A295659 Number of exponents larger than 2 in the prime factorization of n.

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, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 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, 1, 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^e) = 1 if e>2, 0 otherwise.

a(n) = 0 iff A212793(n) = 1.

a(n) = A001221(A053150(n)).

a(n) = A056170(A003557(n)).

a(n) >= A295662(n) = A162642(n) - A056169(n).

a(n) = A295883(n) + A295884(n).

Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = Sum_{p prime} 1/p^3 = 0.174762... (A085541). - Amiram Eldar, Nov 01 2020

Example

For n = 120 = 2^3 * 3^1 * 5^1 there is only one exponent larger than 2, thus a(120) = 1.
For n = 216 = 2^3 * 3^3 there are two exponents larger than 2, thus a(216) = 2.

Mathematica

Array[Count[FactorInteger[#][[All, -1]], _?(# > 2 &)] &, 105] (* Michael De Vlieger, Nov 28 2017 *)

Prog

(Scheme, with memoization-macro definec) (definec (A295659 n) (if (= 1 n) 0 (+ (if (> (A067029 n) 2) 1 0) (A295659 (A028234 n)))))
(PARI) a(n) = { my(v = factor(n)[, 2], i=0); for(x=1, length(v), if(v[x]>2, i++)); i; } \\ Iain Fox, Nov 29 2017

Crossrefs

Cf. A001221, A003557, A053150, A056169, A056170, A085541, A162642, A295657, A295662, A295883, A295884.

Cf. A004709 (positions of zeros), A046099 (of nonzeros), A212793.

Sequence in context: A044938 A072401 A064873 * A188436 A037823 A293162

Adjacent sequences: A295656 A295657 A295658 * A295660 A295661 A295662

Keyword

nonn

Author

Antti Karttunen, Nov 28 2017

Status

approved