A120289 Primes p such that p divides the numerator of Sum_{k=1..n-1} 1/prime(k)^p, where p = prime(n).

5, 19, 47, 79, 109

Offset: 1

Author

Alexander Adamchuk, Jul 08 2006

Keywords

• nonn

Comments

Next term > 1690. - Michael S. Branicky, Jun 27 2022

Examples

			a(1) = 5 because prime 5 divides 275 = numerator(1/2^5 + 1/3^5).
Sum_{k=1..n-1} 1/prime(k)^prime(n) begins:
  n=2: 1/2^3 = 1/8;
  n=3: 1/2^5 + 1/3^5 = 275/7776;
  n=4: 1/2^7 + 1/3^7 + 1/5^7 = 181139311/21870000000;
  n=5: 1/2^11 + 1/3^11 + 1/5^11 + 1/7^11 = 17301861338484245234233/35027750054222100000000000.
		

Crossrefs

Cf. A119722.

Programs

Python

from fractions import Fraction
from sympy import isprime, primerange
def ok(p):
    if p < 3 or not isprime(p): return False
    s = sum(Fraction(1, pk**p) for pk in primerange(2, p))
    return s.numerator%p == 0
print([k for k in range(200) if ok(k)]) # Michael S. Branicky, Jun 26 2022