{
  "index": "1957-B-4",
  "type": "COMB",
  "tag": [
    "COMB",
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "4. Let \\( a(n) \\) be the number of representations of the positive integer \\( n \\) as the sums of 1's and 2's taking order into account. For example, since\n\\[\n\\begin{array}{c}\n4=1+1+2=1+2+1=2+1+1 \\\\\n=2+2=1+1+1+1\n\\end{array}\n\\]\nthen \\( a(4)=5 \\). Let \\( b(n) \\) be the number of representations of \\( n \\) as the sum of integers greater than 1 , again taking order into account and counting the summand \\( n \\). For example, since \\( 6=4+2=2+4=3+3=2+2+2 \\), we have \\( b(6)=5 \\). Show that for each \\( n, a(n)=b(n+2) \\).",
  "solution": "First Solution. By direct counting we can find the first few values of the two sequences:\n\\begin{tabular}{c|rrrrrr}\n\\( n \\) & 1 & 2 & 3 & 4 & 5 & 6 \\\\\n\\hline\\( a(n) \\) & 1 & 2 & 3 & 5 & 8 & 13 \\\\\n\\hline\\( b(n) \\) & 0 & 1 & 1 & 2 & 3 & 5\n\\end{tabular}\n\nThus it appears that the sequences are the well-known Fibonacci sequence satisfying the recursion\n\\[\na(n+1)=a(n)+a(n-1), \\quad n=2,3, \\ldots\n\\]\n\nTo prove that this is indeed the case, note that any representation of \\( n+1 \\) as an ordered sum of 1's and 2's becomes, on removal of the last summand, either a representation of \\( n \\) (if the removed term is a 1 ) or a representation of \\( n-1 \\) (if the removed term is a 2 ). Conversely, any representation of either \\( n \\) or \\( n-1 \\) becomes a representation of \\( n+1 \\) on adjoining either a 1 or 2 , as appropriate. It follows that the \\( a \\)-sequence does indeed satisfy the recursion (1).\n\nNext we show that the b's satisfy the same recursion. Delete the last term from a representation of \\( n+1 \\) as a sum of integers greater than 1. We obtain then either a representation of \\( n-1, n-2, \\ldots, 2 \\), or a vacuous sum. Conversely, any such representation extends to a representation of \\( n+1 \\). Hence for \\( n \\geq 1 \\)\n\\[\nb(n+1)=b(n-1)+b(n-2)+\\cdots+b(2)+1\n\\]\n\nIf \\( \\boldsymbol{n} \\) is at least 2, we have also\n\\[\nb(n)=b(n-2)+\\cdots+b(2)+1\n\\]\n\nComparing (2) and (3), we find\n\\[\nb(n+1)=b(n-1)+b(n), \\quad n \\geq 2\n\\]\n\nSo the \\( b \\)-sequence satisfies the same recursion as the \\( a \\)-sequence.\nThen since \\( b(3)=a(1), b(4)=a(2) \\), it follows by induction that \\( b(n+2)=a(n) \\) for all \\( n \\geq 1 \\).\n\nSecond Solution. We can find the power-series generating function for \\( a(n) \\). The number of ways that \\( n \\) can be represented as an ordered sum of \\( k \\) 1's and 2's is clearly the coefficient of \\( x^{n} \\) in \\( \\left(x+x^{2}\\right)^{k} \\). Hence\n\\[\n1+\\sum_{n-1}^{\\infty} a(n) x^{n}=\\sum_{k=0}^{\\infty}\\left(x+x^{2}\\right)^{k}=\\frac{1}{1-x-x^{2}}\n\\]\n\nOn the other hand, the number of ways that \\( n \\) can be represented as an ordered sum of \\( k \\) integers greater than 1 is the coefficient of \\( x^{n} \\) in\n\\[\n\\left(x^{2}+x^{3}+\\cdots\\right)^{k}=\\left(\\frac{x^{2}}{1-x}\\right)^{k}\n\\]\n\nSo\n\\[\n\\begin{aligned}\n1+\\sum_{n=2}^{\\infty} b(n) x^{n}=\\sum_{k=0}^{\\infty}\\left(\\frac{x^{2}}{1-x}\\right)^{k} & =\\left(1-\\frac{x^{2}}{1-x}\\right)^{-1} \\\\\n& =\\frac{1-x}{1-x-x^{2}} \\\\\n& =1+\\frac{x^{2}}{1-x-x^{2}}\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{n=2}^{\\infty} b(n) x^{n}=x^{2}+x^{2} \\sum_{n=1}^{\\infty} a(n) x^{n}\n\\]\nand the result follows immediately.\n[The formal manipulation of the series can be justified either by considering only small values of \\( x \\), for example \\( |x|<\\frac{1}{10} \\), which makes each series involved absolutely convergent, or by considering the ring of formal power series in which convergence means only the ultimate stabilization of terns of each degree.]\n\nThird Solution. A one-to-one correspondence between representations of \\( n \\) as a sum of 1's and 2's and representations of \\( n+2 \\) as a sum of integers greater than 1 can be established as follows. Representations of \\( n \\) as an ordered sum of 1's and 2's are obviously in one-to-one correspondence with representations of \\( n+2 \\) as a sum of 1 's and 2 's ending with a 2 (and having one more summand). If in such a representation written in linear order we bracket each 2 with the longest string of 1's immediately preceding it, we obtain a representation of \\( n+2 \\) as an ordered sum of integers greater than 1 . Conversely, in any representation of \\( n+2 \\) as such an ordered sum, we can replace each summand by a string of 1 's followed by a single 2 .\n\nExample. The following representation of 9 as an ordered sum of 1 's and 2's,\n\\[\n9=1+1+2+2+2+1\n\\]\ncorresponds to the representation of \\( 11(=9+2) \\)\n\\[\n11=1+1+2+2+2+1+2\n\\]\nwhich corresponds to\n\\[\n\\begin{aligned}\n11 & =(1+1+2)+(2)+(2)+(1+2) \\\\\n& =4+2+2+3\n\\end{aligned}\n\\]\nand also conversely.\nIt follows that \\( b(n+2)=a(n) \\)",
  "vars": [
    "n",
    "k",
    "x"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "indexn",
        "k": "counterk",
        "x": "varxen",
        "a": "aseqid",
        "b": "bseqid"
      },
