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Product of the digits of 3^n.
0

%I #33 Nov 08 2022 07:40:20

%S 1,3,9,14,8,24,126,112,180,1296,0,1372,240,3240,217728,0,0,0,0,24192,

%T 0,0,0,2709504,6635520,0,66355200,8534937600,731566080,0,0,

%U 10369949184,0,0,399983754240,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6243870843076608000

%N Product of the digits of 3^n.

%C a(68) is likely the last nonzero term; see A030700 and A238939. - _Michael S. Branicky_, Nov 06 2022

%F a(n) = A007954(A000244(n)).

%e For a(0), 3^0 = 1 with product of digits 1;

%e for a(3), 3^3 = 27 with product of digits 2*7 = 14;

%e for a(10), 3^10 = 59049 with product of digits 5*9*0*4*9 = 0.

%t a[n_] := Times @@ IntegerDigits[3^n]; Array[a, 69, 0] (* _Amiram Eldar_, Nov 07 2022 *)

%o (Python)

%o from math import prod

%o def a(n): return prod(map(int, str(3**n)))

%o print([a(n) for n in range(69)]) # _Michael S. Branicky_, Nov 06 2022

%o (PARI) a(n) = vecprod(digits(3^n)); \\ _Michel Marcus_, Nov 07 2022

%Y Cf. A007954, A000244.

%Y Cf. A030700, A238939.

%Y Cf. A014257.

%K nonn,base

%O 0,2

%A _Joseph Caliendo_, Nov 06 2022

%E More terms from _Michael S. Branicky_, Nov 06 2022