A225502 Least m > 0 such that prime(n)*triangular(m) is a triangular number, or 0 if no such m exists.

2, 1, 2, 2, 3, 3, 12, 4, 9, 5, 5, 30, 6, 6, 20, 14, 230, 23, 24, 8, 8, 35, 36, 9, 29, 90, 30, 434, 10, 159, 22, 11, 140, 530, 854, 147, 12, 25, 77, 39, 1938509, 13, 41, 69, 182, 70, 14, 104, 105, 60, 30, 15, 15, 47, 240, 65274, 6314, 16, 17009, 33, 50, 68, 17, 264, 371

Offset: 1

Author

Alex Ratushnyak, May 09 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0.

Examples

			n    prime(n)    m     tri(m)   prime(n)*tri(m)
1      2         2       3              6
2      3         1       1              3
3      5         2       3             15
4      7         2       3             21
5     11         3       6             66
6     13         3       6             78
7     17        12      78           1326
8     19         4      10            190
		

Crossrefs

Cf. A000217, A112456, A225503.

Programs

C

#include 
#define TOP 300
typedef unsigned long long U64;
U64 isTriangular(U64 a) {
    U64 sr = 1ULL<<32, s, b, t;
    if (a < (sr/2)*(sr+1))  sr>>=1;
    while (a < sr*(sr+1)/2)  sr>>=1;
    for (b = sr>>1; b; b>>=1) {
        s = sr+b;
        if (s&1) t = s*((s+1)/2);
        else     t = (s/2)*(s+1);
        if (t >= s && a >= t)  sr = s;
    }
    return (sr*(sr+1)/2 == a);
}
int main() {
  U64 i, j, k, m, tm, p, pp = 1, primes[TOP];
  for (primes[0]=2, i = 3; pp < TOP; i+=2) {
    for (p = 1; p < pp; ++p) if (i%primes[p]==0) break;
    if (p==pp) {
        primes[pp++] = i;
        for (j=p=primes[pp-2], m=tm=1; ; j=k, m++, tm+=m) {
           if ((k = p*tm) < j) { m=0; break; }
           if (isTriangular(k)) break;
        }
        printf("%llu, ", m);
    }
  }
  return 0;
}

Mathematica

lm[n_]:=Module[{m=1,p=Prime[n]},While[!OddQ[Sqrt[8(p (m(m+1))/2)+1]], m++];m]; Array[lm,68] (* Harvey P. Dale, Mar 16 2018 *)