A226148 Smallest of three consecutive primes whose sum is a triangular number.

2, 3, 619, 1123, 3373, 4603, 6829, 9067, 18973, 25933, 29179, 29741, 33211, 40583, 48313, 72923, 74923, 117991, 130973, 202201, 231067, 253993, 255217, 267317, 291491, 339139, 346309, 363829, 423191, 449621, 476279, 489337, 519487, 533713, 539093, 592507, 602603, 621133

Offset

1,1

Links

Robert Israel, Table of n, a(n) for n = 1..2300

Maple

R:= 2: count:= 1:
for k from 1 while count < 100 do
 for j from 1 to 2 do
  m:= 4*k+j;
  x:= m*(m+1)/2;
  q= prevprime(ceil(x/3));
  p:= prevprime(q); r:= nextprime(q);
  t:= p+q+r;
  if t < x then while t < x do p:= q; q:= r; r:= nextprime(r); t:=p+q+r od
  elif t > x then while t > x do r:= q; q:= p; p:= prevprime(p); t:= p+q+r od
  fi;
  if t = x then  R:= R, p; count:= count+1; fi
od od :
R; # Robert Israel, Oct 18 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;
      r = sqrt(s*2);
      if (r*(r+1)==s*2) printf("%llu, ", p2);
      p2 = p1, p1 = i;
    }
  return 0;
}

Crossrefs

Cf. A076304, A206279, A226145, A226146, A226147, A226149, A226150.

Cf. A167788 (the resulting triangular numbers).

Sequence in context: A240709 A264577 A165770 * A108332 A256115 A066685

Adjacent sequences: A226145 A226146 A226147 * A226149 A226150 A226151

Keyword

nonn

Author

Alex Ratushnyak, May 28 2013

Status

approved