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

1,32

Comments

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

Links

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

Index entries for sequences computed from indices in prime factorization

Formula

a(p) = 0 for all primes p.

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

Example

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.

Prog

(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));

Crossrefs

Cf. A003961, A048675, A332823, A348717, A353352, A353354, A353355, A353380, A353415 [= a(n^2)].

Cf. also A353303, A353353.

Sequence in context: A102761 A231119 A129558 * A267181 A131185 A307819

Adjacent sequences: A353411 A353412 A353413 * A353415 A353416 A353417

Keyword

nonn

Author

Antti Karttunen, Apr 19 2022

Status

approved