## Divisor Function

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• Sebastián Martín Ruiz Said: 1. Divisor Function Posted by: Sebastián Martín Ruiz sebi_sebi@yahoo.com sebi_sebi Date: Sun Aug 19, 2007 8:31 pm ((PDT))
Message 1 of 7 , Aug 24, 2007
Sebastián Martín Ruiz Said:

1. Divisor Function
Posted by: "Sebastián Martín Ruiz" sebi_sebi@... sebi_sebi
Date: Sun Aug 19, 2007 8:31 pm ((PDT))

Hello all:

I have obtained this result:

DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
if and only if
m is a perfect square

DivisorSigma[k,m]=sum[d^k, d divisor of m]

Can anyone prove this result?

Sincerely

Sebastian Martin Ruiz

Kermit Said:

If m is a perfect square then

m = p1^a1 p2^a2 p3^a3 . . . pj^aj

where each of the a1, a2, .... aj are even.

If d divides m then

d = p1^b1 p2^b2 p3^b3 . . . pj^bj

where each b1,b2,...bj are less than or equal to the corresponding
a1,a2,...aj.

The
DivisorSigma[k,m]=sum[d^k, d divisor of m]

is product of

( 1 + p1^k + p1^(2k) + p1^(3k) + . . . + p1^(a1*k) )
(1 + p2^k + p2^(2k) + p2^(3k) + . . .+ p2^(a2*k) )
. . .
(1 + pj^k + pj^(2k) + pj^(3k) + . . . + pj^(aj * k) )

=

( p1^( a1 * k) - 1) / (p1^k - 1)
(p2^(a2 * k) - 1)/ ( p2^k - 1)
. . .
(pj^(aj*k) - 1) / (pj^k - 1)

When will this product be divisible by

( p1^( a1 * k/2) - 1) / (p1^k - 1)
(p2^(a2 * k/2) - 1)/ ( p2^k - 1)
. . .
(pj^(aj*k/2) - 1) / (pj^k - 1)

?