      "question": "Let \\( aseqid(indexn) \\) be the number of representations of the positive integer \\( indexn \\) as the sums of 1's and 2's taking order into account. For example, since\n\\[\n\\begin{array}{c}\n4=1+1+2=1+2+1=2+1+1 \\\\\n=2+2=1+1+1+1\n\\end{array}\n\\]\nthen \\( aseqid(4)=5 \\). Let \\( bseqid(indexn) \\) be the number of representations of \\( indexn \\) as the sum of integers greater than 1 , again taking order into account and counting the summand \\( indexn \\). For example, since \\( 6=4+2=2+4=3+3=2+2+2 \\), we have \\( bseqid(6)=5 \\). Show that for each \\( indexn, aseqid(indexn)=bseqid(indexn+2) \\).",
      "solution": "First Solution. By direct counting we can find the first few values of the two sequences:\n\\begin{tabular}{c|rrrrrr}\n\\( indexn \\) & 1 & 2 & 3 & 4 & 5 & 6 \\\\\n\\hline\\( aseqid(indexn) \\) & 1 & 2 & 3 & 5 & 8 & 13 \\\\\n\\hline\\( bseqid(indexn) \\) & 0 & 1 & 1 & 2 & 3 & 5\n\\end{tabular}\n\nThus it appears that the sequences are the well-known Fibonacci sequence satisfying the recursion\n\\[\naseqid(indexn+1)=aseqid(indexn)+aseqid(indexn-1), \\quad indexn=2,3, \\ldots\n\\]\n\nTo prove that this is indeed the case, note that any representation of \\( indexn+1 \\) as an ordered sum of 1's and 2's becomes, on removal of the last summand, either a representation of \\( indexn \\) (if the removed term is a 1) or a representation of \\( indexn-1 \\) (if the removed term is a 2). Conversely, any representation of either \\( indexn \\) or \\( indexn-1 \\) becomes a representation of \\( indexn+1 \\) on adjoining either a 1 or 2, as appropriate. It follows that the \\( aseqid \\)-sequence does indeed satisfy the recursion (1).\n\nNext we show that the \\( bseqid \\)-sequence satisfies the same recursion. Delete the last term from a representation of \\( indexn+1 \\) as a sum of integers greater than 1. We obtain then either a representation of \\( indexn-1, indexn-2, \\ldots, 2 \\), or a vacuous sum. Conversely, any such representation extends to a representation of \\( indexn+1 \\). Hence for \\( indexn \\geq 1 \\)\n\\[\nbseqid(indexn+1)=bseqid(indexn-1)+bseqid(indexn-2)+\\cdots+bseqid(2)+1\n\\]\n\nIf \\( \\boldsymbol{indexn} \\) is at least 2, we have also\n\\[\nbseqid(indexn)=bseqid(indexn-2)+\\cdots+bseqid(2)+1\n\\]\n\nComparing (2) and (3), we find\n\\[\nbseqid(indexn+1)=bseqid(indexn-1)+bseqid(indexn), \\quad indexn \\geq 2\n\\]\n\nSo the \\( bseqid \\)-sequence satisfies the same recursion as the \\( aseqid \\)-sequence.\nThen since \\( bseqid(3)=aseqid(1), bseqid(4)=aseqid(2) \\), it follows by induction that \\( bseqid(indexn+2)=aseqid(indexn) \\) for all \\( indexn \\geq 1 \\).\n\nSecond Solution. We can find the power-series generating function for \\( aseqid(indexn) \\). The number of ways that \\( indexn \\) can be represented as an ordered sum of \\( counterk \\) 1's and 2's is clearly the coefficient of \\( varxen^{indexn} \\) in \\( \\left(varxen+varxen^{2}\\right)^{counterk} \\). Hence\n\\[\n1+\\sum_{indexn-1}^{\\infty} aseqid(indexn) varxen^{indexn}=\\sum_{counterk=0}^{\\infty}\\left(varxen+varxen^{2}\\right)^{counterk}=\\frac{1}{1-varxen-varxen^{2}}\n\\]\n\nOn the other hand, the number of ways that \\( indexn \\) can be represented as an ordered sum of \\( counterk \\) integers greater than 1 is the coefficient of \\( varxen^{indexn} \\) in\n\\[\n\\left(varxen^{2}+varxen^{3}+\\cdots\\right)^{counterk}=\\left(\\frac{varxen^{2}}{1-varxen}\\right)^{counterk}\n\\]\n\nSo\n\\[\n\\begin{aligned}\n1+\\sum_{indexn=2}^{\\infty} bseqid(indexn) varxen^{indexn}=\\sum_{counterk=0}^{\\infty}\\left(\\frac{varxen^{2}}{1-varxen}\\right)^{counterk} & =\\left(1-\\frac{varxen^{2}}{1-varxen}\\right)^{-1} \\\\\n& =\\frac{1-varxen}{1-varxen-varxen^{2}} \\\\\n& =1+\\frac{varxen^{2}}{1-varxen-varxen^{2}}\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{indexn=2}^{\\infty} bseqid(indexn) varxen^{indexn}=varxen^{2}+varxen^{2} \\sum_{indexn=1}^{\\infty} aseqid(indexn) varxen^{indexn}\n\\]\nand the result follows immediately.\n[The formal manipulation of the series can be justified either by considering only small values of \\( varxen \\), for example \\( |varxen|<\\frac{1}{10} \\), which makes each series involved absolutely convergent, or by considering the ring of formal power series in which convergence means only the ultimate stabilization of terms of each degree.]\n\nThird Solution. A one-to-one correspondence between representations of \\( indexn \\) as a sum of 1's and 2's and representations of \\( indexn+2 \\) as a sum of integers greater than 1 can be established as follows. Representations of \\( indexn \\) as an ordered sum of 1's and 2's are obviously in one-to-one correspondence with representations of \\( indexn+2 \\) as a sum of 1's and 2's ending with a 2 (and having one more summand). If in such a representation written in linear order we bracket each 2 with the longest string of 1's immediately preceding it, we obtain a representation of \\( indexn+2 \\) as an ordered sum of integers greater than 1. Conversely, in any representation of \\( indexn+2 \\) as such an ordered sum, we can replace each summand by a string of 1's followed by a single 2.\n\nExample. The following representation of 9 as an ordered sum of 1's and 2's,\n\\[\n9=1+1+2+2+2+1\n\\]\ncorresponds to the representation of \\( 11(=9+2) \\)\n\\[\n11=1+1+2+2+2+1+2\n\\]\nwhich corresponds to\n\\[\n\\begin{aligned}\n11 & =(1+1+2)+(2)+(2)+(1+2) \\\\\n& =4+2+2+3\n\\end{aligned}\n\\]\nand also conversely.\nIt follows that \\( bseqid(indexn+2)=aseqid(indexn) \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "watermelon",
