cp's OEIS Frontend

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

%I A134287 #40 Oct 29 2022 04:54:53
%S A134287 1,30,315,1960,8820,31752,97020,261360,637065,1431430,3006003,5962320,
%T A134287 11262160,20391840,35581680,60093504,98590905,157608990,246142435,
%U A134287 376372920,564559380,832117000,1206913500,1724814000,2431508625
%N A134287 Fifth column of triangle A103371 (without leading zeros).
%C A134287 Kekulé numbers for certain benzenoids.
%C A134287 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).
%C A134287 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).
%D A134287 S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.
%H A134287 Reinhard Zumkeller, <a href="/A134287/b134287.txt">Table of n, a(n) for n = 0..1000</a>
%H A134287 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,-45,120,-210,252,-210,120,-45,10,-1).
%F A134287 a(n) = A103371(n+4,4), n >= 0.
%F A134287 a(n) = ((n+1)*(n+2)*(n+3)*(n+4))^2*(n+5)/2880, n >= 0. 2880 = 4!*5! = A010790(4).
%F A134287 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.
%F A134287 a(n) = 5*C(n+5,5)^2/(n+5), n >= 0. - _Zerinvary Lajos_, May 09 2008
%F A134287 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
%F A134287 From _Amiram Eldar_, May 31 2022: (Start)
%F A134287 Sum_{n>=0} 1/a(n) = 350*Pi^2/3 - 13805/12.
%F A134287 Sum_{n>=0} (-1)^n/a(n) = 5*Pi^2 + 640*log(2)/3 - 785/4. (End)
%F A134287 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
%e A134287 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
%p A134287 seq(binomial(n+4,4)^2*(n+5)/5, n=0..24); # _Peter Luschny_, Jan 13 2014
%t A134287 CoefficientList[Series[(1 + 20 x + 60 x^2 + 40 x^3 + 5 x^4)/(1 - x)^10, {x, 0, 24}], x]
%o A134287 (MuPAD) 5*binomial(n+5,5)^2/(n+5) $ n = 0..35; // _Zerinvary Lajos_, May 09 2008
%o A134287 (PARI) a(n) = 5*binomial(n+5, 5)^2/(n+5); \\ _Michel Marcus_, Jan 07 2014
%o A134287 (Haskell)
%o A134287 a134287 = flip a103371 4 . (+ 4)  -- _Reinhard Zumkeller_, Apr 04 2014
%o A134287 (Magma) [5*Binomial(n+5, 5)^2/(n+5): n in [0..30]]; // _G. C. Greubel_, Oct 28 2022
%o A134287 (SageMath) [5*binomial(n+5,5)^2/(n+5) for n in range(31)] # _G. C. Greubel_, Oct 28 2022
%Y A134287 Cf. A108647 (fourth column of triangle A103371).
%K A134287 nonn,easy
%O A134287 0,2
%A A134287 _Wolfdieter Lang_, Nov 13 2007