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A245661
Number of p*pp-divisors of n where a p*pp-number is the product of a prime and perfect power (see A245303).
2
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
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
KEYWORD
nonn
AUTHOR
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