A245661 Number of p*pp-divisors of n where a p*pp-number is the product of a prime and perfect power (see A245303).

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 5, 1, 2, 2, 3, 1, 3, 1, 4, 2, 2, 2, 4, 1, 2, 2, 5, 1, 3, 1, 3, 3, 2, 1, 7, 1, 3, 2, 3, 1, 5, 2, 5, 2, 2, 1, 5, 1, 2, 3, 5, 2, 3, 1, 3, 2, 3, 1, 7, 1, 2, 3, 3, 2, 3, 1, 7, 3, 2, 1, 5, 2, 2, 2, 5, 1, 5, 2, 3, 2, 2, 2, 9, 1, 3, 3, 4, 1, 3, 1, 5, 3, 2, 1, 7, 1, 3, 2, 7, 1, 3, 2, 3, 3, 2, 2, 8

Offset

1,6

Comments

A001222(n) <= a(n) < A000005(n).

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

Example

a(24) = 5 because divisors of 24 of the form "p*pp" are 2 = 2*1, 3 = 3*1, 8 = 2*4, 12 = 3*4 and 24 = 3*8 where 2, 3 are primes and 1, 4, 8 are perfect powers.
For n = 72, because divisors of 72 of the form "p*pp" are 2 = 2*1, 3 = 3*1, 8 = 2*(2^2), 12 = 3*(2^2), 18 = 2*(3^2), 24 = 3*(2^2) and 72 = 2*(6^2), a(72) = 7. The other five divisors of 72: 1, 4, 6, 9 and 36 are not of the required type. - Antti Karttunen, May 28 2017

Prog

(PARI) ispp(n) = (n==1) || ispower(n);
isA245303(n) = {my(f = factor(n)); for (i=1, #f~, p = f[i, 1]; if (ispp(n/p), return(1)); ); return (0); }
a(n) = sumdiv(n, d, isA245303(d)); \\ Michel Marcus, Aug 08 2014

Crossrefs

Cf. A000005 (number of divisors of n), A001222 (number of prime divisors of n), A001597 (perfect powers), A245303 (P*PP-numbers).

Sequence in context: A338649 A337176 A033103 * A302789 A346089 A302776

Adjacent sequences: A245658 A245659 A245660 * A245662 A245663 A245664

Keyword

nonn

Author

Juri-Stepan Gerasimov, Jul 28 2014

Extensions

Data section extended to 120 terms (with corrected term a(72), computed with PARI-program of Michel Marcus) by Antti Karttunen, May 28 2017

Status

approved