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A295883 Number of exponents that are 3 in the prime factorization of n.

%I A295883 #20 Jul 25 2022 22:42:13
%S A295883 0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,
%T A295883 0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,
%U A295883 0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1
%N A295883 Number of exponents that are 3 in the prime factorization of n.
%H A295883 Antti Karttunen, <a href="/A295883/b295883.txt">Table of n, a(n) for n = 1..65537</a>
%H A295883 <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.
%F A295883 Additive with a(p^3) = 1, a(p^e) = 0 when e <> 3.
%F A295883 a(n) = A295659(n) - A295884(n).
%F A295883 a(n) <= A295662(n) <= A295663(n).
%F A295883 Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} (1/p^3 - 1/p^4) = A085541 - A085964 = 0.0977694995... . - _Amiram Eldar_, Jul 25 2022
%e A295883 For n = 8 = 2^3, there is one exponent that is exactly 3, thus a(8) = 1.
%e A295883 For n = 216 = 2^3 * 3^3 there are two exponents that are exactly 3, thus a(216) = 2.
%e A295883 For n = 432 = 2^4 * 3^3, there is one exponent that is exactly 3, thus a(432) = 1.
%t A295883 Array[Total@ Map[Boole[# == 3] &, FactorInteger[#][[All, -1]]] &, 120] (* _Michael De Vlieger_, Nov 29 2017 *)
%t A295883 Count[FactorInteger[#][[All,2]],3]&/@Range[120] (* _Harvey P. Dale_, Apr 13 2019 *)
%o A295883 (Scheme, with memoization-macro definec)
%o A295883 (definec (A295883 n) (if (= 1 n) 0 (+ (if (= 3 (A067029 n)) 1 0) (A295883 (A028234 n)))))
%o A295883 (PARI) a(n) = vecsum(apply(x->(x==3), factor(n)[,2])); \\ _Michel Marcus_, Jul 25 2022
%Y A295883 Cf. A085541, A085964, A295659, A295662, A295663, A295884.
%K A295883 nonn
%O A295883 1,216
%A A295883 _Antti Karttunen_, Nov 29 2017