{
  "index": "1958-2-A-1",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. Let \\( f(m, 1)=f(1, n)=1 \\) for \\( m \\geq 1, n \\geq 1 \\), and let \\( f(m, n)= \\) \\( f(m-1, n)+f(m, n-1)+f(m-1, n-1) \\) for \\( m>1 \\) and \\( n>1 \\). Also let\n\\[\nS(n)=\\sum_{a+b=n} f(a, b), \\quad a \\geq 1 \\text { and } b \\geq 1\n\\]\n\nProve that\n\\[\nS(n+2)=S(n)+2 S(n+1) \\quad \\text { for } n \\geq 2\n\\]",
  "solution": "Solution. If we write the value of \\( f(m, n) \\) at the point \\( \\langle m, n\\rangle \\) in the plane and border the resulting array with zeros as in the diagram,\n\\begin{tabular}{|l|l|l|l|l|l|l|}\n\\hline 0 & 1 & & \\multicolumn{4}{|c|}{\\( f(m-1, n) \\rightarrow f(m, n) \\)} \\\\\n\\hline 0 & 1 & 7 & & \\( f(m \\) & \\( -1, n-1) \\) & \\begin{tabular}{l}\n\\( f(m \\). \\\\\n\\( n-1) \\)\n\\end{tabular} \\\\\n\\hline 0 & 1 & 5 & 13 & 25 & & \\\\\n\\hline & & & & & & \\\\\n\\hline 0 & 1 & 3 & 5 & 7 & & \\\\\n\\hline 0 & 1 & 1 & 1 & 1 & 1 & \\\\\n\\hline 0 & 0 & 0 & 0 & 0 & 0 & \\\\\n\\hline\n\\end{tabular}\nwe see that the recursion relation together with the given values for \\( f(1, n) \\) and \\( f(m, 1) \\) amount to the assertion that every non-zero entry in this array (except \\( f(1,1) \\) ) is the sum of the entry immediately to its left, the entry just below it, and the entry diagonally below it to the left.\n\nNow \\( S(n+2) \\) is the sum of the terms on the \\( (n+2) \\) nd diagonal, \\( x+y \\) \\( =n+2 \\), and it is clear from the diagram that each non-zero term on the \\( (n+1) \\) st diagonal enters this sum twice while each term on the \\( n \\)th diagonal enters once; hence, \\( S(n+2)=2 S(n+1)+S(n) \\).\n\nThis argument can be carried out formally as follows:\n\\[\n\\begin{aligned}\nS(n+2)= & \\sum_{i=1}^{n+1} f(n+2-j, j) \\\\\n= & f(n+1,1)+\\sum_{i=2}^{n}\\{f(n+1-j, j)+f(n+2-j, j-1) \\\\\n& +f(n+1-j, j-1)\\}+f(1, n+1) \\\\\n= & \\left\\{f(n, 1)+\\sum_{i=2}^{n} f(n+1-j, j)\\right\\} \\\\\n+ & \\left\\{\\sum_{k=1}^{n-1} f(n+1-k, k)+f(1, n)\\right\\}+\\sum_{k=1}^{n-1} f(n-k, k) \\\\\n= & S(n+1)+S(n+1)+S(n)\n\\end{aligned}\n\\]\n\nIn the third step we set \\( j=k+1 \\) in two of the sums and used the facts that \\( f(n+1,1)=f(n, 1) \\) and \\( f(1, n+1)=f(1, n) \\).\n\nRemark. This recursion is studied in greater detail by R. G. Stanton and D. D. Cowan, \"Note on a 'Square' Functional Equation,\" SIAM Review, vol. 12, no. 2 (April 1970), pages 277-279. This problem occurs as Lemma 4 in the given paper.",
  "vars": [
    "m",
    "n",
    "a",
    "b",
    "i",
    "j",
    "k",
    "x",
    "y"
  ],
  "params": [
    "f",
    "S"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "m": "rowindex",
        "n": "colindex",
        "a": "sumterm",
        "b": "complement",
        "i": "iterator",
        "j": "subindex",
        "k": "auxindex",
        "x": "horizco",
        "y": "vertico",
        "f": "basefunc",
        "S": "summation"
      },
