cp's OEIS Frontend

A225671 Largest prime p(k) > p(n) such that 1/p(n) + 1/p(n+1) + ... + 1/p(k) < 1, where p(n) is the n-th prime.

%I A225671 #16 May 22 2025 10:21:35
%S A225671 3,23,107,337,853,1621,2971,4919,7757,11657,16103,22193,29251,37699,
%T A225671 48523,61051,75479,91459,110563,131641,155501,183581,214177,248593,
%U A225671 286063,325883,369979,419449,473647,534029,600623,667531,739523,816769,900997,988651,1083613
%N A225671 Largest prime p(k) > p(n) such that 1/p(n) + 1/p(n+1) + ... + 1/p(k) < 1, where p(n) is the n-th prime.
%C A225671 a(n+1) > n^e, by Rosser's theorem p(n) > n*log(n). (In fact, it appears that a(n) > (n*log(n))^e.)
%C A225671 So sum_{n>0} 1/a(n) = 1/3 + 1/23 + 1/107 + ... = 0.39....
%H A225671 Chai Wah Wu, <a href="/A225671/b225671.txt">Table of n, a(n) for n = 1..100</a>
%e A225671 a(1) = 3 because 1/2 + 1/3 < 1 < 1/2 + 1/3 + 1/5 (or because the slowest-growing sequence of primes whose reciprocals sum to 1 is A075442 = 2, 3, 7, ...).
%e A225671 a(2) = 23 because 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/23 < 1 < 1/3 + 1/5 + 1/7 + 1/11 + 1/13 + 1/17 + 1/19 + 1/23 + 1/29 (or because the slowest-growing sequence of odd primes whose reciprocals sum to 1 is A225669 = 3, 5, 7, 11, 13, 17, 19, 23, 967, ...).
%t A225671 L = {1}; n = 0; Do[ k = Last[L]; n++; While[ Sum[ 1/Prime[i], {i, n, k}] < 1, k++]; L = Append[L, k - 1], {22}]; Prime[ Rest[L]]
%o A225671 (Python)
%o A225671 from sympy import prime
%o A225671 def A225671(n):
%o A225671     xn, xd, k, p = 1, prime(n), n, prime(n)
%o A225671     while xn < xd:
%o A225671         k += 1
%o A225671         po, p = p, prime(k)
%o A225671         xn = xn*p + xd
%o A225671         xd *= p
%o A225671     return po # _Chai Wah Wu_, Apr 20 2015
%Y A225671 Cf. A075442, A225669.
%K A225671 nonn
%O A225671 1,1
%A A225671 _Jonathan Sondow_, May 11 2013
%E A225671 a(23)-a(37) from _Chai Wah Wu_, Apr 20 2015