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%I #18 Nov 19 2021 02:37:43
%S 1,3,9,27,72,111,129,148,161,450,762,1233,1260,2052,9153,15840,16067,
%T 16302,16317,16332,16435,74670,74946,125046,208566,347670,347685,
%U 583263,1609667,1610942,1616476,1616532,1616958,2683143,2700261,4480092,7469682,7470432,7492497
%N Numbers k that divide the sum of the digits of 3^k * k!.
%e 9 is a term because the sum of the digits of 3^9 * 9! = 7142567040 is 36 which is divisible by 9.
%t Do[If[Mod[Plus @@ IntegerDigits[3^n * n!], n] == 0, Print[n]], {n, 1, 10000}]
%o (PARI) isok(k) = !(sumdigits(3^k * k!) % k);
%Y Cf. A032031, A007953, A052673, A108861.
%K nonn,base,hard
%O 1,2
%A _Kevin P. Thompson_, Oct 21 2021
%E a(36)-a(39) from _Martin Ehrenstein_, Nov 19 2021