A081142 12th binomial transform of (0,0,1,0,0,0,...).
0, 0, 1, 36, 864, 17280, 311040, 5225472, 83607552, 1289945088, 19349176320, 283787919360, 4086546038784, 57954652913664, 811365140791296, 11234286564802560, 154070215745863680, 2095354934143746048
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..400
- Index entries for linear recurrences with constant coefficients, signature (36,-432,1728).
Crossrefs
Programs
-
GAP
List([0..20],n->12^(n-2)*Binomial(n,2)); # Muniru A Asiru, Nov 24 2018
-
Magma
[12^(n-2)* Binomial(n, 2): n in [0..20]]; // Vincenzo Librandi, Oct 16 2011
-
Maple
seq(coeff(series(x^2/(1-12*x)^3,x,n+1), x, n), n = 0 .. 20); # Muniru A Asiru, Nov 24 2018
-
Mathematica
LinearRecurrence[{36,-432,1728},{0,0,1},30] (* or *) Table[(n-1) (n-2) 3^(n-3) 2^(2n-7),{n,20}] (* Harvey P. Dale, Jul 25 2013 *)
-
PARI
vector(20, n, n--; 2^(2*n-5)*3^(n-2)*n*(n-1)) \\ G. C. Greubel, Nov 23 2018
-
Sage
[2^(2*n-5)*3^(n-2)*n*(n-1) for n in range(20)] # G. C. Greubel, Nov 23 2018
Formula
a(n) = 36*a(n-1) - 432*a(n-2) + 1728*a(n-3), a(0) = a(1) = 0, a(2) = 1.
a(n) = 12^(n-2)*binomial(n, 2).
G.f.: x^2/(1 - 12*x)^3.
a(n) = 2^(2*n-5)*3^(n-2)*n*(n-1). - Harvey P. Dale, Jul 25 2013
E.g.f.: (1/2)*exp(12*x)*x^2. - Franck Maminirina Ramaharo, Nov 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 24 - 264*log(12/11).
Sum_{n>=2} (-1)^n/a(n) = 312*log(13/12) - 24. (End)
Comments