%I #12 Dec 05 2021 00:06:08
%S 2,4,5,9,15,17,20,24,25,26,27,29,31,32,33,34,37,44,45,46,49,51,52,61,
%T 62,63,64,71,74,79,80,81,82,85,87,88,91,95,103,104,105,110,111,112,
%U 115,117,118,119,120,121,127,131,135,137,142,148,150,152,154,158,159,163,165,173,175,177,179,181
%N Numbers k such that A255217(k) divides A002110(k).
%H Martin Ehrenstein, <a href="/A349738/b349738.txt">Table of n, a(n) for n = 1..10000</a>
%e a(3) = 5 is a term because A255217(5) = 2*3*5*7*11 mod (2+3+5+7+11) = 14 divides 2*3*5*7*11.
%p P:= 1: S:= 0: p:= 1:
%p count:= 0: R:= NULL:
%p for n from 1 while count < 100 do
%p p:= nextprime(p);
%p P:= P*p; S:= S+p;
%p r:= P mod S;
%p if r = 0 then next fi;
%p v:= P mod r;
%p if v = 0 then
%p count:= count+1; R:= R,n;
%p fi
%p od:
%p R;
%t Select[Range[200], (m = Mod[Times @@ (p = Prime[Range[#]]), Plus @@ p]) > 0 && Divisible[Times @@ p, m] &] (* _Amiram Eldar_, Nov 28 2021 *)
%Y Cf. A002110, A007504, A255217. Contains A349734.
%K nonn
%O 1,1
%A _J. M. Bergot_ and _Robert Israel_, Nov 28 2021