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Numbers k such that A255217(k) divides A002110(k).
2

%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