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.

3, 23, 107, 337, 853, 1621, 2971, 4919, 7757, 11657, 16103, 22193, 29251, 37699, 48523, 61051, 75479, 91459, 110563, 131641, 155501, 183581, 214177, 248593, 286063, 325883, 369979, 419449, 473647, 534029, 600623, 667531, 739523, 816769, 900997, 988651, 1083613

Offset: 1

Author

Jonathan Sondow, May 11 2013

Keywords

• nonn

Comments

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.)

So sum_{n>0} 1/a(n) = 1/3 + 1/23 + 1/107 + ... = 0.39....

Examples

			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, ...).
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, ...).
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A075442, A225669.

Programs

Mathematica

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]]

Python

from sympy import prime
def A225671(n):
    xn, xd, k, p = 1, prime(n), n, prime(n)
    while xn < xd:
        k += 1
        po, p = p, prime(k)
        xn = xn*p + xd
        xd *= p
    return po # Chai Wah Wu, Apr 20 2015

Extensions

a(23)-a(37) from Chai Wah Wu, Apr 20 2015