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A256605 Least k such that n+1 is the n-th divisor of k.

%I A256605 #21 Jun 01 2025 09:57:56
%S A256605 3,4,20,12,84,120,360,360,3960,2520,32760,27720,27720,55440,942480,
%T A256605 720720,13693680,12252240,12252240,12252240,281801520,232792560,
%U A256605 1163962800,1163962800,3491888400,3491888400,101264763600,80313433200,2489716429200,4658179125600
%N A256605 Least k such that n+1 is the n-th divisor of k.
%C A256605 The case n = 1 is not possible because the number 2 is never the first divisor of k (1 is the first divisor).
%F A256605 a(n) = A003418(n+1)/A007917(n). - _Peter Munn_, May 14 2025
%e A256605 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.
%p A256605 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:
%t A256605 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]
%o A256605 (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
%Y A256605 Cf. A003418, A007917, A225562.
%Y A256605 See A221647 for other sequences giving the smallest number whose n-th divisor satisfies some condition.
%K A256605 nonn,easy
%O A256605 2,1
%A A256605 _Michel Lagneau_, Apr 04 2015