      "question": "1. Let \\( basefunc(rowindex, 1)=basefunc(1, colindex)=1 \\) for \\( rowindex \\geq 1, colindex \\geq 1 \\), and let \\( basefunc(rowindex, colindex)= basefunc(rowindex-1, colindex)+basefunc(rowindex, colindex-1)+basefunc(rowindex-1, colindex-1) \\) for \\( rowindex>1 \\) and \\( colindex>1 \\). Also let\n\\[\nsummation(colindex)=\\sum_{sumterm+complement=colindex} basefunc(sumterm, complement), \\quad sumterm \\geq 1 \\text { and } complement \\geq 1\n\\]\n\nProve that\n\\[\nsummation(colindex+2)=summation(colindex)+2\\,summation(colindex+1) \\quad \\text { for } colindex \\geq 2\n\\]",
      "solution": "Solution. If we write the value of \\( basefunc(rowindex, colindex) \\) at the point \\( \\langle rowindex, colindex\\rangle \\) in the plane and border the resulting array with zeros as in the diagram,\n\\begin{tabular}{|l|l|l|l|l|l|l|}\n\\hline 0 & 1 & & \\multicolumn{4}{|c|}{\\( basefunc(rowindex-1, colindex) \\rightarrow basefunc(rowindex, colindex) \\)} \\\\\n\\hline 0 & 1 & 7 & & \\( basefunc(rowindex \\) & \\( -1, colindex-1) \\) & \\begin{tabular}{l}\n\\( basefunc(rowindex \\). \\\\\n\\( colindex-1) \\)\n\\end{tabular} \\\\\n\\hline 0 & 1 & 5 & 13 & 25 & & \\\\\n\\hline & & & & & & \\\\\n\\hline 0 & 1 & 3 & 5 & 7 & & \\\\\n\\hline 0 & 1 & 1 & 1 & 1 & 1 & \\\\\n\\hline 0 & 0 & 0 & 0 & 0 & 0 & \\\\\n\\hline\n\\end{tabular}\nwe see that the recursion relation together with the given values for \\( basefunc(1, colindex) \\) and \\( basefunc(rowindex, 1) \\) amount to the assertion that every non-zero entry in this array (except \\( basefunc(1,1) \\) ) is the sum of the entry immediately to its left, the entry just below it, and the entry diagonally below it to the left.\n\nNow \\( summation(colindex+2) \\) is the sum of the terms on the \\( (colindex+2) \\) nd diagonal, \\( horizco+vertico = colindex+2 \\), and it is clear from the diagram that each non-zero term on the \\( (colindex+1) \\) st diagonal enters this sum twice while each term on the \\( colindex \\)th diagonal enters once; hence, \\( summation(colindex+2)=2\\,summation(colindex+1)+summation(colindex) \\).\n\nThis argument can be carried out formally as follows:\n\\[\n\\begin{aligned}\nsummation(colindex+2)= & \\sum_{iterator=1}^{colindex+1} basefunc(colindex+2-subindex, subindex) \\\\\n= & basefunc(colindex+1,1)+\\sum_{iterator=2}^{colindex}\\{basefunc(colindex+1-subindex, subindex)+basefunc(colindex+2-subindex, subindex-1) \\\\\n& +basefunc(colindex+1-subindex, subindex-1)\\}+basefunc(1, colindex+1) \\\\\n= & \\left\\{basefunc(colindex, 1)+\\sum_{iterator=2}^{colindex} basefunc(colindex+1-subindex, subindex)\\right\\} \\\\\n+ & \\left\\{\\sum_{auxindex=1}^{colindex-1} basefunc(colindex+1-auxindex, auxindex)+basefunc(1, colindex)\\right\\}+\\sum_{auxindex=1}^{colindex-1} basefunc(colindex-auxindex, auxindex) \\\\\n= & summation(colindex+1)+summation(colindex+1)+summation(colindex)\n\\end{aligned}\n\\]\n\nIn the third step we set \\( subindex=auxindex+1 \\) in two of the sums and used the facts that \\( basefunc(colindex+1,1)=basefunc(colindex, 1) \\) and \\( basefunc(1, colindex+1)=basefunc(1, colindex) \\).\n\nRemark. This recursion is studied in greater detail by R. G. Stanton and D. D. Cowan, \"Note on a 'Square' Functional Equation,\" SIAM Review, vol. 12, no. 2 (April 1970), pages 277-279. This problem occurs as Lemma 4 in the given paper."
