A359323 a(n) is the first prime p such that the average of the n-th powers of n consecutive primes starting with p is prime.

2, 3, 1531, 19, 631, 37, 41, 13, 670231, 614333, 11699, 11, 2447, 3049, 223, 13, 8353, 2531, 2203, 241, 3209, 5023, 52631, 461, 26723, 3307

Offset

1,1

Comments

a(27) > 2*10^8. The 27 consecutive primes starting at a(27) must all be congruent (mod 3).

a(28) = 8677, a(29) = 33349, a(30) = 67103.

a(27) > 10^11. Up to 10^11, the longest runs of consecutive primes that are congruent (mod 3) are all of length 25: 21549657539, ..., 21549658079 (all == 2 (mod 3), 43553959717, ..., 43553960299 (all == 1 (mod 3)), and 55989913757, ..., 55989914177 (all == 2 (mod 3)). - Jon E. Schoenfield, Dec 26 2022

Links

Table of n, a(n) for n=1..26.

Example

a(3) = 1531 because 1531, 1543 and 1549 are consecutive primes and (1531^3 + 1543^3 + 1549^3)/3 = 3659642149 is prime, and 1531 is the least prime that works.

Maple

g:= proc(n) local P, s, i;
   P:= <seq(ithprime(i), i=1..n)>;
   s:= add(P[i]^n, i=1..n)/n;
   do
     if s::integer and isprime(s) then return P[1] fi;
     s:= s - P[1]^n/n;
     P[1..-2] := P[2..-1]; P[n]:= nextprime(P[n]);
     s:= s + P[n]^n/n;
   od
end proc:
map(g, [$1..26]);

Prog

(Python)
from sympy import prime, isprime, nextprime, primerange
def a(n):
    plst = list(primerange(2, prime(n)+1))
    powsum = sum(p**n for p in plst)
    while True:
        q, r = divmod(powsum, n)
        if r == 0 and isprime(q): return plst[0]
        powsum -= plst[0]**n
        plst = plst[1:] + [nextprime(plst[-1])]
        powsum += plst[-1]**n
print([a(n) for n in range(1, 27)]) # Michael S. Branicky, Dec 27 2022

Crossrefs

Cf. A359322.

Sequence in context: A228127 A327660 A036111 * A097549 A323518 A004909

Adjacent sequences: A359320 A359321 A359322 * A359324 A359325 A359326

Keyword

nonn,more

Author

Robert Israel, Dec 25 2022

Status

approved