Sum of prime rereciprocals?

[David W. Wilson]

Why not just sum of reciprocals, e.g:

a(n) = least k with sum(j = n..k; 1/j) >= 1.

It's an interesting sequence, which starts

   1,4,7,10,12,15,...

If you compare it to [en] = A022843

   2,5,8,10,13,16,...

it appears that [en]-a(n) =

   1,1,1,0,1,1,1,1,1,1,0,...

consists of 0's and 1's.

The places where the zeroes occur are

    4, 11, 18, 25, 32, 36, 43, 50, 57, 64, 71, ...

whose differences always seem to be 4, 7 or 11.


> -----Original Message-----
> From: zak seidov [mailto:zakseidov at yahoo.com]
> Sent: Wednesday, April 09, 2008 5:35 AM
> To: seqfan at ext.jussieu.fr
> Subject: Sum of prime rereciprocals?
> 
> 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
> 
> 3,9,27,66,144,253,424,651,977,1392,1866,2479,3169,3981,4978,6137,7420,8
> 829,10477,12279,14295,16613,19124,21906,24904,28056,31494,35320,39486,4
> 4102
> %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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