        "k": "marshmallow",
        "x": "rattlesnake",
        "a": "spacecraft",
        "b": "lighthouse"
      },
      "question": "4. Let \\( spacecraft(watermelon) \\) be the number of representations of the positive integer \\( watermelon \\) as the sums of 1's and 2's taking order into account. For example, since\n\\[\n\\begin{array}{c}\n4=1+1+2=1+2+1=2+1+1 \\\\\n=2+2=1+1+1+1\n\\end{array}\n\\]\nthen \\( spacecraft(4)=5 \\). Let \\( lighthouse(watermelon) \\) be the number of representations of \\( watermelon \\) as the sum of integers greater than 1 , again taking order into account and counting the summand \\( watermelon \\). For example, since \\( 6=4+2=2+4=3+3=2+2+2 \\), we have \\( lighthouse(6)=5 \\). Show that for each \\( watermelon, spacecraft(watermelon)=lighthouse(watermelon+2) \\).",
      "solution": "First Solution. By direct counting we can find the first few values of the two sequences:\n\\begin{tabular}{c|rrrrrr}\n\\( watermelon \\) & 1 & 2 & 3 & 4 & 5 & 6 \\\\\n\\hline\\( spacecraft(watermelon) \\) & 1 & 2 & 3 & 5 & 8 & 13 \\\\\n\\hline\\( lighthouse(watermelon) \\) & 0 & 1 & 1 & 2 & 3 & 5\n\\end{tabular}\n\nThus it appears that the sequences are the well-known Fibonacci sequence satisfying the recursion\n\\[\nspacecraft(watermelon+1)=spacecraft(watermelon)+spacecraft(watermelon-1), \\quad watermelon=2,3, \\ldots\n\\]\n\nTo prove that this is indeed the case, note that any representation of \\( watermelon+1 \\) as an ordered sum of 1's and 2's becomes, on removal of the last summand, either a representation of \\( watermelon \\) (if the removed term is a 1 ) or a representation of \\( watermelon-1 \\) (if the removed term is a 2 ). Conversely, any representation of either \\( watermelon \\) or \\( watermelon-1 \\) becomes a representation of \\( watermelon+1 \\) on adjoining either a 1 or 2 , as appropriate. It follows that the spacecraft-sequence does indeed satisfy the recursion (1).\n\nNext we show that the lighthouse's satisfy the same recursion. Delete the last term from a representation of \\( watermelon+1 \\) as a sum of integers greater than 1. We obtain then either a representation of \\( watermelon-1, watermelon-2, \\ldots, 2 \\), or a vacuous sum. Conversely, any such representation extends to a representation of \\( watermelon+1 \\). Hence for \\( watermelon \\geq 1 \\)\n\\[\nlighthouse(watermelon+1)=lighthouse(watermelon-1)+lighthouse(watermelon-2)+\\cdots+lighthouse(2)+1\n\\]\n\nIf \\( \\boldsymbol{watermelon} \\) is at least 2, we have also\n\\[\nlighthouse(watermelon)=lighthouse(watermelon-2)+\\cdots+lighthouse(2)+1\n\\]\n\nComparing (2) and (3), we find\n\\[\nlighthouse(watermelon+1)=lighthouse(watermelon-1)+lighthouse(watermelon), \\quad watermelon \\geq 2\n\\]\n\nSo the lighthouse-sequence satisfies the same recursion as the spacecraft-sequence.\nThen since \\( lighthouse(3)=spacecraft(1), lighthouse(4)=spacecraft(2) \\), it follows by induction that \\( lighthouse(watermelon+2)=spacecraft(watermelon) \\) for all \\( watermelon \\geq 1 \\).\n\nSecond Solution. We can find the power-series generating function for \\( spacecraft(watermelon) \\). The number of ways that \\( watermelon \\) can be represented as an ordered sum of \\( marshmallow \\) 1's and 2's is clearly the coefficient of \\( rattlesnake^{watermelon} \\) in \\( \\left(rattlesnake+rattlesnake^{2}\\right)^{marshmallow} \\). Hence\n\\[\n1+\\sum_{watermelon-1}^{\\infty} spacecraft(watermelon) rattlesnake^{watermelon}=\\sum_{marshmallow=0}^{\\infty}\\left(rattlesnake+rattlesnake^{2}\\right)^{marshmallow}=\\frac{1}{1-rattlesnake-rattlesnake^{2}}\n\\]\n\nOn the other hand, the number of ways that \\( watermelon \\) can be represented as an ordered sum of \\( marshmallow \\) integers greater than 1 is the coefficient of \\( rattlesnake^{watermelon} \\) in\n\\[\n\\left(rattlesnake^{2}+rattlesnake^{3}+\\cdots\\right)^{marshmallow}=\\left(\\frac{rattlesnake^{2}}{1-rattlesnake}\\right)^{marshmallow}\n\\]\n\nSo\n\\[\n\\begin{aligned}\n1+\\sum_{watermelon=2}^{\\infty} lighthouse(watermelon) rattlesnake^{watermelon}=\\sum_{marshmallow=0}^{\\infty}\\left(\\frac{rattlesnake^{2}}{1-rattlesnake}\\right)^{marshmallow} & =\\left(1-\\frac{rattlesnake^{2}}{1-rattlesnake}\\right)^{-1} \\\\\n& =\\frac{1-rattlesnake}{1-rattlesnake-rattlesnake^{2}} \\\\\n& =1+\\frac{rattlesnake^{2}}{1-rattlesnake-rattlesnake^{2}}\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{watermelon=2}^{\\infty} lighthouse(watermelon) rattlesnake^{watermelon}=rattlesnake^{2}+rattlesnake^{2} \\sum_{watermelon=1}^{\\infty} spacecraft(watermelon) rattlesnake^{watermelon}\n\\]\nand the result follows immediately.\n[The formal manipulation of the series can be justified either by considering only small values of \\( rattlesnake \\), for example \\( |rattlesnake|<\\frac{1}{10} \\), which makes each series involved absolutely convergent, or by considering the ring of formal power series in which convergence means only the ultimate stabilization of terns of each degree.]