A140973 Numbers k such that arithmetic mean of squares of the first k tribonacci numbers is an integer.

1, 2, 8, 15, 16, 18, 22, 32, 47, 48, 53, 58, 64, 70, 77, 78, 80, 94, 95, 96, 103, 106, 128, 138, 163, 199, 206, 256, 257, 266, 269, 311, 326, 330, 352, 358, 385, 397, 398, 401, 419, 421, 499, 512, 514, 538, 587, 599, 617, 622, 640, 672, 683, 757, 768, 770, 773

Offset: 1

Author

Ctibor O. Zizka, Jul 27 2008

Keywords

• easy
• nonn

Comments

Can the arithmetic mean of the tribonacci numbers (T(0)+...+T(k-1)) / k be an integer?

The arithmetic means are integers for the first 1, 2, 47, 53, 94, 103, 106, ... (A141579) tribonacci numbers. - R. J. Mathar, Aug 04 2008

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000073, A141579.

Cf. A107239. - R. J. Mathar, Aug 04 2008

Programs

Mathematica

With[{m = 1000}, Position[Accumulate[LinearRecurrence[{1, 1, 1}, {0, 0, 1}, m]^2] / Range[m], ?IntegerQ] // Flatten] (* _Amiram Eldar, Jul 04 2025 *)

Formula

Numbers k such that (T(0)^2+ T(1)^2+ ... + T(k-1)^2) / k is an integer, where T(i) = i-th tribonacci number.

Extensions

a(1)-a(2) inserted and a(32) onwards added by R. J. Mathar, Aug 04 2008