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

Author

Alex Ratushnyak, May 28 2013

Keywords

• nonn

Comments

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

Crossrefs

Cf. A102655, A051395, A206280.

Programs

C

#include 
#include 
#include 
#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>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;
}

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

Formula

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