A233414 Prime(n), where n is such that (1 + Sum_{i=1..n} prime(i)^15) / n is an integer.

2, 3, 7, 11, 13, 29, 37, 43, 79, 373, 2719, 3767, 4583, 6653, 34919, 83737, 95383, 493523, 741053, 1433689, 1629623, 2254757, 2686819, 2801221, 7283587, 12288799, 49986019, 120365039, 1280220301, 1388048693, 2336739481, 3390500677, 5139223693, 14729858701

Offset: 1

Author

Robert Price, Dec 09 2013

Keywords

• nonn

Comments

a(49) > 661876608760109. - Bruce Garner, Jun 02 2021

Examples

			a(3) = 7, because 7 is the 4th prime and the sum of the first 4 primes^15+1 = 4778093469744 when divided by 4 equals 1194523367436 which is an integer.
		

Links

• Bruce Garner, Table of n, a(n) for n = 1..48 (first 43 terms from Robert Price)
• OEIS Wiki, Sums of powers of primes divisibility sequences

Crossrefs

Cf. A085450 (smallest m > 1 such that m divides Sum_{k=1..m} prime(k)^n).

Cf. A007504, A045345, A171399, A128165, A233523, A050247, A050248.

Cf. A024450, A111441, A217599, A128166, A233862, A217600, A217601.

Programs

Mathematica

t = {}; sm = 1; Do[sm = sm + Prime[n]^15; If[Mod[sm, n] == 0, AppendTo[t, Prime[n]]], {n, 100000}]; t (* Derived from A217599 *)
nn=7000000;With[{pr15=Accumulate[Prime[Range[nn]]^15]+1}, Prime[ #]&/@ Select[ Range[nn],Divisible[pr15[[#]],#]&]] (* This program will generate the first 28 terms of the sequence.  To generate an additional 6 terms terms, nn would have to equal 659 million, and the program would take a long time to run. *) (* Harvey P. Dale, May 01 2014 *)

PARI

is(n)=if(!isprime(n),return(0)); my(t=primepi(n),s); forprime(p=2,n,s+=Mod(p,t)^15); s==0 \\ Charles R Greathouse IV, Nov 30 2013