|
%I #12 Jun 28 2022 15:28:01
%S 5,19,47,79,109
%N Primes p such that p divides the numerator of Sum_{k=1..n-1} 1/prime(k)^p, where p = prime(n).
%C Next term > 1690. - _Michael S. Branicky_, Jun 27 2022
%e a(1) = 5 because prime 5 divides 275 = numerator(1/2^5 + 1/3^5).
%e Sum_{k=1..n-1} 1/prime(k)^prime(n) begins:
%e n=2: 1/2^3 = 1/8;
%e n=3: 1/2^5 + 1/3^5 = 275/7776;
%e n=4: 1/2^7 + 1/3^7 + 1/5^7 = 181139311/21870000000;
%e n=5: 1/2^11 + 1/3^11 + 1/5^11 + 1/7^11 = 17301861338484245234233/35027750054222100000000000.
%o (Python)
%o from fractions import Fraction
%o from sympy import isprime, primerange
%o def ok(p):
%o if p < 3 or not isprime(p): return False
%o s = sum(Fraction(1, pk**p) for pk in primerange(2, p))
%o return s.numerator%p == 0
%o print([k for k in range(200) if ok(k)]) # _Michael S. Branicky_, Jun 26 2022
%Y Cf. A119722.
%K nonn
%O 1,1
%A _Alexander Adamchuk_, Jul 08 2006
| 