    },
    "descriptive_long_confusing": {
      "map": {
        "m": "pinecone",
        "n": "thunderbolt",
        "a": "dinosaur",
        "b": "avalanche",
        "i": "i",
        "j": "unicorns",
        "k": "jellyfish",
        "x": "tornadoes",
        "y": "sandstorm",
        "f": "blueprint",
        "S": "lighthouse"
      },
      "question": "1. Let \\( blueprint(pinecone, 1)=blueprint(1, thunderbolt)=1 \\) for \\( pinecone \\geq 1, thunderbolt \\geq 1 \\), and let \\( blueprint(pinecone, thunderbolt)= \\) \\( blueprint(pinecone-1, thunderbolt)+blueprint(pinecone, thunderbolt-1)+blueprint(pinecone-1, thunderbolt-1) \\) for \\( pinecone>1 \\) and \\( thunderbolt>1 \\). Also let\n\\[\nlighthouse(thunderbolt)=\\sum_{dinosaur+avalanche=thunderbolt} blueprint(dinosaur, avalanche), \\quad dinosaur \\geq 1 \\text { and } avalanche \\geq 1\n\\]\n\nProve that\n\\[\nlighthouse(thunderbolt+2)=lighthouse(thunderbolt)+2\\, lighthouse(thunderbolt+1) \\quad \\text { for } thunderbolt \\geq 2\n\\]",
      "solution": "Solution. If we write the value of \\( blueprint(pinecone, thunderbolt) \\) at the point \\( \\langle pinecone, thunderbolt\\rangle \\) in the plane and border the resulting array with zeros as in the diagram,\n\\begin{tabular}{|l|l|l|l|l|l|l|}\n\\hline 0 & 1 & & \\multicolumn{4}{|c|}{\\( blueprint(pinecone-1, thunderbolt) \\rightarrow blueprint(pinecone, thunderbolt) \\)} \\\\\n\\hline 0 & 1 & 7 & & \\( blueprint(pinecone \\) & \\( -1, thunderbolt-1) \\) & \\begin{tabular}{l}\n\\( blueprint(pinecone \\). \\\\\n\\( thunderbolt-1) \\)\n\\end{tabular} \\\\\n\\hline 0 & 1 & 5 & 13 & 25 & & \\\\\n\\hline & & & & & & \\\\\n\\hline 0 & 1 & 3 & 5 & 7 & & \\\\\n\\hline 0 & 1 & 1 & 1 & 1 & 1 & \\\\\n\\hline 0 & 0 & 0 & 0 & 0 & 0 & \\\\\n\\hline\n\\end{tabular}\nwe see that the recursion relation together with the given values for \\( blueprint(1, thunderbolt) \\) and \\( blueprint(pinecone, 1) \\) amount to the assertion that every non-zero entry in this array (except \\( blueprint(1,1) \\) ) is the sum of the entry immediately to its left, the entry just below it, and the entry diagonally below it to the left.\n\nNow \\( lighthouse(thunderbolt+2) \\) is the sum of the terms on the \\( (thunderbolt+2) \\)nd diagonal, \\( tornadoes+sandstorm \\)\n\\( =thunderbolt+2 \\), and it is clear from the diagram that each non-zero term on the \\( (thunderbolt+1) \\)st diagonal enters this sum twice while each term on the \\( thunderbolt \\)th diagonal enters once; hence, \\( lighthouse(thunderbolt+2)=2\\,lighthouse(thunderbolt+1)+lighthouse(thunderbolt) \\).\n\nThis argument can be carried out formally as follows:\n\\[\n\\begin{aligned}\nlighthouse(thunderbolt+2)= & \\sum_{i=1}^{thunderbolt+1} blueprint(thunderbolt+2-unicorns, unicorns) \\\\\n= & blueprint(thunderbolt+1,1)+\\sum_{i=2}^{thunderbolt}\\{blueprint(thunderbolt+1-unicorns, unicorns)+blueprint(thunderbolt+2-unicorns, unicorns-1) \\\\\n& +blueprint(thunderbolt+1-unicorns, unicorns-1)\\}+blueprint(1, thunderbolt+1) \\\\\n= & \\left\\{blueprint(thunderbolt, 1)+\\sum_{i=2}^{thunderbolt} blueprint(thunderbolt+1-unicorns, unicorns)\\right\\} \\\\\n+ & \\left\\{\\sum_{jellyfish=1}^{thunderbolt-1} blueprint(thunderbolt+1-jellyfish, jellyfish)+blueprint(1, thunderbolt)\\right\\}+\\sum_{jellyfish=1}^{thunderbolt-1} blueprint(thunderbolt-jellyfish, jellyfish) \\\\\n= & lighthouse(thunderbolt+1)+lighthouse(thunderbolt+1)+lighthouse(thunderbolt)\n\\end{aligned}\n\\]\n\nIn the third step we set \\( unicorns=jellyfish+1 \\) in two of the sums and used the facts that \\( blueprint(thunderbolt+1,1)=blueprint(thunderbolt, 1) \\) and \\( blueprint(1, thunderbolt+1)=blueprint(1, thunderbolt) \\).\n\nRemark. This recursion is studied in greater detail by R. G. Stanton and D. D. Cowan, \"Note on a 'Square' Functional Equation,\" SIAM Review, vol. 12, no. 2 (April 1970), pages 277-279. This problem occurs as Lemma 4 in the given paper."
