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

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

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences computed from exponents in factorization of n.

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

Cf. A085541, A085964, A295659, A295662, A295663, A295884.

Sequence in context: A323162 A185017 A359472 * A366124 A295662 A367512

Adjacent sequences: A295880 A295881 A295882 * A295884 A295885 A295886

Keyword

nonn

Author

Antti Karttunen, Nov 29 2017

Status

approved