\n\nThird Solution. A one-to-one correspondence between representations of \\( watermelon \\) as a sum of 1's and 2's and representations of \\( watermelon+2 \\) as a sum of integers greater than 1 can be established as follows. Representations of \\( watermelon \\) as an ordered sum of 1's and 2's are obviously in one-to-one correspondence with representations of \\( watermelon+2 \\) as a sum of 1 's and 2 's ending with a 2 (and having one more summand). If in such a representation written in linear order we bracket each 2 with the longest string of 1's immediately preceding it, we obtain a representation of \\( watermelon+2 \\) as an ordered sum of integers greater than 1 . Conversely, in any representation of \\( watermelon+2 \\) as such an ordered sum, we can replace each summand by a string of 1 's followed by a single 2 .\n\nExample. The following representation of 9 as an ordered sum of 1 's and 2's,\n\\[\n9=1+1+2+2+2+1\n\\]\ncorresponds to the representation of \\( 11(=9+2) \\)\n\\[\n11=1+1+2+2+2+1+2\n\\]\nwhich corresponds to\n\\[\n\\begin{aligned}\n11 & =(1+1+2)+(2)+(2)+(1+2) \\\\\n& =4+2+2+3\n\\end{aligned}\n\\]\nand also conversely.\nIt follows that \\( lighthouse(watermelon+2)=spacecraft(watermelon) \\)"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "nonintegral",
        "k": "continuouscount",
        "x": "constantval",
        "a": "scarcityseries",
        "b": "abundanceseries"
      },
      "question": "4. Let \\( scarcityseries(nonintegral) \\) be the number of representations of the positive integer \\( nonintegral \\) as the sums of 1's and 2's taking order into account. For example, since\n\\[\n\\begin{array}{c}\n4=1+1+2=1+2+1=2+1+1 \\\\\n=2+2=1+1+1+1\n\\end{array}\n\\]\nthen \\( scarcityseries(4)=5 \\). Let \\( abundanceseries(nonintegral) \\) be the number of representations of \\( nonintegral \\) as the sum of integers greater than 1 , again taking order into account and counting the summand \\( nonintegral \\). For example, since \\( 6=4+2=2+4=3+3=2+2+2 \\), we have \\( abundanceseries(6)=5 \\). Show that for each \\( nonintegral, scarcityseries(nonintegral)=abundanceseries(nonintegral+2) \\).",
      "solution": "First Solution. By direct counting we can find the first few values of the two sequences:\n\\begin{tabular}{c|rrrrrr}\n\\( nonintegral \\) & 1 & 2 & 3 & 4 & 5 & 6 \\\\\n\\hline\\( scarcityseries(nonintegral) \\) & 1 & 2 & 3 & 5 & 8 & 13 \\\\\n\\hline\\( abundanceseries(nonintegral) \\) & 0 & 1 & 1 & 2 & 3 & 5\n\\end{tabular}\n\nThus it appears that the sequences are the well-known Fibonacci sequence satisfying the recursion\n\\[\nscarcityseries(nonintegral+1)=scarcityseries(nonintegral)+scarcityseries(nonintegral-1), \\quad nonintegral=2,3, \\ldots\n\\]\n\nTo prove that this is indeed the case, note that any representation of \\( nonintegral+1 \\) as an ordered sum of 1's and 2's becomes, on removal of the last summand, either a representation of \\( nonintegral \\) (if the removed term is a 1) or a representation of \\( nonintegral-1 \\) (if the removed term is a 2). Conversely, any representation of either \\( nonintegral \\) or \\( nonintegral-1 \\) becomes a representation of \\( nonintegral+1 \\) on adjoining either a 1 or 2, as appropriate. It follows that the scarcityseries-sequence does indeed satisfy the recursion (1).\n\nNext we show that the abundanceseries's satisfy the same recursion. Delete the last term from a representation of \\( nonintegral+1 \\) as a sum of integers greater than 1. We obtain then either a representation of \\( nonintegral-1, nonintegral-2, \\ldots, 2 \\), or a vacuous sum. Conversely, any such representation extends to a representation of \\( nonintegral+1 \\). Hence for \\( nonintegral \\geq 1 \\)\n\\[\nabundanceseries(nonintegral+1)=abundanceseries(nonintegral-1)+abundanceseries(nonintegral-2)+\\cdots+abundanceseries(2)+1\n\\]\n\nIf \\( \\mathbf{nonintegral} \\) is at least 2, we have also\n\\[\nabundanceseries(nonintegral)=abundanceseries(nonintegral-2)+\\cdots+abundanceseries(2)+1\n\\]\n\nComparing (2) and (3), we find\n\\[\nabundanceseries(nonintegral+1)=abundanceseries(nonintegral-1)+abundanceseries(nonintegral), \\quad nonintegral \\geq 2\n\\]\n\nSo the abundanceseries-sequence satisfies the same recursion as the scarcityseries-sequence. Then since \\( abundanceseries(3)=scarcityseries(1), abundanceseries(4)=scarcityseries(2) \\), it follows by induction that \\( abundanceseries(nonintegral+2)=scarcityseries(nonintegral) \\) for all \\( nonintegral \\geq 1 \\).\n\nSecond Solution. We can find the power-series generating function for \\( scarcityseries(nonintegral) \\). The number of ways that \\( nonintegral \\) can be represented as an ordered sum of \\( continuouscount \\) 1's and 2's is clearly the coefficient of \\( constantval^{nonintegral} \\) in \\( \\left(constantval+constantval^{2}\\right)^{continuouscount} \\). Hence\n\\[\n1+\\sum_{nonintegral-1}^{\\infty} scarcityseries(nonintegral) constantval^{nonintegral}=\\sum_{continuouscount=0}^{\\infty}\\left(constantval+constantval^{2}\\right)^{continuouscount}=\\frac{1}{1-constantval-constantval^{2}}\n\\]\n\nOn the other hand, the number of ways that \\( nonintegral \\) can be represented as an ordered sum of \\( continuouscount \\) integers greater than 1 is the coefficient of \\( constantval^{nonintegral} \\) in\n\\[\n\\left(constantval^{2}+constantval^{3}+\\cdots\\right)^{continuouscount}=\\left(\\frac{constantval^{2}}{1-constantval}\\right)^{continuouscount}\n\\]\n\nSo\n\\[\n\\begin{aligned}\n1+\\sum_{nonintegral=2}^{\\infty} abundanceseries(nonintegral) constantval^{nonintegral}=\\sum_{continuouscount=0}^{\\infty}\\left(\\frac{constantval^{2}}{1-constantval}\\right)^{continuouscount} & =\\left(1-\\frac{constantval^{2}}{1-constantval}\\right)^{-1} \\\\\n& =\\frac{1-constantval}{1-constantval-constantval^{2}} \\\\\n& =1+\\frac{constantval^{2}}{1-constantval-constantval^{2}}\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{nonintegral=2}^{\\infty} abundanceseries(nonintegral) constantval^{nonintegral}=constantval^{2}+constantval^{2} \\sum_{nonintegral=1}^{\\infty} scarcityseries(nonintegral) constantval^{nonintegral}\n\\]\nand the result follows immediately.