njas:
%N A016088 a(n) = smallest prime p such that Sum_{ primes q = 2, ..., p} 1/q exceeds n. ... %E A016088 Eric Bach comments that the next element in the sequence is about 4.2 * 10^49, so it may remain unknown for all eternity.
Bosh! The paper describes an algorithm which takes around x^1/3 space and x^2/3 time. When x = 4 * 10^49, these are roughly 10^16 and 10^33. We have facilities today that handle 10^16 storage. Current world computing capability is ~10^25 instructions/year. (10^8.5 machines, 10^9.5 inst/sec, 10^7.5 sec/year) I looked over their algorithm; at first blush, it seems very parallelizable. So with current resources, we could have the value for a(5) in a mere 10^8 years. This might seem like a long time, especially if your job is behind this one in the queue, but it's far short of eternity. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of N. J. A. Sloane Sent: Wed 11/8/2006 8:38 PM To: math-fun@mailman.xmission.com; math-fun Cc: njas@research.att.com Subject: Re: [math-fun] First time harmonic series exceeds 100 Victor said:
... a related but much harder problem -- the first time the sum of the reciprocals of the sums of the primes exceed n -- harder just because the latter series grows like log log n.
Me: once again, see the OEIS: %S A016088 5,277,5195977,1801241230056600523 %N A016088 a(n) = smallest prime p such that Sum_{ primes q = 2, ..., p} 1/q exceeds n. %C A016088 The indices of these primes are in A046024: a(n) = A000040(A046024(n)). ... %E A016088 Eric Bach comments that the next element in the sequence is about 4.2 * 10^49, so i t may remain unknown for all eternity. and %S A046024 3,59,361139,43922730588128390 %N A046024 a(n) = smallest k such that Sum_{ i = 1..k } 1/prime(i) exceeds n. Neil