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Numbers k such that A000203(k) and A007955(k) are both divisible by A187680(k).
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%I #16 May 16 2023 10:40:04

%S 4,9,14,16,25,38,42,49,51,55,62,64,70,81,86,92,96,117,121,130,134,138,

%T 140,158,159,161,168,169,182,206,209,234,254,256,266,267,278,282,284,

%U 289,302,322,326,351,361,376,390,398,408,410,422,426,434,446,477,508,529,532,534,542,551,566,590

%N Numbers k such that A000203(k) and A007955(k) are both divisible by A187680(k).

%C Numbers k such that both the sum s and product p of the divisors of k are divisible by (p mod s).

%H Robert Israel, <a href="/A346956/b346956.txt">Table of n, a(n) for n = 1..10000</a>

%e a(3) = 14 is a term because A000203(14) = 1+2+7+14 = 24, A007955(14) = 1*2*7*14 = 196, A187680(14) = 196 mod 24 = 4, and both 24 and 196 are divisible by 4.

%p filter:= proc(n) local d,s,p,r;

%p d:= numtheory:-divisors(n);

%p s:= convert(d,`+`);

%p p:= convert(d,`*`);

%p r:= p mod s;

%p r <> 0 and p mod r = 0 and s mod r = 0

%p end proc:

%p select(filter, [$1..1000]);

%t okQ[n_] := Module[{d, s, p, m},

%t d = Divisors[n];

%t s = Total[d];

%t p = Times @@ d;

%t m = Mod[p, s];

%t If[m == 0, False, Divisible[s, m] && Divisible[p, m]]];

%t Select[Range[1000], okQ] (* _Jean-François Alcover_, May 16 2023 *)

%o (PARI) isok(k) = my(d=divisors(k), s=vecsum(d), p=vecprod(d), m=p % s); (m>0) && !(s%m) && !(p%m); \\ _Michel Marcus_, Aug 09 2021

%Y Cf. A000203, A007955, A187680.

%Y Includes A188061.

%K nonn

%O 1,1

%A _J. M. Bergot_ and _Robert Israel_, Aug 08 2021