\n[The formal manipulation of the series can be justified either by considering only small values of \\( constantval \\), for example \\( |constantval|<\\frac{1}{10} \\), which makes each series involved absolutely convergent, or by considering the ring of formal power series in which convergence means only the ultimate stabilization of terms of each degree.]\n\nThird Solution. A one-to-one correspondence between representations of \\( nonintegral \\) as a sum of 1's and 2's and representations of \\( nonintegral+2 \\) as a sum of integers greater than 1 can be established as follows. Representations of \\( nonintegral \\) as an ordered sum of 1's and 2's are obviously in one-to-one correspondence with representations of \\( nonintegral+2 \\) as a sum of 1's and 2's ending with a 2 (and having one more summand). If in such a representation written in linear order we bracket each 2 with the longest string of 1's immediately preceding it, we obtain a representation of \\( nonintegral+2 \\) as an ordered sum of integers greater than 1. Conversely, in any representation of \\( nonintegral+2 \\) as such an ordered sum, we can replace each summand by a string of 1's followed by a single 2.\n\nExample. The following representation of 9 as an ordered sum of 1's and 2's,\n\\[\n9=1+1+2+2+2+1\n\\]\ncorresponds to the representation of \\( 11(=9+2) \\)\n\\[\n11=1+1+2+2+2+1+2\n\\]\nwhich corresponds to\n\\[\n\\begin{aligned}\n11 & =(1+1+2)+(2)+(2)+(1+2) \\\\\n& =4+2+2+3\n\\end{aligned}\n\\]\nand also conversely.\nIt follows that \\( abundanceseries(nonintegral+2)=scarcityseries(nonintegral) \\)"
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "k": "hjgrksla",
        "x": "mrfplqzi",
        "a": "ltduygse",
        "b": "vbnmwert"
      },
      "question": "4. Let \\( ltduygse(qzxwvtnp) \\) be the number of representations of the positive integer \\( qzxwvtnp \\) as the sums of 1's and 2's taking order into account. For example, since\n\\[\n\\begin{array}{c}\n4=1+1+2=1+2+1=2+1+1 \\\\\n=2+2=1+1+1+1\n\\end{array}\n\\]\nthen \\( ltduygse(4)=5 \\). Let \\( vbnmwert(qzxwvtnp) \\) be the number of representations of \\( qzxwvtnp \\) as the sum of integers greater than 1 , again taking order into account and counting the summand \\( qzxwvtnp \\). For example, since \\( 6=4+2=2+4=3+3=2+2+2 \\), we have \\( vbnmwert(6)=5 \\). Show that for each \\( qzxwvtnp, ltduygse(qzxwvtnp)=vbnmwert(qzxwvtnp+2) \\).",
      "solution": "First Solution. By direct counting we can find the first few values of the two sequences:\n\\begin{tabular}{c|rrrrrr}\n\\( qzxwvtnp \\) & 1 & 2 & 3 & 4 & 5 & 6 \\\\\n\\hline\\( ltduygse(qzxwvtnp) \\) & 1 & 2 & 3 & 5 & 8 & 13 \\\\\n\\hline\\( vbnmwert(qzxwvtnp) \\) & 0 & 1 & 1 & 2 & 3 & 5\n\\end{tabular}\n\nThus it appears that the sequences are the well-known Fibonacci sequence satisfying the recursion\n\\[\nltduygse(qzxwvtnp+1)=ltduygse(qzxwvtnp)+ltduygse(qzxwvtnp-1), \\quad qzxwvtnp=2,3, \\ldots\n\\]\n\nTo prove that this is indeed the case, note that any representation of \\( qzxwvtnp+1 \\) as an ordered sum of 1's and 2's becomes, on removal of the last summand, either a representation of \\( qzxwvtnp \\) (if the removed term is a 1 ) or a representation of \\( qzxwvtnp-1 \\) (if the removed term is a 2 ). Conversely, any representation of either \\( qzxwvtnp \\) or \\( qzxwvtnp-1 \\) becomes a representation of \\( qzxwvtnp+1 \\) on adjoining either a 1 or 2 , as appropriate. It follows that the \\( ltduygse \\)-sequence does indeed satisfy the recursion (1).\n\nNext we show that the vbnmwert's satisfy the same recursion. Delete the last term from a representation of \\( qzxwvtnp+1 \\) as a sum of integers greater than 1. We obtain then either a representation of \\( qzxwvtnp-1, qzxwvtnp-2, \\ldots, 2 \\), or a vacuous sum. Conversely, any such representation extends to a representation of \\( qzxwvtnp+1 \\). Hence for \\( qzxwvtnp \\geq 1 \\)\n\\[\nvbnmwert(qzxwvtnp+1)=vbnmwert(qzxwvtnp-1)+vbnmwert(qzxwvtnp-2)+\\cdots+vbnmwert(2)+1\n\\]\n\nIf \\( \\boldsymbol{qzxwvtnp} \\) is at least 2, we have also\n\\[\nvbnmwert(qzxwvtnp)=vbnmwert(qzxwvtnp-2)+\\cdots+vbnmwert(2)+1\n\\]\n\nComparing (2) and (3), we find\n\\[\nvbnmwert(qzxwvtnp+1)=vbnmwert(qzxwvtnp-1)+vbnmwert(qzxwvtnp), \\quad qzxwvtnp \\geq 2\n\\]\n\nSo the \\( vbnmwert \\)-sequence satisfies the same recursion as the \\( ltduygse \\)-sequence.\nThen since \\( vbnmwert(3)=ltduygse(1), vbnmwert(4)=ltduygse(2) \\), it follows by induction that \\( vbnmwert(qzxwvtnp+2)=ltduygse(qzxwvtnp) \\) for all \\( qzxwvtnp \\geq 1 \\).