{
  "index": "1958-2-B-1",
  "type": "COMB",
  "tag": [
    "COMB",
    "ANA"
  ],
  "difficulty": "",
  "question": "1. Given\n\\[\nb_{n}=\\sum_{k=0}^{n}\\binom{n}{k}^{-1}, \\quad n \\geq 1,\n\\]\nprove that\n\\[\nb_{n}=\\frac{n+1}{2 n} b_{n-1}+1, \\quad n \\geq 2\n\\]\nand hence, as a corollary,\n\\[\n\\lim _{n \\rightarrow \\infty} b_{n}=2\n\\]",
  "solution": "Solution. We have\n\\[\nn!b_{n}=\\sum_{k=0}^{n} k!(n-k)!=\\sum_{k=1}^{n+1}(k-1)!(n-k+1)!\n\\]\n\nSo\n\\[\n\\begin{aligned}\n2 n!b_{n} & =0!n!+\\sum_{k=1}^{n}[k!(n-k)!+(k-1)!(n-k+1)!]+n!0! \\\\\n& =2 n!+(n+1) \\sum_{k=1}^{n}(k-1)!(n-k)! \\\\\n& =2 n!+(n+1)\\left[(n-1)!b_{n-1}\\right] .\n\\end{aligned}\n\\]\n\nDividing by \\( 2 n! \\) we get\n\\[\nb_{n}=1+\\frac{n+1}{2 n} b_{n-1}\n\\]\nas required.\nLet \\( b_{n}=2+C_{n} \\). Then\n\\[\nn C_{n}=1+\\frac{n+1}{2(n-1)}(n-1) C_{n-1} .\n\\]\n\nWe shall prove by induction that \\( n C_{n} \\leq 6 \\) for all \\( n \\). This is clearly true for \\( n=1,2,3\\left(C_{1}=0, C_{2}=\\frac{1}{2}, C_{3}=\\frac{2}{3}\\right) \\). Assume it is true for \\( n=k-1 \\) where \\( k \\geq 4 \\). Then\n\\[\nk C_{k} \\leq 1+\\frac{k+1}{2(k-1)} 6 \\leq 1+\\frac{5}{2 \\cdot 3} 6=6 .\n\\]\nwhere we used the fact that \\( (x+1) /(x-1)=1+2 /(x-1) \\) increases as \\( x \\) decreases for \\( x>1 \\). This completes the induction.\n\nFrom the inequality \\( C_{n} \\leq 6 / n \\) and the obvious fact that \\( C_{n} \\geq 0 \\), it follows that \\( C_{n} \\rightarrow 0 \\), and therefore \\( b_{n} \\rightarrow 2 \\).\n\nRemark. One can establish lim \\( b_{n}=2 \\) directly since\n\\[\n\\begin{array}{c}\nb_{n}=2+\\frac{2}{n}+\\sum_{k=2}^{n-2} \\frac{1}{\\binom{n}{k}} \\\\\n\\leq 2+\\frac{2}{n}+\\frac{n-3}{\\binom{n}{2}}=2+\\frac{4}{n} \\cdot \\frac{n-2}{n-1}\n\\end{array}\n\\]\nfor \\( n \\geq 3 \\).",
  "vars": [
    "b_n",
    "b_n-1",
    "C_n",
    "C_n-1",
    "n",
    "k",
    "x"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "b_n": "termseries",
        "b_n-1": "prevseries",
        "C_n": "offsetseries",
        "C_n-1": "prevoffset",
        "n": "indexvar",
        "k": "summindex",
        "x": "variablex"
      },
      "question": "1. Given\n\\[\ntermseries = \\sum_{summindex=0}^{indexvar} \\binom{indexvar}{summindex}^{-1}, \\quad indexvar \\geq 1,\n\\]\nprove that\n\\[\ntermseries = \\frac{indexvar+1}{2\\, indexvar}\\, prevseries + 1, \\quad indexvar \\geq 2\n\\]\nand hence, as a corollary,\n\\[\n\\lim_{indexvar \\rightarrow \\infty} termseries = 2\n\\]",
