A226149 Smallest of three consecutive primes whose average is a square.

2393, 25913, 47951, 123191, 131759, 219953, 330611, 356387, 450227, 769117, 826271, 870479, 1026143, 1500613, 1515347, 1697797, 1846861, 1907141, 2013541, 2217107, 2486873, 2732383, 3229189, 3294191, 3956101, 4338871, 4481677, 4739297, 5022067, 5239511, 5294591, 5774387

Offset

1,1

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Maple

f:= proc(k) local t, p, q, r, s;
   t:= k^2; q:= prevprime(t);
   p:= prevprime(q);
   r:= nextprime(q);
   s:= nextprime(r);
   if t = (p+q+r)/3 then p
   elif t = (q+r+s)/3 then q
   else NULL
   fi
end proc:
map(f, [seq(i, i=3..10000, 2)]); # Robert Israel, Feb 16 2026

Mathematica

Select[Partition[Prime[Range[400000]], 3, 1], IntegerQ[Sqrt[Mean[#]]]&][[All, 1]] (* Harvey P. Dale, Jan 10 2021 *)

Prog

(C)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define TOP (1ULL<<30)
int main() {
  unsigned long long i, j, p1, p2, r, s;
  unsigned char *c = (unsigned char *)malloc(TOP/8);
  memset(c, 0, TOP/8);
  for (i=3; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0 /*&& i<(1ULL<<32)*/)
        for (j=i*i>>1; j<TOP; j+=i)  c[j>>3] |= 1 << (j&7);
  for (p2=2, p1=3, i=5; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
      s = p2 + p1 + i;
      if ((s%3)==0) {
        s/=3;
        r = sqrt(s);
        if (r*r==s) printf("%llu, ", p2);
      }
      p2 = p1, p1 = i;
    }
  return 0;
}

Crossrefs

Cf. A076304, A206279, A226145, A226146, A226147, A226148, A226150.

Sequence in context: A100974 A279538 A098680 * A115931 A323344 A323341

Adjacent sequences: A226146 A226147 A226148 * A226150 A226151 A226152

Keyword

nonn

Author

Alex Ratushnyak, May 28 2013

Status

approved