A226145 Numbers n such that triangular(n) is a sum of three successive primes.

4, 5, 61, 82, 142, 166, 202, 233, 337, 394, 418, 422, 446, 493, 538, 661, 670, 841, 886, 1101, 1177, 1234, 1237, 1266, 1322, 1426, 1441, 1477, 1593, 1642, 1690, 1713, 1765, 1789, 1798, 1885, 1901, 1930, 1941, 2041, 2061, 2098, 2101, 2161, 2218, 2277, 2305, 2350, 2614

Offset

1,1

Links

David A. Corneth, Table of n, a(n) for n = 1..10433 (first 800 terms from Harvey P. Dale, terms <= 10^6)

Example

For k = 5, triangular(k) = triangular(5) = 15. 15/3 = 5. The next prime larger or equal to 5 is 5. The prime before 5 is 3. If there is a triple of consecutive primes that sum to 15 then 3 and 5 are two of them. Then the third one must be 15 - 3 - 5 = 7. 7 is prime and 3, 5 and 7 are consecutive primes (as 7 is the next larger prime than 5 or the previous prime to 3). Therefore, k = 5 is in the sequence. - David A. Corneth, Sep 18 2019

Mathematica

(Sqrt[8#+1]-1)/2&/@Select[Total/@Partition[Prime[Range[ 100000]], 3, 1], OddQ[ Sqrt[8#+1]]&] (* Harvey P. Dale, Sep 18 2019 *)

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, ", r);
      p2 = p1, p1 = i;
    }
  return 0;
}
(PARI) upto(n) = {my(res = List(), t = 10); for(i = 5, n, c = t/3; p = nextprime(ceil(c)); q = precprime(p - 1); r = t - p - q; if(isprime(r) && nextprime(r + 1) == q || nextprime(p + 1) == r, listput(res, i - 1)); t+=i); res}

Crossrefs

Cf. A076304, A206279, A226146, A226147, A226148, A226149, A226150.

Cf. A167788 (the corresponding triangular numbers).

Sequence in context: A368018 A341575 A013331 * A042425 A109348 A192084

Adjacent sequences: A226142 A226143 A226144 * A226146 A226147 A226148

Keyword

nonn

Author

Alex Ratushnyak, May 28 2013

Status

approved