A226154 Smallest of four consecutive primes whose sum is a triangular number.

5, 23, 191, 389, 449, 2593, 3011, 5167, 5639, 5851, 8669, 18839, 25463, 26953, 28843, 38561, 47963, 51907, 57859, 65419, 67789, 77711, 78301, 79889, 96013, 102023, 128119, 139501, 163417, 192037, 206233, 234083, 273601, 299329, 324593, 354677, 359323, 362723, 417451

Offset

1,1

Links

Table of n, a(n) for n=1..39.

Maple

A000217inv:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return t1 ; else return -1; end if; end;
isA226154 := proc(n)
    local p1, p2, p3, a034963 ;
    if isprime(n) then
        p1 := nextprime(n) ;
        p2 := nextprime(p1) ;
        p3 := nextprime(p2) ;
        a034963 := n+p1+p2+p3 ;
        if A000217inv(a034963) >= 0 then
            return true;
        else
            return false;
        end if;
    else
        false;
    end if;
end proc:
for n from 1 do
    p := ithprime(n) ;
    if isA226154(p) then
        printf("%d, \n", p) ;
    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;
      r = sqrt(s*2);
      if (r*(r+1)==s*2) printf("%llu, ", p3);
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

Crossrefs

Cf. A034963, A226151.

Sequence in context: A197965 A198041 A197922 * A203223 A173041 A156187

Adjacent sequences: A226151 A226152 A226153 * A226155 A226156 A226157

Keyword

nonn

Author

Alex Ratushnyak, May 28 2013

Status

approved