A226151 Numbers n such that triangular(n) is a sum of 4 consecutive primes.

8, 15, 39, 56, 60, 144, 155, 203, 212, 216, 263, 388, 451, 464, 480, 555, 619, 644, 680, 723, 736, 788, 791, 799, 876, 903, 1012, 1056, 1143, 1239, 1284, 1368, 1479, 1547, 1611, 1684, 1695, 1703, 1827, 1859, 1908, 1939, 2100, 2108, 2135, 2148, 2152, 2187, 2199, 2216

Offset

1,1

Links

Harvey P. Dale, Table of n, a(n) for n = 1..500

Maple

istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return t1 ; else return -1; end if; end;
A034963 := proc(n)
    add(ithprime(i), i=n..n+3) ;
end proc:
for n from 1 to 90000 do
    ist := istriangular(A034963(n)) ;
    if ist >= 0 then
        printf("%d, ", ist) ;
    end if;
end do: # R. J. Mathar, Jun 04 2013

Mathematica

(Sqrt[8#+1]-1)/2&/@Select[Total/@Partition[Prime[Range[ 60000]], 4, 1], OddQ[ Sqrt[8#+1]]&] (* Harvey P. Dale, Apr 06 2016 *)

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, ", r);
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

Crossrefs

Cf. A000217, A034963, A051395, A206280, A226154.

Sequence in context: A343141 A197602 A306599 * A253767 A254541 A137658

Adjacent sequences: A226148 A226149 A226150 * A226152 A226153 A226154

Keyword

nonn

Author

Alex Ratushnyak, May 28 2013

Status

approved