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

1,36

Comments

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

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences computed from indices in prime factorization

Formula

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

a(p) = 0 for all primes p.

Example

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.

Prog

(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));

Crossrefs

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

Cf. also A353303, A353333.

Sequence in context: A206479 A219484 A060396 * A195470 A324902 A252370

Adjacent sequences: A353350 A353351 A353352 * A353354 A353355 A353356

Keyword

nonn,easy

Author

Antti Karttunen, Apr 15 2022

Status

approved