A226147 -id:A226147 - cp's OEIS frontend

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

Alex Ratushnyak, May 28 2013

nonn

			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

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

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

Cf. A167788 (the corresponding triangular numbers).

#include 
#include 
#include 
#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>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;
}

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

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}

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

Original entry on oeis.org

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

Alex Ratushnyak, May 28 2013

nonn

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

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

Cf. A167788 (the resulting triangular numbers).

#include 
#include 
#include 
#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>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;
}

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

cp's OEIS Frontend

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

C

Mathematica

PARI