      "solution": "Solution. We have\n\\[\nindexvar!\\, termseries = \\sum_{summindex=0}^{indexvar} summindex!\\,(indexvar-summindex)! = \\sum_{summindex=1}^{indexvar+1} (summindex-1)!\\,(indexvar-summindex+1)!\n\\]\nSo\n\\[\n\\begin{aligned}\n2\\, indexvar!\\, termseries &= 0!\\, indexvar! + \\sum_{summindex=1}^{indexvar} [\\, summindex!\\,(indexvar-summindex)! + (summindex-1)!\\,(indexvar-summindex+1)!\\,] + indexvar!\\,0! \\\\ &= 2\\, indexvar! + (indexvar+1) \\sum_{summindex=1}^{indexvar} (summindex-1)!\\,(indexvar-summindex)! \\\\ &= 2\\, indexvar! + (indexvar+1)\\left[(indexvar-1)!\\, prevseries\\right].\n\\end{aligned}\n\\]\nDividing by \\( 2\\, indexvar! \\) we get\n\\[\ntermseries = 1 + \\frac{indexvar+1}{2\\, indexvar}\\, prevseries\n\\]\nas required.\nLet \\( termseries = 2 + offsetseries \\). Then\n\\[\nindexvar\\, offsetseries = 1 + \\frac{indexvar+1}{2(indexvar-1)} (indexvar-1)\\, prevoffset.\n\\]\nWe shall prove by induction that \\( indexvar\\, offsetseries \\le 6 \\) for all \\( indexvar \\). This is clearly true for \\( indexvar = 1,2,3\\left(offsetseries_{1}=0,\\; offsetseries_{2}=\\tfrac{1}{2},\\; offsetseries_{3}=\\tfrac{2}{3}\\right) \\). Assume it is true for \\( indexvar = summindex-1 \\) where \\( summindex \\ge 4 \\). Then\n\\[\nsummindex\\, offsetseries_{summindex} \\le 1 + \\frac{summindex+1}{2(summindex-1)}\\, 6 \\le 1 + \\frac{5}{2\\cdot 3}\\, 6 = 6.\n\\]\nwhere we used the fact that \\( (variablex+1)/(variablex-1) = 1 + 2/(variablex-1) \\) increases as \\( variablex \\) decreases for \\( variablex>1 \\). This completes the induction.\n\nFrom the inequality \\( offsetseries \\le 6 / indexvar \\) and the obvious fact that \\( offsetseries \\ge 0 \\), it follows that \\( offsetseries \\to 0 \\), and therefore \\( termseries \\to 2 \\).\n\nRemark. One can establish \\( \\lim termseries = 2 \\) directly since\n\\[\n\\begin{array}{c}\ntermseries = 2 + \\frac{2}{indexvar} + \\sum_{summindex=2}^{indexvar-2} \\frac{1}{\\binom{indexvar}{summindex}} \\\\\n\\le 2 + \\frac{2}{indexvar} + \\frac{indexvar-3}{\\binom{indexvar}{2}} = 2 + \\frac{4}{indexvar} \\cdot \\frac{indexvar-2}{indexvar-1}\n\\end{array}\n\\]\nfor \\( indexvar \\ge 3 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "b_n": "meadowlark",
        "b_n-1": "hummingjay",
        "C_n": "buttercup",
        "C_n-1": "dandelion",
        "n": "caterpillar",
        "k": "grasshopper",
        "x": "dragonfly"
      },
      "question": "1. Given\n\\[\nmeadowlark=\\sum_{grasshopper=0}^{caterpillar}\\binom{caterpillar}{grasshopper}^{-1}, \\quad caterpillar \\geq 1,\n\\]\nprove that\n\\[\nmeadowlark=\\frac{caterpillar+1}{2 caterpillar} hummingjay+1, \\quad caterpillar \\geq 2\n\\]\nand hence, as a corollary,\n\\[\n\\lim _{caterpillar \\rightarrow \\infty} meadowlark=2\n\\]\n",