    },
    "descriptive_long_misleading": {
      "map": {
        "m": "constantone",
        "n": "constanttwo",
        "a": "constantthr",
        "b": "constantfour",
        "i": "constantfive",
        "j": "constantsix",
        "k": "constantseven",
        "x": "constanteight",
        "y": "constantnine",
        "f": "staticfunc",
        "S": "differencer"
      },
      "question": "1. Let \\( staticfunc(constantone, 1)=staticfunc(1, constanttwo)=1 \\) for \\( constantone \\geq 1, constanttwo \\geq 1 \\), and let \\( staticfunc(constantone, constanttwo)= \\) \\( staticfunc(constantone-1, constanttwo)+staticfunc(constantone, constanttwo-1)+staticfunc(constantone-1, constanttwo-1) \\) for \\( constantone>1 \\) and \\( constanttwo>1 \\). Also let\n\\[\ndifferencer(constanttwo)=\\sum_{constantthr+constantfour=constanttwo} staticfunc(constantthr, constantfour), \\quad constantthr \\geq 1 \\text { and } constantfour \\geq 1\n\\]\n\nProve that\n\\[\ndifferencer(constanttwo+2)=differencer(constanttwo)+2 differencer(constanttwo+1) \\quad \\text { for } constanttwo \\geq 2\n\\]",
      "solution": "Solution. If we write the value of \\( staticfunc(constantone, constanttwo) \\) at the point \\( \\langle constantone, constanttwo\\rangle \\) in the plane and border the resulting array with zeros as in the diagram,\n\\begin{tabular}{|l|l|l|l|l|l|l|}\n\\hline 0 & 1 & & \\multicolumn{4}{|c|}{\\( staticfunc(constantone-1, constanttwo) \\rightarrow staticfunc(constantone, constanttwo) \\)} \\\\\n\\hline 0 & 1 & 7 & & \\( staticfunc(constantone \\) & \\( -1, constanttwo-1) \\) & \\begin{tabular}{l}\n\\( staticfunc(constantone \\). \\\\\n\\( constanttwo-1) \\)\n\\end{tabular} \\\\\n\\hline 0 & 1 & 5 & 13 & 25 & & \\\\\n\\hline & & & & & & \\\\\n\\hline 0 & 1 & 3 & 5 & 7 & & \\\\\n\\hline 0 & 1 & 1 & 1 & 1 & 1 & \\\\\n\\hline 0 & 0 & 0 & 0 & 0 & 0 & \\\\\n\\hline\n\\end{tabular}\nwe see that the recursion relation together with the given values for \\( staticfunc(1, constanttwo) \\) and \\( staticfunc(constantone, 1) \\) amount to the assertion that every non-zero entry in this array (except \\( staticfunc(1,1) \\) ) is the sum of the entry immediately to its left, the entry just below it, and the entry diagonally below it to the left.\n\nNow \\( differencer(constanttwo+2) \\) is the sum of the terms on the \\( (constanttwo+2) \\) nd diagonal, \\( constanteight+constantnine \\) \\( =constanttwo+2 \\), and it is clear from the diagram that each non-zero term on the \\( (constanttwo+1) \\) st diagonal enters this sum twice while each term on the \\( constanttwo \\)th diagonal enters once; hence, \\( differencer(constanttwo+2)=2\\,differencer(constanttwo+1)+differencer(constanttwo) \\).