A086602 a(n) = A000217(A000217(n))-n^2.

0, 0, 2, 12, 39, 95, 195, 357, 602, 954, 1440, 2090, 2937, 4017, 5369, 7035, 9060, 11492, 14382, 17784, 21755, 26355, 31647, 37697, 44574, 52350, 61100, 70902, 81837, 93989, 107445, 122295, 138632, 156552, 176154, 197540, 220815, 246087

Offset

0,3

Links

Table of n, a(n) for n=0..37.

Q. T. Bach, R. Paudyal, J. B. Remmel, A Fibonacci analogue of Stirling numbers, arXiv preprint arXiv:1510.04310 [math.CO], 2015-2016.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = A000330(n-1)+A001295(n-1). - Alford Arnold, Jun 29 2005

a(n) = 3*C(n+2,4) - C(n,2). - Zerinvary Lajos, May 02 2007, corrected Jun 12 2018

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) = n*(n-1)*(n^2+3*n-2)/8. [R. J. Mathar, Oct 30 2009]

G.f.: x^2*(-2-2*x+x^2)/(x-1)^5. [R. J. Mathar, Oct 30 2009]

a(n) = (n-1)*A005581(n) - Sum_{i=0..n-1} A005581(i). [Bruno Berselli, Aug 27 2014]

Example

a(3) = t(t(3))-3^2 = t(6)-9 = 21-9 = 12.

Maple

seq(3*binomial(n+2, 4)-binomial(n, 2), n=0..35); # Zerinvary Lajos, May 02 2007

Mathematica

Table[n (n - 1) (n^2 + 3 n - 2)/8, {n, 0, 40}] (* Bruno Berselli, Aug 27 2014 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 0, 2, 12, 39}, 60] (* Harvey P. Dale, Apr 04 2023 *)

Prog

(PARI) t(i)=i*(i+1)/2
vector(40, i, t(t(i))-i^2)
(Magma) [n*(n-1)*(n^2+3*n-2)/8: n in [0..40]]; // Vincenzo Librandi, Jun 26 2016

Crossrefs

Cf. A000330, A001296, A005581.

Sequence in context: A048349 A009632 A190022 * A019006 A363123 A168057

Adjacent sequences: A086599 A086600 A086601 * A086603 A086604 A086605

Keyword

nonn,easy

Author

Jon Perry, Jul 23 2003

Status

approved