      "solution": "Solution. We have\n\\[\ncaterpillar!meadowlark=\\sum_{grasshopper=0}^{caterpillar} grasshopper!(caterpillar-grasshopper)!=\\sum_{grasshopper=1}^{caterpillar+1}(grasshopper-1)!(caterpillar-grasshopper+1)!\n\\]\n\nSo\n\\[\n\\begin{aligned}\n2 caterpillar!meadowlark & =0!caterpillar!+\\sum_{grasshopper=1}^{caterpillar}[grasshopper!(caterpillar-grasshopper)!+(grasshopper-1)!(caterpillar-grasshopper+1)!]+caterpillar!0! \\\\\n& =2 caterpillar!+(caterpillar+1) \\sum_{grasshopper=1}^{caterpillar}(grasshopper-1)!(caterpillar-grasshopper)! \\\\\n& =2 caterpillar!+(caterpillar+1)\\left[(caterpillar-1)! hummingjay\\right] .\n\\end{aligned}\n\\]\n\nDividing by \\( 2 caterpillar! \\) we get\n\\[\nmeadowlark=1+\\frac{caterpillar+1}{2 caterpillar} hummingjay\n\\]\nas required.\nLet \\( meadowlark=2+buttercup \\). Then\n\\[\ncaterpillar buttercup=1+\\frac{caterpillar+1}{2(caterpillar-1)}(caterpillar-1) dandelion .\n\\]\n\nWe shall prove by induction that \\( caterpillar buttercup \\leq 6 \\) for all \\( caterpillar \\). This is clearly true for \\( caterpillar=1,2,3\\left(buttercup_{1}=0, buttercup_{2}=\\frac{1}{2}, buttercup_{3}=\\frac{2}{3}\\right) \\). Assume it is true for \\( caterpillar=grasshopper-1 \\) where \\( grasshopper \\geq 4 \\). Then\n\\[\ngras shopper C_{grasshopper} \\leq 1+\\frac{grasshopper+1}{2(grasshopper-1)} 6 \\leq 1+\\frac{5}{2 \\cdot 3} 6=6 .\n\\]\nwhere we used the fact that \\( (dragonfly+1) /(dragonfly-1)=1+2 /(dragonfly-1) \\) increases as \\( dragonfly \\) decreases for \\( dragonfly>1 \\). This completes the induction.\n\nFrom the inequality \\( buttercup \\leq 6 / caterpillar \\) and the obvious fact that \\( buttercup \\geq 0 \\), it follows that \\( buttercup \\rightarrow 0 \\), and therefore \\( meadowlark \\rightarrow 2 \\).\n\nRemark. One can establish lim meadowlark=2 directly since\n\\[\n\\begin{array}{c}\nmeadowlark=2+\\frac{2}{caterpillar}+\\sum_{grasshopper=2}^{caterpillar-2} \\frac{1}{\\binom{caterpillar}{grasshopper}} \\\\\n\\leq 2+\\frac{2}{caterpillar}+\\frac{caterpillar-3}{\\binom{caterpillar}{2}}=2+\\frac{4}{caterpillar} \\cdot \\frac{caterpillar-2}{caterpillar-1}\n\\end{array}\n\\]\nfor \\( caterpillar \\geq 3 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "b_n": "staticvalue",
        "b_n-1": "staticformer",
        "C_n": "chaosvalue",
        "C_n-1": "chaosformer",
        "n": "continuum",
        "k": "fixedpoint",
        "x": "certainty"
      },
      "question": "1. Given\n\\[\nstaticvalue=\\sum_{fixedpoint=0}^{continuum}\\binom{continuum}{fixedpoint}^{-1}, \\quad continuum \\geq 1,\n\\]\nprove that\n\\[\nstaticvalue=\\frac{continuum+1}{2 continuum} staticformer+1, \\quad continuum \\geq 2\n\\]\nand hence, as a corollary,\n\\[\n\\lim _{continuum \\rightarrow \\infty} staticvalue=2\n\\]",