\n\nThis argument can be carried out formally as follows:\n\\[\n\\begin{aligned}\ndifferencer(constanttwo+2)= & \\sum_{constantfive=1}^{constanttwo+1} staticfunc(constanttwo+2-constantsix, constantsix) \\\\\n= & staticfunc(constanttwo+1,1)+\\sum_{constantfive=2}^{constanttwo}\\{staticfunc(constanttwo+1-constantsix, constantsix)+staticfunc(constanttwo+2-constantsix, constantsix-1) \\\\\n& +staticfunc(constanttwo+1-constantsix, constantsix-1)\\}+staticfunc(1, constanttwo+1) \\\\\n= & \\left\\{staticfunc(constanttwo, 1)+\\sum_{constantfive=2}^{constanttwo} staticfunc(constanttwo+1-constantsix, constantsix)\\right\\} \\\\\n+ & \\left\\{\\sum_{constantseven=1}^{constanttwo-1} staticfunc(constanttwo+1-constantseven, constantseven)+staticfunc(1, constanttwo)\\right\\}+\\sum_{constantseven=1}^{constanttwo-1} staticfunc(constanttwo-constantseven, constantseven) \\\\\n= & differencer(constanttwo+1)+differencer(constanttwo+1)+differencer(constanttwo)\n\\end{aligned}\n\\]\n\nIn the third step we set \\( constantsix=constantseven+1 \\) in two of the sums and used the facts that \\( staticfunc(constanttwo+1,1)=staticfunc(constanttwo, 1) \\) and \\( staticfunc(1, constanttwo+1)=staticfunc(1, constanttwo) \\).\n\nRemark. This recursion is studied in greater detail by R. G. Stanton and D. D. Cowan, \"Note on a 'Square' Functional Equation,\" SIAM Review, vol. 12, no. 2 (April 1970), pages 277-279. This problem occurs as Lemma 4 in the given paper."
    },
    "garbled_string": {
      "map": {
        "m": "qzxwvtnp",
        "n": "hjgrksla",
        "a": "vlmnyqrs",
        "b": "tcdkufpe",
        "i": "sowrtbmk",
        "j": "pfkzleqh",
        "k": "dyrpwmse",
        "x": "agbcsnmd",
        "y": "zlhvkrue",
        "f": "xowbeltr",
        "S": "udftqzni"
      },
      "question": "1. Let \\( xowbeltr(qzxwvtnp, 1)=xowbeltr(1, hjgrksla)=1 \\) for \\( qzxwvtnp \\geq 1, hjgrksla \\geq 1 \\), and let \\( xowbeltr(qzxwvtnp, hjgrksla)= \\) \\( xowbeltr(qzxwvtnp-1, hjgrksla)+xowbeltr(qzxwvtnp, hjgrksla-1)+xowbeltr(qzxwvtnp-1, hjgrksla-1) \\) for \\( qzxwvtnp>1 \\) and \\( hjgrksla>1 \\). Also let\n\\[\nudftqzni(hjgrksla)=\\sum_{vlmnyqrs+tcdkufpe=hjgrksla} xowbeltr(vlmnyqrs, tcdkufpe), \\quad vlmnyqrs \\geq 1 \\text { and } tcdkufpe \\geq 1\n\\]\n\nProve that\n\\[\nudftqzni(hjgrksla+2)=udftqzni(hjgrksla)+2 \\, udftqzni(hjgrksla+1) \\quad \\text { for } hjgrksla \\geq 2\n\\]",
