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

Author

Alex Ratushnyak, May 28 2013

Keywords

• nonn

Links

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

Crossrefs

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

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;
      r = sqrt(s*2);
      if (r*(r+1)==s*2) printf("%llu, ", r);
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

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 *)