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

1,216

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) = 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

Example

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.

Mathematica

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

Prog

(Scheme, with memoization-macro definec)
(definec (A295662 n) (if (= 1 n) 0 (+ (if (= 1 (A067029 n)) 0 (A000035 (A067029 n))) (A295662 (A028234 n)))))
(PARI) a(n) = vecsum(apply(x -> x%2 - (x==1), factor(n)[, 2])); \\ Amiram Eldar, Sep 28 2023

Crossrefs

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

Cf. A295661 (positions of nonzero terms).

Sequence in context: A359472 A295883 A366124 * A367512 A187946 A330549

Adjacent sequences: A295659 A295660 A295661 * A295663 A295664 A295665

Keyword

nonn,easy

Author

Antti Karttunen, Nov 28 2017

Status

approved