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A379130
a(n) is the number of unitary divisors d of n for which A048720(A065621(sigma(d)),sigma(n/d)) is equal to sigma(n).
2
1, 1, 2, 1, 1, 2, 2, 1, 1, 3, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 4, 3, 1, 2, 1, 2, 1, 2, 1, 6, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 2, 2, 3, 1, 1, 3, 2, 2, 2, 1, 4, 1, 2, 2, 1, 2, 6, 1, 1, 2, 6, 1, 2, 1, 1, 2, 2, 2, 4, 1, 2, 1, 1, 1, 4, 1, 2, 2, 2, 1, 1, 2, 2, 4, 3, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 4
OFFSET
1,3
COMMENTS
It seems that A046528 gives all numbers k for which a(k) = A034444(k).
FORMULA
a(n) = Sum_{d|n, gcd(d,n/d)=1} [A048720(A065621(sigma(d)),sigma(n/d)) == sigma(n)], where [ ] is the Iverson bracket.
A379129(n) <= a(n) <= A034444(n).
EXAMPLE
For every n, a(n) >= 1, because A048720(A065621(sigma(1)), sigma(n)) = A048720(A065621(1), sigma(n)) = A048720(1, sigma(n)) = sigma(n).
For n = 21 = 3*7, after the divisor pair [1,21], all other divisor pairs also satisfy the condition: A048720(A065621(sigma(3)),sigma(7)) [= A048720(4,8)] and A048720(A065621(sigma(7)),sigma(3)) [= A048720(8,4)] and A048720(A065621(sigma(21)),sigma(1)) [= A048720(32,1)] all yield the decided result, 32 = sigma(21), therefore a(21) = 4.
See also examples in A379129.
PROG
(PARI)
A048720(b, c) = fromdigits(Vec(Pol(binary(b))*Pol(binary(c)))%2, 2);
A065621(n) = bitxor(n-1, n+n-1);
A379130(n) = { my(s=sigma(n)); sumdiv(n, d, if(1!=gcd(d, n/d), 0, A048720(A065621(sigma(n/d)), sigma(d))==s)); };
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Dec 18 2024
STATUS
approved