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A256605
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Least k such that n+1 is the n-th divisor of k.
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0
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3, 4, 20, 12, 84, 120, 360, 360, 3960, 2520, 32760, 27720, 27720, 55440, 942480, 720720, 13693680, 12252240, 12252240, 12252240, 281801520, 232792560, 1163962800, 1163962800, 3491888400, 3491888400, 101264763600, 80313433200, 2489716429200, 4658179125600
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OFFSET
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2,1
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COMMENTS
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The case n = 1 is not possible because the number 2 is never the first divisor of k (1 is the first divisor).
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LINKS
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EXAMPLE
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a(6) = 84 because the divisors of 84 are {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84} and 7 is the 6th divisor of 84.
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MAPLE
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with(numtheory):for n from 2 to 31 do:ii:=0:for k from 1 to 10^9 while(ii=0) do:x:=divisors(k):n1:=nops(x):if n<=n1 and x[n]=n+1 then ii:=1: printf ( "%d %d \n", n, k):else fi:od:od:
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MATHEMATICA
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nn=20; t=Table[0, {nn}]; found=1; n=2; While[found<nn, n++; d=Divisors[n]; Do[If[i<=nn&&d[[i]]==i+1&&t[[i]]==0, t[[i]]=n; found++], {i, Length[d]}]]; Rest[t]
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PROG
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(PARI) a(n) = {k = 1; ok = 0; while (!ok, d = divisors(k); if ((#d >= n) && (d[n] == n+1), ok = 1, k++); ); k; } \\ Michel Marcus, Apr 04 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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