      "solution": "Solution. If we write the value of \\( xowbeltr(qzxwvtnp, hjgrksla) \\) at the point \\( \\langle qzxwvtnp, hjgrksla\\rangle \\) in the plane and border the resulting array with zeros as in the diagram,\n\\begin{tabular}{|l|l|l|l|l|l|l|}\n\\hline 0 & 1 & & \\multicolumn{4}{|c|}{\\( xowbeltr(qzxwvtnp-1, hjgrksla) \\rightarrow xowbeltr(qzxwvtnp, hjgrksla) \\)} \\\\\n\\hline 0 & 1 & 7 & & \\( xowbeltr(qzxwvtnp \\) & \\( -1, hjgrksla-1) \\) & \\begin{tabular}{l}\n\\( xowbeltr(qzxwvtnp \\). \\\\\n\\( hjgrksla-1) \\)\n\\end{tabular} \\\\\n\\hline 0 & 1 & 5 & 13 & 25 & & \\\\\n\\hline & & & & & & \\\\\n\\hline 0 & 1 & 3 & 5 & 7 & & \\\\\n\\hline 0 & 1 & 1 & 1 & 1 & 1 & \\\\\n\\hline 0 & 0 & 0 & 0 & 0 & 0 & \\\\\n\\hline\n\\end{tabular}\nwe see that the recursion relation together with the given values for \\( xowbeltr(1, hjgrksla) \\) and \\( xowbeltr(qzxwvtnp, 1) \\) amount to the assertion that every non-zero entry in this array (except \\( xowbeltr(1,1) \\) ) is the sum of the entry immediately to its left, the entry just below it, and the entry diagonally below it to the left.\n\nNow \\( udftqzni(hjgrksla+2) \\) is the sum of the terms on the \\( (hjgrksla+2) \\) nd diagonal, \\( agbcsnmd+zlhvkrue =hjgrksla+2 \\), and it is clear from the diagram that each non-zero term on the \\( (hjgrksla+1) \\) st diagonal enters this sum twice while each term on the \\( hjgrksla \\)th diagonal enters once; hence, \\( udftqzni(hjgrksla+2)=2 \\, udftqzni(hjgrksla+1)+udftqzni(hjgrksla) \\).\n\nThis argument can be carried out formally as follows:\n\\[\n\\begin{aligned}\nudftqzni(hjgrksla+2)= & \\sum_{sowrtbmk=1}^{hjgrksla+1} xowbeltr(hjgrksla+2-pfkzleqh, pfkzleqh) \\\\\n= & xowbeltr(hjgrksla+1,1)+\\sum_{sowrtbmk=2}^{hjgrksla}\\{xowbeltr(hjgrksla+1-pfkzleqh, pfkzleqh)+xowbeltr(hjgrksla+2-pfkzleqh, pfkzleqh-1) \\\\\n& +xowbeltr(hjgrksla+1-pfkzleqh, pfkzleqh-1)\\}+xowbeltr(1, hjgrksla+1) \\\\\n= & \\left\\{xowbeltr(hjgrksla, 1)+\\sum_{sowrtbmk=2}^{hjgrksla} xowbeltr(hjgrksla+1-pfkzleqh, pfkzleqh)\\right\\} \\\\\n+ & \\left\\{\\sum_{dyrpwmse=1}^{hjgrksla-1} xowbeltr(hjgrksla+1-dyrpwmse, dyrpwmse)+xowbeltr(1, hjgrksla)\\right\\}+\\sum_{dyrpwmse=1}^{hjgrksla-1} xowbeltr(hjgrksla-dyrpwmse, dyrpwmse) \\\\\n= & udftqzni(hjgrksla+1)+udftqzni(hjgrksla+1)+udftqzni(hjgrksla)\n\\end{aligned}\n\\]\n\nIn the third step we set \\( pfkzleqh=dyrpwmse+1 \\) in two of the sums and used the facts that \\( xowbeltr(hjgrksla+1,1)=xowbeltr(hjgrksla, 1) \\) and \\( xowbeltr(1, hjgrksla+1)=xowbeltr(1, hjgrksla) \\).\n\nRemark. This recursion is studied in greater detail by R. G. Stanton and D. D. Cowan, \"Note on a 'Square' Functional Equation,\" SIAM Review, vol. 12, no. 2 (April 1970), pages 277-279. This problem occurs as Lemma 4 in the given paper."
