Sum of prime rereciprocals?

[sellersj at math.psu.edu]

Dear Zak,

I just calculated the first 40 values of your sequence in Maple and
obtained the following:

3,9,27,66,144,253,424,651,977,1392,1866,2479,3169,3981,4978,6137,7420,
8829,10477,12279,14295,16613,19124,21906,24904,28056,31494,35320,39486,
44102,49116,54103,59468,65143,71315,77649,84504,91720,99303,107365

I note that the ratio of the values a(n+1)/a(n) dips below 1.12 and is
near 1.08 when one considers a(40)/a(39).

I wish you the very best.

James


> Dear seqfans,
>
> Just submitted:
>
> a(n) = least m such that sum of m reciprocal primes
> starting  with n-th prime is >1.
>
> Q is:
> What is the asymptotic ratio a(n+1)/a(n)?
> Numerically, a(n+1)/a(n)  => 1.12.
> More terms?
>
> Thanks, zak
>
>
> (n+1)/a(n)?
> %I A000001
> %S A000001
>
> 3,9,27,66,144,253,424,651,977,1392,1866,2479,3169,3981,4978,6137,7420,8829,10477,12279,14295,16613,19124,21906,24904,28056,31494,35320,39486,44102
> %N A000001 a(n) = least m such that sum of m
> reciprocal primes starting
>  with n-th prime is >1.
> %C A000001 What is asymptotic formula for a(n+1)/a(n)?
>
> Numerically, a(n+1)/a(n)  => 1.12.
> %F A000001 a(n)=m: sum(1/prime(i), i=n,n+m-1))>1,
> while
>  sum(1/prime(i), i=n,n+m-2))>1.
> %e A000001 a(1)=3 because 1/2+1/3+1/5=31/30 (3 terms),
> while  1/2+1/3<1,
> a(2)=9 because
> 1/3+1/5+1/7+1/11+1/13+1/17+1/19+1/23+1/29
>  =3343015913/3234846615 (9 terms), while
> 1/3+...+1/23<1,
> a(3)=27 because 1/p(3)+...1/p(29)>1 (27 terms) while
>  1/p(3)+...1/p(28)<1;
> in general
> a(n)=m because sum(1/p(i), i=n..n+m-1)>1 while
> sum(1/p(i), i=n..n+m-2)<1.
> %t A000001
>
> ss={};Do[s=1/Prime[n];k=1;While[s<1,k++;s+=1/Prime[n+k-1]];AppendTo[ss,k],{n,1,30}]
> %O A000001 1
> %K A000001 ,more,nonn,
> %A A000001 Zak Seidov (zakseidov at yahoo.com), Apr 09
> 2008
>
>
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