Queneau-Daniel permutation and type-2 ONB

[N. J. A. Sloane]

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


__________________________________________________
Do You Yahoo!?
Tired of spam?  Yahoo! Mail has the best spam protection around 
http://mail.yahoo.com