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A324574
a(1) = 0; for n > 1, a(n) = A033879(A087207(n)).
7
0, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 5, 0, 1, 1, 2, 1, 4, 2, 16, 1, 2, 1, 18, 1, 5, 1, 6, 1, 1, -3, 46, -4, 2, 1, 82, 14, 4, 1, 10, 1, 16, 0, 256, 1, 2, 1, 4, -12, 18, 1, 2, -2, 5, 8, 226, 1, 6, 1, 748, 2, 1, -19, 18, 1, 46, -12, 12, 1, 2, 1, 1362, 0, 82, -12, 22, 1, 4, 1, 3838, 1, 10, 10, 5458, 254, 16, 1, 6, -10, 256, -348, 12250
OFFSET
1,6
COMMENTS
As A087207 is a surjective function that toggles the parity, it follows that if it can be proved/disproved that a(n) = 0 for some/any even n, then it also proves/disproves the existence of odd perfect numbers.
The positions (n > 1) of zeros in squarefree n, 15, 385, ..., can be obtained as A019565(A000396(n)).
FORMULA
a(1) = 0; for n > 1, a(n) = A033879(A087207(n)).
a(n) = a(A007947(n)) = A324575(A007947(n)).
PROG
(PARI)
A033879(n) = (2*n-sigma(n));
A087207(n) = vecsum(apply(p->1<<primepi(p-1), factor(n)[, 1])); \\ From A087207
A324574(n) = if(1==n, 0, A033879(A087207(n)));
KEYWORD
sign
AUTHOR
Antti Karttunen, Mar 08 2019
STATUS
approved