\n\nSecond Solution. We can find the power-series generating function for \\( ltduygse(qzxwvtnp) \\). The number of ways that \\( qzxwvtnp \\) can be represented as an ordered sum of 1's and 2's is clearly the coefficient of \\( mrfplqzi^{qzxwvtnp} \\) in \\( \\left(mrfplqzi+mrfplqzi^{2}\\right)^{hjgrksla} \\). Hence\n\\[\n1+\\sum_{qzxwvtnp-1}^{\\infty} ltduygse(qzxwvtnp) mrfplqzi^{qzxwvtnp}=\\sum_{hjgrksla=0}^{\\infty}\\left(mrfplqzi+mrfplqzi^{2}\\right)^{hjgrksla}=\\frac{1}{1-mrfplqzi-mrfplqzi^{2}}\n\\]\n\nOn the other hand, the number of ways that \\( qzxwvtnp \\) can be represented as an ordered sum of \\( hjgrksla \\) integers greater than 1 is the coefficient of \\( mrfplqzi^{qzxwvtnp} \\) in\n\\[\n\\left(mrfplqzi^{2}+mrfplqzi^{3}+\\cdots\\right)^{hjgrksla}=\\left(\\frac{mrfplqzi^{2}}{1-mrfplqzi}\\right)^{hjgrksla}\n\\]\n\nSo\n\\[\n\\begin{aligned}\n1+\\sum_{qzxwvtnp=2}^{\\infty} vbnmwert(qzxwvtnp) mrfplqzi^{qzxwvtnp}=\\sum_{hjgrksla=0}^{\\infty}\\left(\\frac{mrfplqzi^{2}}{1-mrfplqzi}\\right)^{hjgrksla} & =\\left(1-\\frac{mrfplqzi^{2}}{1-mrfplqzi}\\right)^{-1} \\\\\n& =\\frac{1-mrfplqzi}{1-mrfplqzi-mrfplqzi^{2}} \\\\\n& =1+\\frac{mrfplqzi^{2}}{1-mrfplqzi-mrfplqzi^{2}}\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\sum_{qzxwvtnp=2}^{\\infty} vbnmwert(qzxwvtnp) mrfplqzi^{qzxwvtnp}=mrfplqzi^{2}+mrfplqzi^{2} \\sum_{qzxwvtnp=1}^{\\infty} ltduygse(qzxwvtnp) mrfplqzi^{qzxwvtnp}\n\\]\nand the result follows immediately.\n[The formal manipulation of the series can be justified either by considering only small values of \\( mrfplqzi \\), for example \\( |mrfplqzi|<\\frac{1}{10} \\), which makes each series involved absolutely convergent, or by considering the ring of formal power series in which convergence means only the ultimate stabilization of terns of each degree.]\n\nThird Solution. A one-to-one correspondence between representations of \\( qzxwvtnp \\) as a sum of 1's and 2's and representations of \\( qzxwvtnp+2 \\) as a sum of integers greater than 1 can be established as follows. Representations of \\( qzxwvtnp \\) as an ordered sum of 1's and 2's are obviously in one-to-one correspondence with representations of \\( qzxwvtnp+2 \\) as a sum of 1's and 2's ending with a 2 (and having one more summand). If in such a representation written in linear order we bracket each 2 with the longest string of 1's immediately preceding it, we obtain a representation of \\( qzxwvtnp+2 \\) as an ordered sum of integers greater than 1 . Conversely, in any representation of \\( qzxwvtnp+2 \\) as such an ordered sum, we can replace each summand by a string of 1's followed by a single 2 .\n\nExample. The following representation of 9 as an ordered sum of 1's and 2's,\n\\[\n9=1+1+2+2+2+1\n\\]\ncorresponds to the representation of \\( 11(=9+2) \\)\n\\[\n11=1+1+2+2+2+1+2\n\\]\nwhich corresponds to\n\\[\n\\begin{aligned}\n11 & =(1+1+2)+(2)+(2)+(1+2) \\\\\n& =4+2+2+3\n\\end{aligned}\n\\]\nand also conversely.\nIt follows that \\( vbnmwert(qzxwvtnp+2)=ltduygse(qzxwvtnp) \\)"
    },
    "kernel_variant": {
      "question": "Fix an integer $r\\ge 0$.  For integers $n,k\\ge 0$ define  \n\n\\[\n\\begin{aligned}\na_r(n,k)\\; &=\\;\\text{number of ordered representations of }n\\text{ in the form}\\\\[-2pt]\n&\\qquad 1+\\dots +1+2+1+\\dots +1+2+\\cdots +1+\\dots +1 \\\\[-2pt]\n&\\phantom{=\\ }\\bigl(\\text{exactly }k\\text{ occurrences of the summand }2,\\text{ hence }k+1\\text{ blocks of }1\\bigr)\\\\[4pt]\n&\\phantom{=\\ }\\text{with every block of consecutive }1\\text{’s having length }\\le r;\n\\\\[6pt]\nb_r(n,k)\\; &=\\;\\text{number of ordered representations of }n\\text{ as a sum of exactly}\\\\[-2pt]\n&\\qquad k+1\\text{ integers, each lying in the interval }\\{2,3,\\dots ,r+2\\}.\n\\end{aligned}\n\\]\n\nProve the refined identity  \n\\[\n\\boxed{\\,a_r(n,k)=b_r\\bigl(n+2,k\\bigr)\\qquad\\text{for all }r,n,k\\ge 0.}\n\\]\n\nConsequently, after summing over $k$ the total numbers  \n\n\\[\nA_r(n):=\\sum_{k\\ge 0}a_r(n,k), \n\\qquad \nB_r(n):=\\sum_{k\\ge 0}b_r(n,k)\n\\]\n\nsatisfy $A_r(n)=B_r(n+2)$ for every $n\\ge 0$, and their ordinary generating function is  \n\n\\[\n\\boxed{\\;\n\\sum_{n\\ge 0}A_r(n)x^{\\,n}= \n\\frac{1+x+x^{2}+\\dots +x^{\\,r}}{1-x^{2}-x^{3}-\\dots -x^{\\,r+2}}\n\\;}\n\\]\n\n(as formal power series in $x$).",
      "solution": "\\textbf{Step 1.  A weight–preserving bijection $\\Phi$ from $\\{a\\}$–objects to $\\{b\\}$–objects.}  \n\nStart with a word counted by $a_r(n,k)$, i.e. a string on $\\{1,2\\}$ that has total weight $n$, contains exactly $k$ symbols $2$, and whose $k+1$ intervening blocks of consecutive $1$’s have lengths $t_0,t_1,\\dots ,t_k$ satisfying $0\\le t_j\\le r$.  \n\nAppend an extra symbol $2$ at the \\emph{right} end.  The augmented word has weight $n+2$ and contains $k+1$ symbols $2$.  Scan it from left to right and cut \\emph{just before} every $2$; each piece therefore equals  \n\n\\[\n1^{\\,t_j}\\,2\\qquad(0\\le t_j\\le r).\n\\]\n\nReplace every block $1^{\\,t_j}2$ by the single integer $t_j+2\\in\\{2,3,\\dots ,r+2\\}$.  \nThe $k+1$ blocks yield an ordered $(k+1)$–tuple of integers whose sum is $n+2$, whence an element of $b_r(n+2,k)$.  Denote this map by $\\Phi$.\n\n\\textbf{Step 2.  Invertibility of $\\Phi$.