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A295883
Number of exponents that are 3 in the prime factorization of n.
9
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, 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, 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
OFFSET
1,216
FORMULA
Additive with a(p^3) = 1, a(p^e) = 0 when e <> 3.
a(n) = A295659(n) - A295884(n).
a(n) <= A295662(n) <= A295663(n).
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
EXAMPLE
For n = 8 = 2^3, there is one exponent that is exactly 3, thus a(8) = 1.
For n = 216 = 2^3 * 3^3 there are two exponents that are exactly 3, thus a(216) = 2.
For n = 432 = 2^4 * 3^3, there is one exponent that is exactly 3, thus a(432) = 1.
MATHEMATICA
Array[Total@ Map[Boole[# == 3] &, FactorInteger[#][[All, -1]]] &, 120] (* Michael De Vlieger, Nov 29 2017 *)
Count[FactorInteger[#][[All, 2]], 3]&/@Range[120] (* Harvey P. Dale, Apr 13 2019 *)
PROG
(Scheme, with memoization-macro definec)
(definec (A295883 n) (if (= 1 n) 0 (+ (if (= 3 (A067029 n)) 1 0) (A295883 (A028234 n)))))
(PARI) a(n) = vecsum(apply(x->(x==3), factor(n)[, 2])); \\ Michel Marcus, Jul 25 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Nov 29 2017
STATUS
approved