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A353373
Number of ways to write n as a product of the terms of A325698 larger than 1; a(1) = 1 by convention (an empty product).
3
1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 2, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0
OFFSET
1,36
COMMENTS
Number of factorizations of n into factors k > 1 for which there are equal number of odd-indexed primes (A031368) as there are even-indexed primes (A031215) in their 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 23 divisors of 936 larger than 1, only [6, 26, 36, 156, 936] are in A325698. These can be combined to yield a product yielding 936 in four possible ways as 936 = 156*6 = 36*26 = 26*6*6, therefore a(936) = 4.
Of the 31 divisors of 2184 larger than 1, only [6, 14, 26, 84, 156, 364, 2184] are in A325698. These can be combined to yield a product yielding 2184 in five possible ways as 2184 = 364*6 = 156*14 = 84*26, therefore a(2184) = 5.
PROG
(PARI)
A353370(n) = { my(f = factor(n)); (0==sum(i=1, #f~, f[i, 2]*((-1)^(primepi(f[i, 1])%2)))); }; \\ Antti Karttunen, Apr 16 2022
A353373(n, m=n) = if(1==n, 1, my(s=0); fordiv(n, d, if((d>1)&&(d<=m)&&A353370(d), s += A353373(n/d, d))); (s));
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Apr 16 2022
STATUS
approved