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%I #13 May 14 2023 06:56:41
%S 1,2,5,7,20,24,45,54,286,1942,30771000,71149819,106438598,668274063
%N Numbers k such that A001414(k^3+1) is divisible by k.
%C a(15) > 1.5*10^9. - _Giovanni Resta_, Aug 07 2019
%e 5 is a member because the prime factorization of 5^3+1=126 is 2*3^2*7 and 2+3+3+7=15 is divisible by 5.
%p filter:= proc(n) local F,t;
%p F:= ifactors(n^3+1)[2];
%p add(t[1]*t[2], t=F) mod n = 0
%p end proc:
%p select(f, [$1..10000]);
%t sopfr[n_] := Total[Times @@@ FactorInteger[n]];
%t okQ[n_] := Divisible[sopfr[n^3+1], n];
%t Select[Range[10^5], okQ] (* _Jean-François Alcover_, May 14 2023 *)
%Y Cf. A001414, A309534, A309544.
%K nonn,more
%O 1,2
%A _J. M. Bergot_ and _Robert Israel_, Aug 06 2019
%E a(11)-a(14) from _Giovanni Resta_, Aug 07 2019
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