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%I #37 Jul 11 2023 15:13:57
%S 1,2,3,5,6,9,27,81,126,159,205,252,254,267,285,675,1053,1086,1125,
%T 1146,2007,5088,5382,5448,14652,23401,23574,24009,41004,66789,67482,
%U 111480,866538,1447875,2413152,2414019,2417828,2421360,4045482,6713982
%N Numbers k that divide the sum of the digits of 2^k * k!.
%C Next term after 5448 is greater than 10000.
%C a(34) > 10^6. - _D. S. McNeil_, Mar 03 2009
%C a(39) > 2.54 * 10^6, if it exists. - _Kevin P. Thompson_, Oct 20 2021
%C a(41) > 7*10^6, if it exists. - _Kevin P. Thompson_, Dec 08 2021
%e 9 is a term because the sum of the digits of 2^9 * 9! = 185794560 is 45 which is divisible by 9.
%t Do[If[Mod[Plus @@ IntegerDigits[2^n * n! ], n] == 0, Print[n]], {n, 1, 10000}]
%t Select[Range[6714000],Mod[Total[IntegerDigits[2^# #!]],#]==0&] (* _Harvey P. Dale_, Jul 11 2023 *)
%o (PARI) isok(k) = !(sumdigits(2^k * k!) % k); \\ _Michel Marcus_, Oct 20 2021
%o (Python)
%o from itertools import islice
%o def A108861(): # generator of terms
%o k, k2, kf = 1, 2, 1
%o while True:
%o c = sum(int(d) for d in str(k2*kf))
%o if not c % k: yield k
%o k += 1
%o k2 *= 2
%o kf *= k
%o A108861_list = list(islice(A108861(),10)) # _Chai Wah Wu_, Oct 26 2021
%Y Cf. A000165, A007953, A052582.
%K nonn,base,hard,more
%O 1,2
%A _Ryan Propper_, Jul 11 2005
%E a(25)-a(33) from _D. S. McNeil_, Mar 03 2009
%E a(34)-a(38) from _Kevin P. Thompson_, Oct 20 2021
%E a(39)-a(40) from _Kevin P. Thompson_, Dec 08 2021
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