A226152 Numbers n such that n^2 is an average of 4 consecutive primes.

3, 9, 12, 21, 24, 35, 72, 126, 129, 179, 189, 194, 198, 214, 243, 253, 255, 279, 304, 322, 432, 443, 480, 487, 511, 523, 663, 681, 696, 699, 711, 717, 721, 734, 738, 796, 802, 838, 910, 975, 1008, 1034, 1070, 1144, 1215, 1230, 1237, 1265, 1276, 1370, 1375, 1386, 1469

Offset

1,1

Comments

Integers of the form sqrt(A102655(k)) for any k. - R. J. Mathar, Jun 06 2013

Links

Table of n, a(n) for n=1..53.

Formula

a(n) = A051395(n)/2.

Maple

A034963 := proc(n)
    add(ithprime(i), i=n..n+3) ;
end proc:
for n from 1 to 90000 do
    s := A034963(n)/4 ;
    if type(s, 'integer') then
    if issqr(s) then
        printf("%d, ", sqrt(s)) ;
    end if;
    end if;
end do: # R. J. Mathar, Jun 06 2013

Prog

(C)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define TOP (1ULL<<30)
int main() {
  unsigned long long i, j, p1, p2, p3, 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 (p3=2, p2=3, p1=5, i=7; i < TOP; i+=2)
    if ((c[i>>4] & (1<<((i>>1) & 7)))==0) {
      s = p3 + p2 + p1 + i;
      if (s%4==0) {
        s/=4;
        r = sqrt(s);
        if (r*r==s) printf("%llu, ", r);
      }
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

Crossrefs

Cf. A102655, A051395, A206280.

Sequence in context: A272027 A310323 A212059 * A155504 A270672 A211217

Adjacent sequences: A226149 A226150 A226151 * A226153 A226154 A226155

Keyword

nonn

Author

Alex Ratushnyak, May 28 2013

Status

approved