A091361 - OEIS

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

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OFFSET

1,2

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)

LINKS

Table of n, a(n) for n=1..57.

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

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

EXAMPLE

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

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 *)

CROSSREFS

Cf. A001840.

Sequence in context: A109663 A056702 A294126 * A092352 A061933 A251239

Adjacent sequences: A091358 A091359 A091360 * A091362 A091363 A091364

KEYWORD

nonn

AUTHOR

Jon Perry, Mar 01 2004

STATUS

approved