Joseph Caliendo - cp's OEIS frontend

User: Joseph Caliendo

Joseph Caliendo has authored 2 sequences.

A358271 Product of the digits of 3^n.

1, 3, 9, 14, 8, 24, 126, 112, 180, 1296, 0, 1372, 240, 3240, 217728, 0, 0, 0, 0, 24192, 0, 0, 0, 2709504, 6635520, 0, 66355200, 8534937600, 731566080, 0, 0, 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

Offset: 0

Joseph Caliendo, Nov 06 2022

a(68) is likely the last nonzero term; see A030700 and A238939. - Michael S. Branicky, Nov 06 2022

			For a(0), 3^0 = 1 with product of digits 1;
for a(3), 3^3 = 27 with product of digits 2*7 = 14;
for a(10), 3^10 = 59049 with product of digits 5*9*0*4*9 = 0.

Cf. A007954, A000244.
Cf. A030700, A238939.
Cf. A014257.

a[n_] := Times @@ IntegerDigits[3^n]; Array[a, 69, 0] (* Amiram Eldar, Nov 07 2022 *)

a(n) = vecprod(digits(3^n)); \\ Michel Marcus, Nov 07 2022

from math import prod
def a(n): return prod(map(int, str(3**n)))
print([a(n) for n in range(69)]) # Michael S. Branicky, Nov 06 2022

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

More terms from Michael S. Branicky, Nov 06 2022

A337759 Squares that are the sum of 3 distinct nonzero squares.

Original entry on oeis.org

49, 81, 121, 169, 196, 225, 289, 324, 361, 441, 484, 529, 625, 676, 729, 784, 841, 900, 961, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969

Offset: 1

Joseph Caliendo, Sep 18 2020

nonn

These are the squares in A004432. - Omar E. Pol, Sep 18 2020

			49 is a term because 6^2(36) + 3^2(9) + 2^2(4) = 7^2(49).
81 is a term because 8^2(64) + 4^2(16) + 1^2(1) = 9^2(81).
121 is a term because 9^2(81) + 6^2(36) + 2^2(4) = 11^2(121).
625 is a term because 9^2(81) + 12^2(144) + 20^2(400) = 25^2(625).

Cf. A004432, A161992.

Select[Range[63]^2, Length @ Reduce[x^2 + y^2 + z^2 == # && 0 < x < y < z, {x, y, z}, Integers] > 0 &] (* Amiram Eldar, Sep 18 2020 *)

a(n) = A161992(n)^2. - Andrew Howroyd, Sep 18 2020

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User: Joseph Caliendo

A358271 Product of the digits of 3^n.

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