zeta(r)^zeta(r)
Leroy Quet
qq-quet at mindspring.com
Wed Mar 10 02:38:02 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,...
I gave the b-sequence from k = 1, of course.
And b(k)*k! is:
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?A
num=A087761
>
>(Is this sequence, or anything related to the a-sequence, in the EIS??)
>
thanks,
Leroy Quet
More information about the SeqFan
mailing list