%I #17 Jan 21 2023 09:33:20
%S 0,1,2,3,4,5,6,9,11,13,14,15,16,18,19,21,22,27,28,30,31,32,34,35,36,
%T 38,39,46,47,49,50,52,54,55,57,58,60,61,62,63,64,65,66,69,70,74,75,77,
%U 78,79,80,82,83,84,85,86,87,88,90,91,92,93,94,95,96,97,98,99,100
%N Integers k such that the product of the first k primes is a Niven number.
%C Integers k such that A002110(k) belongs to A005349.
%C So, the sequence A002110(a(n)) is a subsequence of A359960. - _Bernard Schott_, Jan 21 2023
%e A002110(5) = 2310 and 2310 is divisible by 2+3+1+0=6, so 5 is a term.
%t a={}; For[k=0, k<=100, k++, p=Product[Prime[i],{i,k}]; If[Mod[p,Total[IntegerDigits[p]]]==0, AppendTo[a,k]]]; a (* _Stefano Spezia_, Jan 21 2023 *)
%o (PARI) isok(k) = my(p=factorback(primes(k))); !(p % sumdigits(p));
%Y Cf. A002110, A005349, A359960.
%K nonn,base
%O 1,3
%A _Michel Marcus_, Jan 21 2023