A295662 Number of odd exponents larger than one in the canonical 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, 1, 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, 1, 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

Author

Antti Karttunen, Nov 28 2017

Keywords

• nonn
• easy

Examples

			For n = 24 = 2^3 * 3^1 there are two odd exponents, but only the other is larger than 1, thus a(24) = 1.
For n = 216 = 2^3 * 3^3 there are two odd exponents larger than 1, thus a(216) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences computed from exponents in factorization of n.

Crossrefs

Cf. A056169, A162642, A295659, A295663, A295664.

Cf. A295661 (positions of nonzero terms).

Programs

Mathematica

Array[Count[FactorInteger[#][[All, -1]], ?(And[OddQ@ #, # > 1] &)] &, 105] (* _Michael De Vlieger, Nov 28 2017 *)

PARI

a(n) = vecsum(apply(x -> x%2 - (x==1), factor(n)[, 2])); \\ Amiram Eldar, Sep 28 2023

Formula

Additive with a(p) = 0, a(p^e) = A000035(e) if e > 1.

a(1) = 0; and for n > 1, if A067029(n) = 1, a(n) = a(A028234(n)), otherwise A000035(A067029(n)) + a(A028234(n)).

a(n) = A162642(n) - A056169(n).

a(n) <= A295659(n).

a(n) = 0 iff A295663(n) = 0, and when A295663(n) > 0, a(n) <= A295663(n).

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{p prime} 1/(p^2*(p+1)) = 0.122017493776862257491... . - Amiram Eldar, Sep 28 2023