A353353 Number of ways to write n as a product of the terms of A332820 larger than 1; a(1) = 1 by convention (an empty product).

1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 2, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 2, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1

Offset: 1

Author

Antti Karttunen, Apr 15 2022

Keywords

• nonn
• easy

Comments

Number of factorizations of n into factors k > 1 for which A048675(k) is a multiple of three.

Examples

			Of the eight divisors of 36 larger than 1, [2, 3, 4, 6, 9, 12, 18, 36], only 6 and 36 are in A332820, and because these allow two different factorizations as 36 = 6*6, we have a(36) = 2.
		

Links

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

Crossrefs

Cf. A003961, A048675, A332820, A348717, A353350, A353359 [= a(n^3)].

Cf. also A353303, A353333.

Programs

PARI

A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
A353350(n) = (0==(A048675(n)%3));
A353353(n, m=n) = if(1==n, 1, my(s=0); fordiv(n, d, if((d>1)&&(d<=m)&&A353350(d), s += A353353(n/d, d))); (s));

Formula

a(n) = a(A003961(n)) = a(A348717(n)), for all n >= 1.

a(p) = 0 for all primes p.