    },
    "kernel_variant": {
      "question": "Let a function g: ℤ_{>0} × ℤ_{>0} → ℝ be defined by the boundary values\n  g(m,1) = g(1,n) = 5      (m ≥ 1, n ≥ 1)\nand the interior recursion\n  g(m,n) = g(m−1,n) + g(m,n−1) + g(m−1,n−1)      (m>1, n>1).\nFor every integer k ≥ 1 put\n  T(k) = ∑_{\\substack{a+b = k\\\\ a,b ≥ 1}} g(a,b).\n(The sum is empty when k = 1, so T(1) = 0.)\nProve that the diagonal sums satisfy the recurrence relation\n  T(k+2) = 2 T(k+1) + T(k)      for all k ≥ 1.",
      "solution": "Step 1 – Extending g with a zero border.\nIntroduce the auxiliary values\n      g(0,n) = g(m,0) = 0   (m,n ≥ 0).\nWith this convention the relation\n      g(m,n) = g(m−1,n) + g(m,n−1) + g(m−1,n−1)            (1)\nnow holds for every pair (m,n) with m,n ≥ 1 except for the single point (1,1).  Indeed:\n• If m = 1 and n > 1, the right–hand side is 0 + 5 + 0 = 5 = g(1,n).\n• If n = 1 and m > 1, it is 5 + 0 + 0 = 5 = g(m,1).\n• If m,n > 1, (1) is exactly the given interior rule.\n(The case m = n = 1 gives 0 ≠ 5, so (1) is not claimed there.)\n\nStep 2 – Antidiagonal notation.\nFor r ≥ 1 define\n      D_r = { (a,b) ∈ ℤ_{>0}² : a + b = r },\n      T(r) = ∑_{(a,b)∈D_r} g(a,b).\nNote that D_1 = ∅, hence T(1) = 0 as required.\n\nStep 3 – Summing the recursion along D_{k+2}.\nFix k ≥ 1.  Since every point of D_{k+2} satisfies a+b = k+2 ≥ 3, it is different from (1,1); therefore formula (1) is valid for all points of D_{k+2}.  Summing (1) over that diagonal yields\n      T(k+2) = ∑_{a+b=k+2} g(a,b)\n             = ∑_{a+b=k+2} g(a−1,b)   +   ∑_{a+b=k+2} g(a,b−1)\n               + ∑_{a+b=k+2} g(a−1,b−1).          (2)\n\nStep 4 – Re-indexing the three sums.\n• First sum: put u = a−1, v = b.  Then u+v = k+1 and v ≥ 1.  When u = 0 the term is g(0,v)=0, so only u ≥ 1 contributes.  Thus the sum equals T(k+1).\n\n• Second sum: put u = a, v = b−1.  Again u+v = k+1 with u ≥ 1, and possible v = 0 gives g(u,0)=0.  Hence this sum is another T(k+1).\n\n• Third sum: put u = a−1, v = b−1.  Then u+v = k with u,v ≥ 0.  Terms with u = 0 or v = 0 vanish, so the sum equals T(k).\n\nSubstituting these evaluations into (2) gives\n      T(k+2) = T(k+1) + T(k+1) + T(k) = 2 T(k+1) + T(k)\nfor every k ≥ 1.\n\nStep 5 – Initial values and conclusion.\nBecause T(1)=0 and T(2)=g(1,1)=5, the recurrence determines all further values (for instance T(3)=g(1,2)+g(2,1)=10).  The identity T(k+2)=2T(k+1)+T(k) is therefore proved.",
      "_meta": {
        "core_steps": [
          "View the values f(m,n) as entries of an infinite array bordered by constant edge–values.",
          "Note that the given recursion makes every interior entry the sum of its left, lower and lower-left neighbours.",
          "Realise that S(k) is precisely the sum of the entries lying on the diagonal m+n = k.",
          "Apply the recursion to every entry of the (n+2)-diagonal; each term of the (n+1)-diagonal is created twice and each term of the n-diagonal once.",
          "Collect coefficients to obtain S(n+2)=2·S(n+1)+S(n)."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Common constant prescribed on the two coordinate axes (the ‘boundary value’  for f(m,1) and f(1,n)).  Because all later steps are linear, replacing 1 by any common constant merely scales every f–value and every S(k) by that factor, leaving the final relation unchanged.",
            "original": "1"
          },
          "slot2": {
            "description": "The lower index from which the identity is asserted; any starting index for which the three involved diagonals exist (e.g. n≥1 instead of n≥2) works equally well.",
            "original": "n ≥ 2"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}