OFFSET
1,1
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....
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..100
EXAMPLE
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, ...).
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]]
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, May 11 2013
EXTENSIONS
a(23)-a(37) from Chai Wah Wu, Apr 20 2015
STATUS
approved