A038845 3-fold convolution of A000302 (powers of 4).

1, 12, 96, 640, 3840, 21504, 114688, 589824, 2949120, 14417920, 69206016, 327155712, 1526726656, 7046430720, 32212254720, 146028888064, 657129996288, 2937757630464, 13056700579840, 57724360458240, 253987186016256, 1112705767309312, 4855443348258816

Offset: 0

Author

Wolfdieter Lang

Keywords

• easy
• nonn

Comments

Also convolution of A002802 with A000984 (central binomial coefficients).

With a different offset, number of n-permutations of 5 objects u, v, w, z, x with repetition allowed, containing exactly two u's. - Zerinvary Lajos, Dec 29 2007

Also convolution of A000302 with A002697, also convolution of A002457 with itself. - Rui Duarte, Oct 08 2011

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..400
• Adam Ehrenberg, Joseph T. Iosue, Abhinav Deshpande, Dominik Hangleiter, and Alexey V. Gorshkov, The Second Moment of Hafnians in Gaussian Boson Sampling, arXiv:2403.13878 [quant-ph], 2024. See p. 30.
• Index entries for linear recurrences with constant coefficients, signature (12,-48,64).

Crossrefs

Cf. A000302, A000984, A002457, A002697, A002802, A038231, A052780.

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), this sequence (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).

Programs

GAP

List([0..30], n-> 4^n*Binomial(n+2,n) ); # G. C. Greubel, Jul 20 2019

Magma

[4^n*Binomial(n+2, 2): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011

Maple

seq((n+2)*(n+1)*4^n/2, n=0..30); # Zerinvary Lajos, Apr 25 2007

Mathematica

Table[4^n*Binomial[n+2,n], {n,0,30}] (* G. C. Greubel, Jul 20 2019 *)

PARI

a(n)=(n+2)*(n+1)<<(2*n-1) \\ Charles R Greathouse IV, Aug 21 2015

Sage

[4^(n-2)*binomial(n,2) for n in range(2, 30)] # Zerinvary Lajos, Mar 11 2009

Formula

a(n) = (n+2)*(n+1)*2^(2*n-1).

G.f.: 1/(1-4*x)^3.

a(n) = Sum_{u+v+w+x+y+z=n} f(u)*f(v)*f(w)*f(x)*f(y)*f(z) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004

a(n) = binomial(n+2,n) * 4^n. - Rui Duarte, Oct 08 2011

E.g.f.: (1 + 8*x + 8*x^2)*exp(4*x). - G. C. Greubel, Jul 20 2019

From Amiram Eldar, Jan 05 2022: (Start)

Sum_{n>=0} 1/a(n) = 8 - 24*log(4/3).

Sum_{n>=0} (-1)^n/a(n) = 40*log(5/4) - 8. (End)