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A233863
Prime(n), where n is such that (1 + Sum_{i=1..n} prime(i)^3) / n is an integer.
1
2, 3, 7, 11, 13, 29, 37, 43, 257, 421, 449, 7333, 7673, 9433, 9539, 12163, 53551, 74759, 119429, 199909, 295703, 2494781, 6941633, 39150679, 50026091, 165204709, 410054731, 724768817, 1282680871, 1777452847, 2923304383, 6053209493, 7423469173, 35896955599, 46936773853
OFFSET
1,1
COMMENTS
a(50) > 730228645826551. - Bruce Garner, Apr 04 2021
a(55) > 7824556240506443. - Bruce Garner, Mar 28 2022
EXAMPLE
a(5) = 13, because 13 is the 6th prime and the sum of the first 6 primes^3+1 = 4032 when divided by 6 equals 672 which is an integer.
MATHEMATICA
t = {}; sm = 1; Do[sm = sm + Prime[n]^3; If[Mod[sm, n] == 0, AppendTo[t, Prime[n]]], {n, 100000}]; t (* Derived from A217599 *)
Module[{nn=7500, pt}, pt=1+Accumulate[Prime[Range[nn]]^3]; Prime[#]&/@ Select[ Thread[{pt, Range[nn]}], Divisible[#[[1]], #[[2]]]&]][[All, 2]] (* The program generates the first 18 terms of the sequence. It is not suitable for generating many more. *) (* Harvey P. Dale, Mar 17 2022 *)
PROG
(PARI) is(n)=if(!isprime(n), return(0)); my(t=primepi(n), s); forprime(p=2, n, s+=Mod(p, t)^3); s==0 \\ Charles R Greathouse IV, Nov 30 2013
CROSSREFS
Cf. A085450 = smallest m > 1 such that m divides Sum_{k=1..m} prime(k)^n.
Sequence in context: A117112 A145032 A233414 * A371065 A233194 A233040
KEYWORD
nonn
AUTHOR
Robert Price, Dec 16 2013
STATUS
approved