A226154 -id:A226154 - cp's OEIS frontend

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

Alex Ratushnyak, May 28 2013

nonn

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

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

#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;
}

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

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

A226155 Smallest of four consecutive primes whose average is a triangular number.

Original entry on oeis.org

11, 47, 101, 109, 241, 587, 1217, 1481, 2069, 2203, 3313, 4357, 5443, 6779, 7351, 7489, 10723, 11927, 12239, 16267, 18911, 24517, 24733, 27953, 36571, 44839, 51347, 55249, 55931, 56941, 60017, 64951, 68239, 68993, 70117, 75041, 86719, 87133, 92227, 94819, 98773, 111611

Offset: 1

Alex Ratushnyak, May 28 2013

nonn

Cf. A102655, A226151 A226153, A226154.

#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;
      if (s%4==0) {
        s/=4;
        r = sqrt(s*2);
        if (r*(r+1)==s*2) printf("%llu, ", p3);
      }
      p3 = p2, p2 = p1, p1 = i;
    }
  return 0;
}

A000217inv:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return t1 ; else return -1; end if; end;
isA226155 := proc(n)
    local p1,p2,p3,a102655 ;
    if isprime(n) then
        p1 := nextprime(n) ;
        p2 := nextprime(p1) ;
        p3 := nextprime(p2) ;
        a102655 := (n+p1+p2+p3)/4 ;
        if type(a102655,'integer') then
            if A000217inv(a102655) >= 0 then
                return true;
            else
                return false;
            end if;
        else
            return false;
        end if;
    else
        false;
    end if;
end proc:
for n from 1 do
    p := ithprime(n) ;
    if isA226155(p) then
        printf("%d,\n",p) ;
    end if;
end do: # R. J. Mathar, Jun 06 2013

cp's OEIS Frontend

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

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