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A225502
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Least m > 0 such that prime(n)*triangular(m) is a triangular number, or 0 if no such m exists.
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3
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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
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OFFSET
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1,1
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COMMENTS
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Conjecture: a(n) > 0.
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LINKS
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Table of n, a(n) for n=1..65.
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EXAMPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(C)
#include <stdio.h>
#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;
}
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CROSSREFS
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Cf. A000217, A112456, A225503.
Sequence in context: A161281 A226916 A003113 * A152227 A246583 A244788
Adjacent sequences: A225499 A225500 A225501 * A225503 A225504 A225505
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KEYWORD
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nonn
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AUTHOR
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Alex Ratushnyak, May 09 2013
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STATUS
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approved
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