8 Nov
2006
8 Nov
'06
8:38 p.m.
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