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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).
8
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
1,16
COMMENTS
Number of factorizations of n into factors k > 1 for which both A001222(k) and A056239(k) are even.
FORMULA
a(p) = 0 for all primes p.
a(n) = a(A003961(n)) = a(A348717(n)), for all n >= 1.
EXAMPLE
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.
PROG
(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));
CROSSREFS
Differs from A353303 for the first time at n=30, where a(30) = 0, while A353303(30) = 1.
Sequence in context: A331902 A333817 A270417 * A353303 A307666 A319995
KEYWORD
nonn
AUTHOR
Antti Karttunen, Apr 14 2022
STATUS
approved