A353414 Number of ways to write n as a product of the terms of A353355 larger than 1; a(1) = 1 by convention (an empty product).

1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 2, 1, 0, 1, 3, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 3, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 3, 1, 0, 0, 1, 1, 0, 0, 4, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 0, 1, 0, 0, 0, 2, 0, 1, 1, 0, 1, 3, 0, 1, 1, 2, 0, 0, 0, 1, 0

Offset: 1

Author

Antti Karttunen, Apr 19 2022

Keywords

• nonn

Comments

Number of factorizations of n into factors k > 1 for which A353354(k) = 0.

Examples

			Of the fourteen divisors of 144 larger than 1, only [4, 6, 8, 9, 12, 18, 36, 48, 72, 144] are in A353355. Using combinations of these, 144 can be factored in six different ways as 144 = 36*4 = 18*8 = 12*12 = 9*4*4 = 6*6*4, therefore a(144) = 6.
		

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, A332823, A348717, A353352, A353354, A353355, A353380, A353415 [= a(n^2)].

Cf. also A353303, A353353.

Programs

PARI

A332823(n) = { my(f = factor(n),u=(sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2)%3); if(2==u,-1,u); };
A353354(n) = sumdiv(n,d,A332823(d));
A353380(n) = (0==A353354(n));
A353414(n, m=n) = if(1==n, 1, my(s=0); fordiv(n, d, if((d>1)&&(d<=m)&&A353380(d), s += A353414(n/d, d))); (s));

Formula

a(p) = 0 for all primes p.

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