      "solution": "Solution. We have\n\\[\ncontinuum!staticvalue=\\sum_{fixedpoint=0}^{continuum} fixedpoint!(continuum-fixedpoint)!=\\sum_{fixedpoint=1}^{continuum+1}(fixedpoint-1)!(continuum-fixedpoint+1)!\n\\]\n\nSo\n\\[\n\\begin{aligned}\n2\\,continuum!staticvalue & =0!\\,continuum!+\\sum_{fixedpoint=1}^{continuum}[fixedpoint!(continuum-fixedpoint)!+(fixedpoint-1)!(continuum-fixedpoint+1)!]+continuum!0! \\\\\n& =2\\,continuum!+(continuum+1) \\sum_{fixedpoint=1}^{continuum}(fixedpoint-1)!(continuum-fixedpoint)! \\\\\n& =2\\,continuum!+(continuum+1)\\left[(continuum-1)!staticformer\\right] .\n\\end{aligned}\n\\]\n\nDividing by \\( 2\\,continuum! \\) we get\n\\[\nstaticvalue=1+\\frac{continuum+1}{2 continuum} staticformer\n\\]\nas required.\nLet \\( staticvalue=2+chaosvalue \\). Then\n\\[\ncontinuum\\,chaosvalue=1+\\frac{continuum+1}{2(continuum-1)}(continuum-1)\\,chaosformer .\n\\]\n\nWe shall prove by induction that \\( continuum\\,chaosvalue \\le 6 \\) for all \\( continuum \\). This is clearly true for \\( continuum=1,2,3\\left(chaosvalue_{1}=0,\\, chaosvalue_{2}=\\frac{1}{2},\\, chaosvalue_{3}=\\frac{2}{3}\\right) \\). Assume it is true for \\( continuum=fixedpoint-1 \\) where \\( fixedpoint \\ge 4 \\). Then\n\\[\nfixedpoint\\,chaosvalue_{fixedpoint} \\le 1+\\frac{fixedpoint+1}{2(fixedpoint-1)} 6 \\le 1+\\frac{5}{2\\cdot3} 6=6 .\n\\]\nwhere we used the fact that \\( (certainty+1)/(certainty-1)=1+2/(certainty-1) \\) increases as \\( certainty \\) decreases for \\( certainty>1 \\). This completes the induction.\n\nFrom the inequality \\( chaosvalue \\le 6 / continuum \\) and the obvious fact that \\( chaosvalue \\ge 0 \\), it follows that \\( chaosvalue \\rightarrow 0 \\), and therefore \\( staticvalue \\rightarrow 2 \\).\n\nRemark. One can establish \\(\\lim staticvalue=2\\) directly since\n\\[\n\\begin{array}{c}\nstaticvalue=2+\\frac{2}{continuum}+\\sum_{fixedpoint=2}^{continuum-2} \\frac{1}{\\binom{continuum}{fixedpoint}} \\\\\n\\le 2+\\frac{2}{continuum}+\\frac{continuum-3}{\\binom{continuum}{2}}=2+\\frac{4}{continuum}\\cdot\\frac{continuum-2}{continuum-1}\n\\end{array}\n\\]\nfor \\( continuum \\ge 3 \\)."
    },
    "garbled_string": {
      "map": {
        "b_n": "qzxwvtnp",
        "b_n-1": "hjgrksla",
        "C_n": "mydpltha",
        "C_n-1": "vfzsnrqe",
        "n": "lkdjcqye",
        "k": "yqztlvms",
        "x": "brhcvwpo"
      },
      "question": "1. Given\n\\[\nqzxwvtnp=\\sum_{yqztlvms=0}^{lkdjcqye}\\binom{lkdjcqye}{yqztlvms}^{-1}, \\quad lkdjcqye \\geq 1,\n\\]\nprove that\n\\[\nqzxwvtnp=\\frac{lkdjcqye+1}{2 lkdjcqye} hjgrksla+1, \\quad lkdjcqye \\geq 2\n\\]\nand hence, as a corollary,\n\\[\n\\lim _{lkdjcqye \\rightarrow \\infty} qzxwvtnp=2\n\\]",
      "solution": "Solution. We have\n\\[\nlkdjcqye!qzxwvtnp=\\sum_{yqztlvms=0}^{lkdjcqye} yqztlvms!(lkdjcqye-yqztlvms)!=\\sum_{yqztlvms=1}^{lkdjcqye+1}(yqztlvms-1)!(lkdjcqye-yqztlvms+1)!\n\\]\n\nSo\n\\[\n\\begin{aligned}\n2 lkdjcqye!qzxwvtnp & =0!lkdjcqye!+\\sum_{yqztlvms=1}^{lkdjcqye}[yqztlvms!(lkdjcqye-yqztlvms)!+(yqztlvms-1)!(lkdjcqye-yqztlvms+1)!]+lkdjcqye!0! \\\\\n& =2 lkdjcqye!+(lkdjcqye+1) \\sum_{yqztlvms=1}^{lkdjcqye}(yqztlvms-1)!(lkdjcqye-yqztlvms)! \\\\\n& =2 lkdjcqye!+(lkdjcqye+1)\\left[(lkdjcqye-1)!hjgrksla\\right] .