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%I #33 Oct 29 2022 04:54:47
%S 1,20,150,700,2450,7056,17640,39600,81675,157300,286286,496860,828100,
%T 1332800,2080800,3162816,4694805,6822900,9728950,13636700,18818646,
%U 25603600,34385000,45630000,59889375,77808276,100137870,127747900
%N a(n) = (n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144.
%C Kekulé numbers for certain benzenoids.
%C 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
%D 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).
%H Reinhard Zumkeller, <a href="/A108647/b108647.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8,-28,56,-70,56,-28,8,-1).
%F a(n) = C(n+4,4)*C(n+3,2)(n+1)/3. - _Paul Barry_, May 13 2006
%F G.f.: (1+12*x+18*x^2+4*x^3)/(1-x)^8.
%F a(n) = 4*C(n,4)^2/n, n >= 4. - _Zerinvary Lajos_, May 09 2008
%F From _Amiram Eldar_, May 29 2022: (Start)
%F Sum_{n>=0} 1/a(n) = 20*Pi^2 - 589/3.
%F Sum_{n>=0} (-1)^n/a(n) = 64*log(2) - 2*Pi^2 - 71/3. (End)
%F 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
%e 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
%p a:=(n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144: seq(a(n),n=0..30);
%t Array[Binomial[# + 4, 4] Binomial[# + 3, 2] (# + 1)/3 &, 28, 0] (* or *)
%t CoefficientList[Series[(1 + 12 x + 18 x^2 + 4 x^3)/(1 - x)^8, {x, 0, 27}], x] (* _Michael De Vlieger_, Dec 17 2017 *)
%o (MuPAD) 4*binomial(n,4)^2/n $ n = 4..35; // _Zerinvary Lajos_, May 09 2008
%o (Haskell)
%o a108647 = flip a103371 3 . (+ 3) -- _Reinhard Zumkeller_, Apr 04 2014
%o (Magma) [4*Binomial(n+4, 4)^2/(n+4): n in [0..30]]; // _G. C. Greubel_, Oct 28 2022
%o (SageMath) [4*binomial(n+4, 4)^2/(n+4) for n in (0..30)] # _G. C. Greubel_, Oct 28 2022
%Y Fourth column of triangle A103371.
%K nonn
%O 0,2
%A _Emeric Deutsch_, Jun 13 2005
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