|
| |
| |
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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 *)
|
|
|
PROG
|
(PARI) t(i)=i*(i+1)/2 w=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 A168057 A290131
Adjacent sequences: A086599 A086600 A086601 * A086603 A086604 A086605
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
Jon Perry, Jul 23 2003
|
|
|
STATUS
|
approved
|
| |
|
|