\n\\end{aligned}\n\\]\n\nDividing by \\( 2 lkdjcqye! \\) we get\n\\[\nqzxwvtnp=1+\\frac{lkdjcqye+1}{2 lkdjcqye} hjgrksla\n\\]\nas required.\nLet \\( qzxwvtnp=2+mydpltha \\). Then\n\\[\nlkdjcqye mydpltha=1+\\frac{lkdjcqye+1}{2(lkdjcqye-1)}(lkdjcqye-1) vfzsnrqe .\n\\]\n\nWe shall prove by induction that \\( lkdjcqye mydpltha \\leq 6 \\) for all \\( lkdjcqye \\). This is clearly true for \\( lkdjcqye=1,2,3\\left(mydpltha=0, mydpltha=\\frac{1}{2}, mydpltha=\\frac{2}{3}\\right) \\). Assume it is true for \\( lkdjcqye=yqztlvms-1 \\) where \\( yqztlvms \\geq 4 \\). Then\n\\[\nyqztlvms mydpltha \\leq 1+\\frac{yqztlvms+1}{2(yqztlvms-1)} 6 \\leq 1+\\frac{5}{2 \\cdot 3} 6=6 .\n\\]\nwhere we used the fact that \\( (brhcvwpo+1) /(brhcvwpo-1)=1+2 /(brhcvwpo-1) \\) increases as \\( brhcvwpo \\) decreases for \\( brhcvwpo>1 \\). This completes the induction.\n\nFrom the inequality \\( mydpltha \\leq 6 / lkdjcqye \\) and the obvious fact that \\( mydpltha \\geq 0 \\), it follows that \\( mydpltha \\rightarrow 0 \\), and therefore \\( qzxwvtnp \\rightarrow 2 \\).\n\nRemark. One can establish lim \\( qzxwvtnp=2 \\) directly since\n\\[\n\\begin{array}{c}\nqzxwvtnp=2+\\frac{2}{lkdjcqye}+\\sum_{yqztlvms=2}^{lkdjcqye-2} \\frac{1}{\\binom{lkdjcqye}{yqztlvms}} \\\\\n\\leq 2+\\frac{2}{lkdjcqye}+\\frac{lkdjcqye-3}{\\binom{lkdjcqye}{2}}=2+\\frac{4}{lkdjcqye} \\cdot \\frac{lkdjcqye-2}{lkdjcqye-1}\n\\end{array}\n\\]\nfor \\( lkdjcqye \\geq 3 \\)."
    },
    "kernel_variant": {
      "question": "For every integer n ≥ 1 set\n\na_n = \\displaystyle\\sum_{k=0}^{n}\\frac{2}{\\binom{n}{k}}.\n\n(a)  Prove that for every n ≥ 2\n\na_n = 2 + \\frac{n+1}{2n}\\,a_{n-1}.\n\n(b)  Prove that the sequence (a_n) converges and find its limit.",
      "solution": "Solution.\n\nThroughout we put\n\nS_n := \\sum_{k=0}^{n} k!(n-k)!  \\qquad (n\\ge 0).\n\nObserve that\n\nn!\\,a_n = 2S_n.                                                    (1)\n\nPart (a).\n\nStep 1  –  A second expression for S_n.\nShift the index in the definition of S_n by writing k ↦ k-1:\n\nS_n = \\sum_{k=1}^{n+1} (k-1)!(n-k+1)!.                            (2)\n\nStep 2  –  Adding (1) and (2) without double counting the\nend–terms.\n\nBoth series (1) and (2) contain the terms 0!·n! and n!·0!.  We add\nthe two sums and keep each of those boundary terms only once:\n\n2S_n = 0!\\,n! + n!\\,0! + \\sum_{k=1}^{n} \\bigl[k!(n-k)! + (k-1)!(n-k+1)!\\bigr]. (3)\n\nSince 0!=1, the first two summands equal 2n!.\n\nStep 3  –  A factorisation inside the bracket.\nFor 1 ≤ k ≤ n we have\n\nk!(n-k)! + (k-1)!(n-k+1)! = (k + n-k +1)(k-1)!(n-k)! = (n+1)(k-1)!(n-k)!. (4)\n\nInserting (4) into (3) gives\n\n2S_n = 2n! + (n+1) \\sum_{k=1}^{n} (k-1)!(n-k)!.                  (5)\n\nStep 4  –  Identifying the remaining sum.\nPut j = k-1 in the last sum of (5):\n\n\\sum_{k=1}^{n} (k-1)!