}  \n\nConversely, let $(s_0,\\dots ,s_k)\\in b_r(n+2,k)$; thus each $s_j\\in\\{2,\\dots ,r+2\\}$ and $\\sum_{j=0}^{k}s_j=n+2$.  Replace $s_j$ by $s_j-2$ copies of the symbol $1$ followed by a single $2$.  The resulting word on $\\{1,2\\}$ contains $k+1$ symbols $2$, and every string of consecutive $1$’s has length $\\le r$.  Deleting the \\emph{final} $2$ leaves a representation of $n$ counted by $a_r(n,k)$.  Hence $\\Phi$ is a bijection and  \n\n\\[\na_r(n,k)=b_r(n+2,k)\\qquad\\forall\\,r,n,k\\ge 0.\n\\]\n\n\\textbf{Step 3.  Generating function for $A_r(n)$.}  \n\nAny word counted by $a_r(\\cdot,\\cdot)$ is uniquely written as  \n\n\\[\n1^{\\,t_0}\\,2\\,1^{\\,t_1}\\,2\\cdots 2\\,1^{\\,t_k},\n\\qquad 0\\le t_j\\le r.\n\\]\n\nInsert the auxiliary terminal $2$ that appears in the bijection to obtain  \n\n\\[\n1^{\\,t_0}\\,2\\,1^{\\,t_1}\\,2\\cdots 2\\,1^{\\,t_k}\\,2.\n\\]\n\nIts weight in the variable $x$ is  \n\n\\[\nx^{\\,t_0+2}\\,x^{\\,t_1+2}\\cdots x^{\\,t_k+2}\n   =\\bigl[x^{2}(1+x+\\dots +x^{\\,r})\\bigr]^{k+1}.\n\\]\n\nSumming independently over all $k\\ge 0$ and all choices of $(t_0,\\dots ,t_k)$ gives  \n\n\\[\n\\sum_{k\\ge 0}\\bigl[x^{2}(1+x+\\dots +x^{\\,r})\\bigr]^{k+1}\n   \\;=\\;\\frac{x^{2}(1+x+\\dots +x^{\\,r})}{1-x^{2}(1+x+\\dots +x^{\\,r})}.\n\\]\n\nDividing the \\emph{entire fraction} by $x^{2}$ (equivalently, multiplying numerator and denominator by $x^{-2}$) yields\n\n\\[\n\\sum_{n\\ge 0}A_r(n)x^{\\,n}\n   =\\frac{1+x+\\dots +x^{\\,r}}{1-x^{2}-x^{3}-\\dots -x^{\\,r+2}}.\n\\]\n\n\\textbf{Step 4.  Linear recurrence for $A_r(n)$.}  \n\nMultiplying the last identity by its denominator and equating coefficients of $x^{\\,n}$ gives, for every $n\\ge r+2$, the homogeneous linear recurrence  \n\n\\[\nA_r(n)=A_r(n-2)+A_r(n-3)+\\dots +A_r\\bigl(n-(r+2)\\bigr).\n\\]\n\nAll initial values $A_r(0),A_r(1),\\dots ,A_r(r+1)$ are extracted from the power series on the left-hand side.  \n\n\\textit{Check for }$r\\to\\infty$.  Write  \n\n\\[\n\\frac{1}{1-x^{2}-x^{3}-\\dots }\n   \\;=\\;\\frac{1}{1-\\dfrac{x^{2}}{1-x}}\n   \\;=\\;\\frac{1-x}{1-x-x^{2}},\n\\]\n\nso that the denominator becomes $1-x-x^{2}$ and the recurrence specializes to the classical Fibonacci relation  \n\n\\[\nA_{\\infty}(n)=A_{\\infty}(n-1)+A_{\\infty}(n-2)\\qquad(n\\ge 2).\n\\]\n\nThe refined bijection therefore generalises the well–known correspondence underlying the Fibonacci numbers.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.490358",
        "was_fixed": false,
        "difficulty_analysis": "1.  Vector–valued enumeration:  The original problem involves only a single sequence. Here we refine the count simultaneously in two variables (total sum n and number of 2’s k), forcing the solver to keep track of an additional parameter throughout the argument.\n\n2.  Extra combinatorial constraint:  A cap r on the length of every block of 1’s introduces a local restriction that drastically complicates both the bijective proof and the analytic (generating-function) derivation.  The solver must verify that the map Φ respects this limitation and that its inverse does so as well.\n\n3.  Parametric family:  All identities must hold uniformly for every non–negative integer r, so the argument must treat an infinite family of structural constraints in one stroke.  Algebraically this yields a denominator 1 − x² − x³ − … − x^{r+2}, whose manipulation demands comfort with formal power-series techniques well beyond the simple Fibonacci kernel of the original problem.\n\n4.  Higher-order recurrence:  Instead of the 2-term Fibonacci relation, the enhanced sequence satisfies an (r+1)-term recurrence, reflecting a much richer combinatorial structure; identifying and proving this requires a deeper understanding of linear recurrences with variable order.\n\n5.  Bivariate and univariate synthesis:  The solver must show not only the refined identity (★) but also how it collapses, upon summation over k, to an identity between unrefined sequences and how the corresponding generating functions emerge.  This cumulative reasoning demands coordination of bijective, algebraic and analytic methods, far beyond the “remove the last summand” trick sufficient for the original exercise."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer $r\\ge 0$.  For integers $n,k\\ge 0$ define  \n\n\\[\n\\begin{aligned}\na_r(n,k)\\; &=\\;\\text{number of ordered representations of }n\\text{ in the form}\\\\[-2pt]\n&\\qquad 1+\\dots +1+2+1+\\dots +1+2+\\cdots +1+\\dots +1 \\\\[-2pt]\n&\\phantom{=\\ }\\bigl(\\text{exactly }k\\text{ occurrences of the summand }2,\\text{ hence }k+1\\text{ blocks of }1\\bigr)\\\\[4pt]\n&\\phantom{=\\ }\\text{with every block of consecutive }1\\text{’s having length }\\le r;\n\\\\[6pt]\nb_r(n,k)\\; &=\\;\\text{number of ordered representations of }n\\text{ as a sum of exactly}\\\\[-2pt]\n&\\qquad k+1\\text{ integers, each lying in the interval }\\{2,3,\\dots ,r+2\\}.\n\\end{aligned}\n\\]\n\nProve the refined identity  \n\\[\n\\boxed{\\,a_r(n,k)=b_r\\bigl(n+2,k\\bigr)\\qquad\\text{for all }r,n,k\\ge 0.}\n\\]\n\nConsequently, after summing over $k$ the total numbers  \n\n\\[\nA_r(n):=\\sum_{k\\ge 0}a_r(n,k), \n\\qquad \nB_r(n):=\\sum_{k\\ge 0}b_r(n,k)\n\\]\n\nsatisfy $A_r(n)=B_r(n+2)$ for every $n\\ge 0$, and their ordinary generating function is  \n\n\\[\n\\boxed{\\;\n\\sum_{n\\ge 0}A_r(n)x^{\\,n}= \n\\frac{1+x+x^{2}+\\dots +x^{\\,r}}{1-x^{2}-x^{3}-\\dots -x^{\\,r+2}}\n\\;}\n\\]\n\n(as formal power series in $x$).",
