A257090 -id:A257090 - cp's OEIS frontend

A257089 a(n) = log_3 (A256689(n)).

0, 1, 1, 2, 1, 2, 1, 4, 2, 2, 1, 3, 1, 2, 2, 5, 1, 3, 1, 3, 2, 2, 1, 5, 2, 2, 4, 3, 1, 3, 1, 6, 2, 2, 2, 4, 1, 2, 2, 5, 1, 3, 1, 3, 3, 2, 1, 6, 2, 3, 2, 3, 1, 5, 2, 5, 2, 2, 1, 4, 1, 2, 3, 8, 2, 3, 1, 3, 2, 3, 1, 6, 1, 2, 3, 3, 2, 3, 1, 6, 5, 2, 1, 4, 2, 2, 2, 5, 1, 4, 2, 3, 2, 2, 2, 7, 1, 3, 3, 4

Offset: 1

Wolfgang Hintze, Apr 16 2015

nonn

a(n) is the logarithm to the base 3 of the denominator of the Dirichlet series of zeta(s)^(1/3). For details, see A256689.

Wolfgang Hintze, Table of n, a(n) for n = 1..500

Cf. A046645 (k = 2, log_2), A257089 (k = 3, log_3), A257090 (k = 4, log_2), A257091 (k = 5, log_5).

3^a(n) = A256689(n). a(n) = A007949(A256689(n)).

A257091 a(n) = log_5 (A256693(n)).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 4, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 3, 3, 1, 3, 1, 6, 2, 2, 2, 4, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 5, 2, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 4, 1, 2, 3, 7, 2, 3, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 2, 3, 1, 5, 4, 2, 1, 4, 2, 2, 2, 4, 1, 4, 2, 3, 2, 2, 2, 7, 1, 3, 3, 4

Offset: 1

Wolfgang Hintze, Apr 16 2015

nonn

a(n) is the logarithm to the base 5 of the denominator of the Dirichlet series of zeta(s)^(1/5). For details, see A256693.

Robert Israel and Wolfgang Hintze, Table of n, a(n) for n = 1..10000 (up to 500 from Wolfgang Hintze)
MathOverflow, The number of prime factors of a natural number.

Cf. A046645 (k = 2, log_2), A257089 (k = 3, log_3), A257090 (k = 4, log_2), A257091 (k = 5, log_5).

F:= proc(n) local e,m;
add(add(floor(e/5^m),m=0..floor(log[5](e))),e=map(t-> t[2],ifactors(n)[2]));
end proc:
seq(F(i),i=1..100);

F[n_] := Sum[Sum[Floor[e/5^m], {m, 0, Floor[Log[5, e]]}], {e, FactorInteger[n][[All, 2]]}];
F[1] = 0;
Array[F, 100] (* Jean-François Alcover, Jun 18 2020, after Maple *)

5^a(n) = A256693(n).
For n<=10000, if n = Product_i p_i^(e_i) is the prime factorization of n, a(n) = A001222(n) + Sum_i floor(e_i/5). - Robert Israel, May 13 2016
If n = Product_i p_i^(e_i) is the prime factorization of n, a(n) = Sum_{j >= 0} Sum_i floor(e_i/5^j). - Robert Israel, May 16 2016

cp's OEIS Frontend

A257089 a(n) = log_3 (A256689(n)).

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