A205007 a(n) = (1/n)*A205006(n), where A205006(n) = s(k)-s(j), with (s(k),s(j)) the least pair of distinct triangular numbers for which n divides their difference.

2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Clark Kimberling, Jan 21 2012

Keywords

• nonn

Comments

For a guide to related sequences, see A204892.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..23515

Crossrefs

Cf. A000217, A205002, A205006, A204892.

Cf. A318894 (gives the positions terms larger than one).

Programs

Mathematica

(See the program at A205002.)

PARI

A205007(n) = for(k=2,oo,my(sk=binomial(k+1,2)); for(j=1,k-1,if(!((sk-binomial(j+1,2))%n),return((sk-binomial(j+1,2))/n)))); \\ Antti Karttunen, Sep 27 2018

Extensions

More terms from Antti Karttunen, Sep 27 2018