zeta(r)^zeta(r)
Leroy Quet
qq-quet at mindspring.com
Wed Mar 10 01:38:22 CET 2004
[my original post and reply to sci.math]
qqquet at mindspring.com (Leroy Quet) wrote in message
news:<b4be2fdf.0403081540.7d8e0625 at posting.google.com>...
> Let a(1) = 1;
>
> Let, for m >= 2,
>
> a(m) =
>
> (1/ln(m)) sum{p=primes} sum{k|m,k>=2} a(m/k) ln(k) H(c(p,k)),
>
> where H(n) = 1+1/2+1/3+...+1/n, the n_th harmonic number,
>
> and c(p,k) is a nonnegative integer where
> p^c(p,k) is the highest power of the prime p which divides k.
>
> (And, oh yeah, H(0) =0.)
>
>
> So, we have then:
>
> sum{m=1 to oo} a(m)/m^r =
>
> zeta(r)^zeta(r),
>
> unless I made a mistake.
> (My math was not rigorous.)
>
> But what I am wondering is,
> what is a closed-form (non-recursive definition) for
> a(m) ??
>
I might as well give the first few terms of this sequence:
a(m): 1, 1, 1, 2, 1, 3, 1, 7/2, 2,...
Is suspect that each term is a rational which depends only on the
exponents in the prime-factorization of that term's index, but I am not
absolutely certain.
If, for example, b(k) = a(p^k), p = prime, then:
sum{k=0 to oo} x^k b(k) =
(1/(1-x))^(1/(1-x)),
I believe.
And b(k)(k-1)! forms the sequence:
1, 2, 7, 35,...
(Is this sequence, or anything related to the a-sequence, in the EIS??)
thanks,
Leroy Quet
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