A353333 Number of ways to write n as a product of the terms of A340784 larger than 1; a(1) = 1 by convention (an empty product).

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

Offset: 1

Author

Antti Karttunen, Apr 14 2022

Keywords

• nonn

Comments

Number of factorizations of n into factors k > 1 for which both A001222(k) and A056239(k) are even.

Examples

			Of the eleven divisors of 220 larger than one, only [4, 10, 22, 55, 220] are in A340784, as both the number of their prime factors (with repetition, A001222), [2, 2, 2, 2, 4], and their integer pseudo logarithms (A056239), [2, 4, 6, 8, 10], are even. Using these factors gives the following possible factorizations: 220 = 22*10 = 55*4, therefore a(220) = 3.
Of the eight divisors of 256 larger than one, only [1, 4, 16, 64, 256] are in A340784. Using these factors, we obtain the following factorizations: 256 = 64*4 = 16*16 = 16*4*4 = 4*4*4*4, therefore a(256) = 5.
Of the 23 divisors of 792 larger than one, only [4, 9, 22, 36, 88, 198, 792] are in A340784. Using these factors gives the following possible factorizations: 792 = 198*4 = 88*9 = 36*22 = 22*4*9, therefore a(792) = 5.
		

Links

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

Crossrefs

Cf. A001222, A003961, A056239, A340784, A348717, A353331, A353334 [= a(n^2)].

Differs from A353303 for the first time at n=30, where a(30) = 0, while A353303(30) = 1.

Programs

PARI

A056239(n) = { my(f); if(1==n, 0, f=factor(n); sum(i=1, #f~, f[i, 2] * primepi(f[i, 1]))); }
A353331(n) = ((!(bigomega(n)%2)) && (!(A056239(n)%2)));
A353333(n, m=n) = if(1==n, 1, my(s=0); fordiv(n, d, if((d>1) && (d<=m) && A353331(d), s += A353333(n/d, d))); (s));

Formula

a(p) = 0 for all primes p.

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