A134287 Fifth column of triangle A103371 (without leading zeros).

1, 30, 315, 1960, 8820, 31752, 97020, 261360, 637065, 1431430, 3006003, 5962320, 11262160, 20391840, 35581680, 60093504, 98590905, 157608990, 246142435, 376372920, 564559380, 832117000, 1206913500, 1724814000, 2431508625

Offset: 0

Author

Wolfdieter Lang, Nov 13 2007

Keywords

• nonn
• easy

Comments

Kekulé numbers for certain benzenoids.

a(n) = K(L(n))*K(O(2,4,n)) with the Cyvin and Gutman Kekulé number notation. See p. 62 for the L(n) structure with K(L(n))=n+1 and p. 105 (i) for the O(k,m,n) structure and its Kekulé number. This corresponds to an essentially disconnected 7-tier benzenoid structure similar to the 6-tier structure shown on p. 230, nr. 23 (see A108647).

a(n-5), n >= 5, is the number of ways to put n identical objects into m=5 of altogether n distinguishable boxes (n-5 boxes stay empty).

Examples

			a(2)=315 because n=7 identical balls can be put into m=5 of n=7 distinguishable boxes in binomial(7,5)*(5!/(4!*1!)+ 5!/(3!*2!)) = 21*(5+10) = 315 ways. The m=5 part partitions of 7, namely (1^4,3) and (1^3,2^2) specify the filling of each of the 21 possible five 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.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Crossrefs

Cf. A108647 (fourth column of triangle A103371).

Programs

Haskell

a134287 = flip a103371 4 . (+ 4)  -- Reinhard Zumkeller, Apr 04 2014

Magma

[5*Binomial(n+5, 5)^2/(n+5): n in [0..30]]; // G. C. Greubel, Oct 28 2022

Maple

seq(binomial(n+4,4)^2*(n+5)/5, n=0..24); # Peter Luschny, Jan 13 2014

Mathematica

CoefficientList[Series[(1 + 20 x + 60 x^2 + 40 x^3 + 5 x^4)/(1 - x)^10, {x, 0, 24}], x]

MuPAD

5*binomial(n+5,5)^2/(n+5) $ n = 0..35; // Zerinvary Lajos, May 09 2008

PARI

a(n) = 5*binomial(n+5, 5)^2/(n+5); \\ Michel Marcus, Jan 07 2014

SageMath

[5*binomial(n+5,5)^2/(n+5) for n in range(31)] # G. C. Greubel, Oct 28 2022

Formula

a(n) = A103371(n+4,4), n >= 0.

a(n) = ((n+1)*(n+2)*(n+3)*(n+4))^2*(n+5)/2880, n >= 0. 2880 = 4!*5! = A010790(4).

G.f.: (1+20*x+60*x^2+40*x^3+5*x^4)/(1-x)^10. Numerator polynomial from fifth row of triangle A132813.

a(n) = 5*C(n+5,5)^2/(n+5), n >= 0. - Zerinvary Lajos, May 09 2008

a(n) = (C(n+6,6)*C(n+5,4)+5*C(n+5,6)*C(n+5,4))/(n+5). - Gary Detlefs, Jan 06 2014

From Amiram Eldar, May 31 2022: (Start)

Sum_{n>=0} 1/a(n) = 350*Pi^2/3 - 13805/12.

Sum_{n>=0} (-1)^n/a(n) = 5*Pi^2 + 640*log(2)/3 - 785/4. (End)

E.g.f.: (2880 + 83520*x + 368640*x^2 + 529920*x^3 + 330120*x^4 + 102024*x^5 + 16616*x^6 + 1432*x^7 + 61*x^8 + x^9)*exp(x)/2880. - G. C. Greubel, Oct 28 2022