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• ... As Adam already disproved the only if part, let s concentrate on the if one. First, there are two simple observations to make: 1) If p is a prime,
Message 1 of 7 , Aug 21, 2007
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> 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]

As Adam already disproved the "only if" part, let's concentrate on the
"if" one. First, there are two simple observations to make:
1) If p is a prime,
DivisorSigma[k, p^e] = 1 + p^k + ... + p^ek = (p^[k(e+1)] - 1)/(p^k-1)
2) If c and d are coprime,
DivisorSigma[k, c*d] = DivisorSigma[k, c] * DivisorSigma[k, d]

Thus, if m = p1^e2 * p2^e2 ... * pn^en is the prime factorization, we have
DivisorSigma[k, m] =
(p1^[k(e1+1)] - 1) / (p1^k - 1) *
(p2^[k(e2+1)] - 1) / (p2^k - 1) *
...
(pn^[k(en+1)] - 1) / (pn^k - 1).
and DivisorSigma[2k, m] =
(p1^[2k(e1+1)] - 1) / (p1^(2k) - 1) *
(p2^[2k(e2+1)] - 1) / (p2^(2k) - 1) *
...
(pn^[2k(en+1)] - 1) / (pn^(2k) - 1).

The ratio of these two is ten equal to
DivisorSigma[2k, m] / DivisorSigma[k, m] =
(p1^[k(e1+1)] + 1) / (p1^k + 1) *
(p2^[k(e2+1)] + 1) / (p2^k + 1) *
...
(pn^[k(en+1)] + 1) / (pn^k + 1).

Obviously, if all the exponents are even, each fraction of the product
is an integer (the numerator factors algebraically in such case). This
proves the "if" direction of your statement.

The smallest counterexample to the "only if" part seems to be m=20, k=1:
DivisorSigma[20, 2] = 1+4+16+25+100+400 = 546
DivisorSigma[20, 1] = 1+2+4+5+10+20 = 42
546 / 42 = 13

Peter

PS. Now I only have to hope that Yahoogroups do not mangle this mail
beyond comprehensibility :-)

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
• 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 2 of 7 , Aug 24, 2007
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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)

?