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%I #26 Jun 18 2020 08:30:24
%S 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,
%T 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,
%U 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
%N a(n) = log_5 (A256693(n)).
%C 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.
%H Robert Israel and Wolfgang Hintze, <a href="/A257091/b257091.txt">Table of n, a(n) for n = 1..10000</a> (up to 500 from Wolfgang Hintze)
%H MathOverflow, <a href="http://mathoverflow.net/questions/238725/the-number-of-prime-factors-of-a-natural-number/238726#238726">The number of prime factors of a natural number</a>.
%F 5^a(n) = A256693(n).
%F 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
%F 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
%p F:= proc(n) local e,m;
%p add(add(floor(e/5^m),m=0..floor(log[5](e))),e=map(t-> t[2],ifactors(n)[2]));
%p end proc:
%p seq(F(i),i=1..100);
%t F[n_] := Sum[Sum[Floor[e/5^m], {m, 0, Floor[Log[5, e]]}], {e, FactorInteger[n][[All, 2]]}];
%t F[1] = 0;
%t Array[F, 100] (* _Jean-François Alcover_, Jun 18 2020, after Maple *)
%Y Cf. A046645 (k = 2, log_2), A257089 (k = 3, log_3), A257090 (k = 4, log_2), A257091 (k = 5, log_5).
%K nonn
%O 1,4
%A _Wolfgang Hintze_, Apr 16 2015
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