(n-k)! = \\sum_{j=0}^{n-1} j!(n-1-j)! = S_{n-1}. (6)\n\nWith (1), (5) and (6) we obtain\n\nn!\\,a_n = 2n! + (n+1) S_{n-1}.                                    (7)\n\nFrom (1) with n replaced by n−1 we have S_{n-1} = (n-1)!\\,a_{n-1}/2.\nInsert this into (7):\n\nn!\\,a_n = 2n! + \\frac{n+1}{2}\\,(n-1)!\\,a_{n-1}.\n\nFinally divide by n!:\n\na_n = 2 + \\frac{n+1}{2n}\\,a_{n-1}, \\qquad n\\ge 2.                 (8)\n\nThat proves part (a).\n\nPart (b).\n\nStep 1  –  Passing to the limit.\nWrite (8) as\n\na_n = 2 + c_n\\,a_{n-1} \\quad\\text{with}\\quad c_n = \\frac{n+1}{2n} \\xrightarrow[n\\to\\infty]{} \\frac12.\n\nIf (a_n) converges to a limit L we must have\n\nL = 2 + \\frac12\\,L \\;\\Longrightarrow\\; L = 4.                      (9)\n\nIt remains to prove convergence.\n\nStep 2  –  A remainder recursion.\nSet C_n := a_n - 4.  Subtract 4 from both sides of (8):\n\nC_n = \\frac{n+1}{2n}\\,C_{n-1} + \\frac{2}{n}.                      (10)\n\nStep 3  –  Bounding C_n.\nWe show by induction that\n\n0 \\le C_n \\le \\frac{12}{n} \\qquad(n\\ge 1).                       (11)\n\nBase n = 1 :  a_1 = 2/\\binom{1}{0} + 2/\\binom{1}{1} = 4, so C_1 = 0 and (11) holds.\n\nInduction step.  Assume (n−1)C_{n-1} ≤ 12 for some n ≥ 2.  From\n(10) we obtain\n\nnC_n = \\frac{n+1}{2}\\,C_{n-1} + 2 \\le \\frac{n+1}{2}\\,\\frac{12}{n-1} + 2 = 6\\,\\frac{n+1}{n-1} + 2.\n\nFor n ≥ 4 one has (n+1)/(n-1) ≤ 5/3, and hence\n\nnC_n ≤ 6·(5/3) + 2 = 12.\n\nThe cases n = 2,3 can be checked directly and also satisfy (11).\nThus (11) holds for every n.\n\nStep 4  –  Convergence.\nFrom (11) we get |C_n| ≤ 12/n → 0, hence a_n = 4 + C_n → 4.\nBy (9) this is the only possible limit, so (a_n) indeed converges\nand\n\n\\boxed{\\displaystyle \\lim_{n\\to\\infty} a_n = 4}.\n\nThis completes the solution.",
      "_meta": {
        "core_steps": [
          "Multiply by n! so that 1/\\binom{n}{k} becomes k!(n-k)!, turning the sum into a factorial sum",
          "Combine/shift the terms in pairs to rewrite 2·n!·b_n as 2·n! + (n+1)(n−1)!·b_{n−1}",
          "Divide by 2 n! to obtain the linear recurrence b_n = 1 + (n+1)/(2n)·b_{n−1}",
          "Introduce C_n = b_n − 2, translate the recurrence to n·C_n = 1 + (n+1)/(2(n−1))·(n−1)·C_{n−1}",
          "Choose a fixed constant and prove by induction that n·C_n stays below it; this forces C_n → 0, hence b_n → 2"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The numerical constant selected as an upper bound in the induction argument",
            "original": "6"
          },
          "slot2": {
            "description": "The treatment of the endpoint terms k = 0 and k = n in the pairing step, which produces the standalone '+1' in the recurrence",
            "original": "addition of 0!·n! and n!·0! (weight 1 each)"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}