Some Sums

[Franklin T. Adams-Watters]

I have made a little more progress on these.

Since the reciprocal of a multiplicative function is also multiplicative, these sums can be factored, just like the Riemann zeta function.  Thus we have product_p sum_n 1/f(p^n).

When this is done, and the product is taken for primes through N, the remaining terms have a product of
(1+1/(N*log(N))-1/(N*log(N)^2)+O(1/N*log(N)^3),
which means that applying a factor of
(1+1/(N*log(N)))
will produce an upper bound on the answer (for large N).

Running PARI with a prime limit of 1e9, I get:

I wrote:
>I'm trying to evaluate the following sums.  Can anybody produce more accurate results?
>
>Number 1: sum_n=1^inf 1/sigma(n)^2
>I get 1.306456[50]
>(the digits in brackets I am not sure of).

1.30645651197... => 1.30645651204...
which makes me fairly sure that 1.30645651204 is correct.

>Number 2: sum n=1^inf 1/sigma_2(n)
>I get 1.537812[88]

1.53781289175... => 1.53781289183
for 1.5378128918

>Number 3: sum n=1^inf 1/phi(n)^2
>In this case, I'm not entirely sure of any digits, but I think it's
>3.39064.

In this case, the sum inside the product has a closed form:
1.+p^2/((p-1)^2*(p^2-1))
3.39064200541... => 3.39064200557...
so that I'm fairly sure of 3.3906420055

>For the first two, when you sum to N, the error is bounded by 1/N.
>
>None of these is in the OEIS.

I'm going to submit these at this point, but I would still like more accuracy.
-- 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645


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