      "solution": "\\textbf{Step 1.  A weight–preserving bijection $\\Phi$ from $\\{a\\}$–objects to $\\{b\\}$–objects.}  \n\nStart with a word counted by $a_r(n,k)$, i.e. a string on $\\{1,2\\}$ that has total weight $n$, contains exactly $k$ symbols $2$, and whose $k+1$ intervening blocks of consecutive $1$’s have lengths $t_0,t_1,\\dots ,t_k$ satisfying $0\\le t_j\\le r$.  \n\nAppend an extra symbol $2$ at the \\emph{right} end.  The augmented word has weight $n+2$ and contains $k+1$ symbols $2$.  Scan it from left to right and cut \\emph{just before} every $2$; each piece therefore equals  \n\n\\[\n1^{\\,t_j}\\,2\\qquad(0\\le t_j\\le r).\n\\]\n\nReplace every block $1^{\\,t_j}2$ by the single integer $t_j+2\\in\\{2,3,\\dots ,r+2\\}$.  \nThe $k+1$ blocks yield an ordered $(k+1)$–tuple of integers whose sum is $n+2$, whence an element of $b_r(n+2,k)$.  Denote this map by $\\Phi$.\n\n\\textbf{Step 2.  Invertibility of $\\Phi$.}  \n\nConversely, let $(s_0,\\dots ,s_k)\\in b_r(n+2,k)$; thus each $s_j\\in\\{2,\\dots ,r+2\\}$ and $\\sum_{j=0}^{k}s_j=n+2$.  Replace $s_j$ by $s_j-2$ copies of the symbol $1$ followed by a single $2$.  The resulting word on $\\{1,2\\}$ contains $k+1$ symbols $2$, and every string of consecutive $1$’s has length $\\le r$.  Deleting the \\emph{final} $2$ leaves a representation of $n$ counted by $a_r(n,k)$.  Hence $\\Phi$ is a bijection and  \n\n\\[\na_r(n,k)=b_r(n+2,k)\\qquad\\forall\\,r,n,k\\ge 0.\n\\]\n\n\\textbf{Step 3.  Generating function for $A_r(n)$.}  \n\nAny word counted by $a_r(\\cdot,\\cdot)$ is uniquely written as  \n\n\\[\n1^{\\,t_0}\\,2\\,1^{\\,t_1}\\,2\\cdots 2\\,1^{\\,t_k},\n\\qquad 0\\le t_j\\le r.\n\\]\n\nInsert the auxiliary terminal $2$ that appears in the bijection to obtain  \n\n\\[\n1^{\\,t_0}\\,2\\,1^{\\,t_1}\\,2\\cdots 2\\,1^{\\,t_k}\\,2.\n\\]\n\nIts weight in the variable $x$ is  \n\n\\[\nx^{\\,t_0+2}\\,x^{\\,t_1+2}\\cdots x^{\\,t_k+2}\n   =\\bigl[x^{2}(1+x+\\dots +x^{\\,r})\\bigr]^{k+1}.\n\\]\n\nSumming independently over all $k\\ge 0$ and all choices of $(t_0,\\dots ,t_k)$ gives  \n\n\\[\n\\sum_{k\\ge 0}\\bigl[x^{2}(1+x+\\dots +x^{\\,r})\\bigr]^{k+1}\n   \\;=\\;\\frac{x^{2}(1+x+\\dots +x^{\\,r})}{1-x^{2}(1+x+\\dots +x^{\\,r})}.\n\\]\n\nDividing the \\emph{entire fraction} by $x^{2}$ (equivalently, multiplying numerator and denominator by $x^{-2}$) yields\n\n\\[\n\\sum_{n\\ge 0}A_r(n)x^{\\,n}\n   =\\frac{1+x+\\dots +x^{\\,r}}{1-x^{2}-x^{3}-\\dots -x^{\\,r+2}}.\n\\]\n\n\\textbf{Step 4.  Linear recurrence for $A_r(n)$.}  \n\nMultiplying the last identity by its denominator and equating coefficients of $x^{\\,n}$ gives, for every $n\\ge r+2$, the homogeneous linear recurrence  \n\n\\[\nA_r(n)=A_r(n-2)+A_r(n-3)+\\dots +A_r\\bigl(n-(r+2)\\bigr).\n\\]\n\nAll initial values $A_r(0),A_r(1),\\dots ,A_r(r+1)$ are extracted from the power series on the left-hand side.  \n\n\\textit{Check for }$r\\to\\infty$.  Write  \n\n\\[\n\\frac{1}{1-x^{2}-x^{3}-\\dots }\n   \\;=\\;\\frac{1}{1-\\dfrac{x^{2}}{1-x}}\n   \\;=\\;\\frac{1-x}{1-x-x^{2}},\n\\]\n\nso that the denominator becomes $1-x-x^{2}$ and the recurrence specializes to the classical Fibonacci relation  \n\n\\[\nA_{\\infty}(n)=A_{\\infty}(n-1)+A_{\\infty}(n-2)\\qquad(n\\ge 2).\n\\]\n\nThe refined bijection therefore generalises the well–known correspondence underlying the Fibonacci numbers.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.410374",
        "was_fixed": false,
        "difficulty_analysis": "1.  Vector–valued enumeration:  The original problem involves only a single sequence. Here we refine the count simultaneously in two variables (total sum n and number of 2’s k), forcing the solver to keep track of an additional parameter throughout the argument.\n\n2.  Extra combinatorial constraint:  A cap r on the length of every block of 1’s introduces a local restriction that drastically complicates both the bijective proof and the analytic (generating-function) derivation.  The solver must verify that the map Φ respects this limitation and that its inverse does so as well.\n\n3.  Parametric family:  All identities must hold uniformly for every non–negative integer r, so the argument must treat an infinite family of structural constraints in one stroke.  Algebraically this yields a denominator 1 − x² − x³ − … − x^{r+2}, whose manipulation demands comfort with formal power-series techniques well beyond the simple Fibonacci kernel of the original problem.\n\n4.  Higher-order recurrence:  Instead of the 2-term Fibonacci relation, the enhanced sequence satisfies an (r+1)-term recurrence, reflecting a much richer combinatorial structure; identifying and proving this requires a deeper understanding of linear recurrences with variable order.\n\n5.  Bivariate and univariate synthesis:  The solver must show not only the refined identity (★) but also how it collapses, upon summation over k, to an identity between unrefined sequences and how the corresponding generating functions emerge.  This cumulative reasoning demands coordination of bijective, algebraic and analytic methods, far beyond the “remove the last summand” trick sufficient for the original exercise."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}