A108647 a(n) = (n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144.

1, 20, 150, 700, 2450, 7056, 17640, 39600, 81675, 157300, 286286, 496860, 828100, 1332800, 2080800, 3162816, 4694805, 6822900, 9728950, 13636700, 18818646, 25603600, 34385000, 45630000, 59889375, 77808276, 100137870, 127747900

Offset: 0

Author

Emeric Deutsch, Jun 13 2005

Keywords

• nonn

Comments

Kekulé numbers for certain benzenoids.

a(n-4), n>=4, is the number of ways to have n identical objects in m=4 of altogether n distinguishable boxes (n-4 boxes stay empty). - Wolfdieter Lang, Nov 13 2007

Examples

			a(2) = 150 because n=6 identical balls can be put into m=4 of n=6 distinguishable boxes in binomial(6,4)*(4!/(3!*1!)+ 4!/(2!*2!)) = 15*(4 + 6) = 150 ways. The m=4 part partitions of 6, namely (1^3,3) and (1^2,2^2) specify the filling of each of the 15 possible four box choices. - _Wolfdieter Lang_, Nov 13 2007
		

References

• S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, no. 23).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Crossrefs

Fourth column of triangle A103371.

Programs

Haskell

a108647 = flip a103371 3 . (+ 3)  -- Reinhard Zumkeller, Apr 04 2014

Magma

[4*Binomial(n+4, 4)^2/(n+4): n in [0..30]]; // G. C. Greubel, Oct 28 2022

Maple

a:=(n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144: seq(a(n),n=0..30);

Mathematica

Array[Binomial[# + 4, 4] Binomial[# + 3, 2] (# + 1)/3 &, 28, 0] (* or *)
CoefficientList[Series[(1 + 12 x + 18 x^2 + 4 x^3)/(1 - x)^8, {x, 0, 27}], x] (* Michael De Vlieger, Dec 17 2017 *)

MuPAD

4*binomial(n,4)^2/n $ n = 4..35; // Zerinvary Lajos, May 09 2008

SageMath

[4*binomial(n+4, 4)^2/(n+4) for n in (0..30)] # G. C. Greubel, Oct 28 2022

Formula

a(n) = C(n+4,4)*C(n+3,2)(n+1)/3. - Paul Barry, May 13 2006

G.f.: (1+12*x+18*x^2+4*x^3)/(1-x)^8.

a(n) = 4*C(n,4)^2/n, n >= 4. - Zerinvary Lajos, May 09 2008

From Amiram Eldar, May 29 2022: (Start)

Sum_{n>=0} 1/a(n) = 20*Pi^2 - 589/3.

Sum_{n>=0} (-1)^n/a(n) = 64*log(2) - 2*Pi^2 - 71/3. (End)

E.g.f.: (144 + 2736*x + 7992*x^2 + 7416*x^3 + 2826*x^4 + 486*x^5 + 37*x^6 + x^7)*exp(x)/144. - G. C. Greubel, Oct 28 2022