Biggest Prime-Factorization Exponent

[Leroy Quet]

If m, a positive integer, equals

product{p=primes} p^a(p,m),

where the a(p,m)'s are nonnegative integers,

then let
A(m) = the largest of the a(p,m)'s, given an m.

{A(m)} is EIS sequence A051903:
http://www.research.att.com/projects/OEIS?Anum=A051903


Now, let {b(k)} be any sequence where the sums below converge absolutely 
(just to be safe) for some range of r's.

Does, for those r's where the sum converges:

sum{j=2 to oo} b(A(j))/j^r

=

sum{k=1 to oo} sum{p=primes} 
    (b(k)/p^(kr))*
        product{q=primes<p} ((1 -q^(-r(k+1)))/(1 -1/q^r)) *
          product{q=primes>p} ((1 -q^(-rk))/(1 -1/q^r))

(which should also =)

zeta(r)*
  sum{k=1 to oo} sum{p=primes} 
    (b(k)/p^(kr))*(1 -1/p^r)
        product{q=primes<p} (1 -q^(-r(k+1))) *
          product{q=primes>p} (1 -q^(-rk))

???

(So, for example, for each b(k) = 1, we have both sides equal to
zeta(r) - 1.)


thanks,
Leroy Quet