising that no\n   lower-order terms contribute, is essential.\n\n5. Increased combinatorial insight:  understanding that the whole s-fold sum can still be\n   collapsed to a single generating-function calculation requires seeing through several\n   layers of combinatorial encoding.\n\nThese layers of complexity make the enhanced problem substantially harder than both the\noriginal B-5 and the “current kernel” variant, while still remaining solvable with a\ncarefully orchestrated combination of generating-function techniques, asymptotic\nexpansion at a singularity, and operator algebra."
      }
    },
    "original_kernel_variant": {
      "question": "Let n and s be positive integers.  \nFor an integer k ≥ 0 write k in binary and let  \nd(k) = (number of 1’s in the binary expansion of k).  \nDefine  \n\n  T(n,s)=  ∑_{k_1=0}^{2^{\\,n}-1}  …  ∑_{k_s=0}^{2^{\\,n}-1} (-1)^{\\,d(k_1)+\\dots +d(k_s)}\\,\n           (k_1+k_2+\\dots +k_s)^{\\,n\\,s} .\n\nShow that T(n,s) can be written in the form  \n\n  T(n,s)=(-1)^{\\,n s}\\,2^{\\,\\frac12 s\\,n\\,(n-1)}\\,(n s)! .\n\n(That is, b = 2 and q(n,s)=s n(n−1)/2 in the stipulated notation.)\n\n--------------------------------------------------------------------",
      "solution": "Step 1.  A multivariate generating function.  \nFor one variable set  \n\n  D(x)=∑_{k=0}^{2^{\\,n}-1}(-1)^{d(k)}x^{\\,k}\n       =∏_{j=0}^{\\,n-1}(1-x^{2^{\\,j}})                         (1)\n\n(the product representation is classical: every 0–1 string of length n occurs exactly once).\n\nFor s variables let  \n\n  G_s(x)=D(x)^{\\,s}=∑_{m≥0}c_m\\,x^{\\,m},  c_m:=∑_{k_1+\\dots+k_s=m}(-1)^{d(k_1)+\\dots+d(k_s)}.   (2)\n\nStep 2.  Converting the required sum to a derivative.  \nIntroduce the Euler operator L:=x(d/dx).  Because\n L^{\\,r}x^{\\,m}=m^{\\,r}x^{\\,m}, we have\n\n  L^{\\,n s} G_s(x)=∑_{m≥0}c_m\\,m^{\\,n s}x^{\\,m}.                 (3)\n\nEvaluating at x=1 therefore yields exactly the desired sum:\n\n  T(n,s)= (L^{\\,n s} G_s)(1).                                (4)\n\nStep 3.  The local behaviour of D(x) at x=1.  \nEach factor in (1) vanishes simply at x=1, so x=1 is a zero of order n of D(x).  \nSet y=x−1; then, for small y,\n\n  D(1+y)=(-1)^{\\,n}\\,2^{\\,n(n-1)/2}\\,y^{\\,n}+O\\!\\bigl(y^{\\,n+1}\\bigr).   (5)\n\n(The coefficient is obtained exactly as in the original B-5 solution, using\n 1−(1+y)^{2^{j}}=−2^{j}y+O(y^{2}).)\n\nRaising (5) to the s-th power gives\n\n  G_s(1+y)=(-1)^{\\,n s}\\,2^{\\,\\frac12 s\\,n(n-1)}\\,y^{\\,n s}+O\\!\\bigl(y^{\\,n s+1}\\bigr).   (6)\n\nStep 4.  Action of L^{ns} at x=1.  \nWrite x=1+y, so that L=(1+y)d/dy.  \nIf F(y)=A\\,y^{\\,n s}+O\\!\\bigl(y^{\\,n s+1}\\bigr), the elementary identity  \n\n  [(1+y)d/dy]^{\\,n s} \\bigl( A\\,y^{\\,n s}\\bigr)\\bigl|_{y=0}= (n s)! \\,A              (7)\n\nfollows immediately from [(1+y)d/dy]\\,y^{r}=r\\,y^{r}+r\\,y^{r-1}.  \nApplying (7) with A from (6) gives  \n\n  T(n,s)=(-1)^{\\,n s}\\,2^{\\,\\frac12 s\\,n(n-1)}\\,(n s)!.              (8)\n\nThis matches the required form with  \n\n  b=2 and q(n,s)=½ s n (n−1). □\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.533556",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional structure:  the original single sum has been replaced by an s-fold\n   sum over an s-dimensional discrete cube; the exponent also scales to n s, coupling the\n   two parameters.\n\n2. Additional parameters and interactions:  the answer now depends simultaneously and\n   non-trivially on both n and s, and one must keep track of how the zero of order n in\n   D(x) propagates to a zero of order n s in G_s(x)=D(x)^{s}.\n\n3. More elaborate generating-function manipulations:  one has to raise D(x) to the s-th\n   power, expand near x=1, and control the leading term, all while\n   handling the multivariate sum implicitly encoded in (2).\n\n4. Deeper operator calculus:  the derivative operator L^{n s} must be applied to a\n   function with a high-order zero; proving (7) in this context, and recognising that no\n   lower-order terms contribute, is essential.\n\n5. Increased combinatorial insight:  understanding that the whole s-fold sum can still be\n   collapsed to a single generating-function calculation requires seeing through several\n   layers of combinatorial encoding.\n\nThese layers of complexity make the enhanced problem substantially harder than both the\noriginal B-5 and the “current kernel” variant, while still remaining solvable with a\ncarefully orchestrated combination of generating-function techniques, asymptotic\nexpansion at a singularity, and operator algebra."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}