A091361 Numbers n such that A001840(n) == 0 (mod n).

1, 2, 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165, 171, 177, 183, 189, 195, 201, 207, 213, 219, 225, 231, 237, 243, 249, 255, 261, 267, 273, 279, 285, 291, 297, 303, 309, 315, 321, 327

Offset: 1

Author

Jon Perry, Mar 01 2004

Keywords

• nonn

Comments

Apart from 1 and 2 it is conjectured that the only values present are congruent to 3 mod 6 (all these values are present).

From R. J. Mathar, Feb 25 2008: (Start)

Proof of the conjecture that this is 1 and 2 followed by A016945 follows by considering the 6 cases n=6k-1, 6k, 6k+1, 6k+2, 6k+3 or 6k+4, individual evaluation of A001840(n) with their corresponding 3 formulas quoted in A001840 in each case and searching for solutions of the form A001840(n) = t*n for integer t.

Example: A001840[6k+4]=A001840[3(2k+1)+1]=(2k+2)(6k+5)/2=t*(6k+4) implies t=k+7/6+1/[6(3k+2)] which cannot be solved in integers t and k. So numbers of the form 6k+4 are not members here. (End)

Examples

			A001840(9)=18, so 9 is in the sequence.
		

Links

• M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550, 2013

Crossrefs

Cf. A001840.

Programs

Mathematica

(* b = A001840 *) b[0] = 0; b[1] = 1; b[n_] := b[n] = n (n + 1)/2 - b[n - 1] - b[n - 2]; Reap[For[n = 1, n <= 400, n++, If[Mod[b[n], n] == 0, Sow[n]]]][[2, 1]] (* Jean-François Alcover, Feb 09 2019 *)

Formula

G.f.: conjecture: 2*(1+x)/(1-x)/G(0) +x, where G(k)= 1 + 1/(1 - x*(3*k+1)/(x*(3*k+4) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013