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A114242 a(n) = (n+1)(n+2)^2*(n+3)^2*(n+4)(2n+5)/720. 6

%I #22 May 29 2022 03:07:10

%S 1,14,90,385,1274,3528,8568,18810,38115,72358,130130,223587,369460,

%T 590240,915552,1383732,2043621,2956590,4198810,5863781,8065134,

%U 10939720,14651000,19392750,25393095,32918886,42280434,53836615,68000360

%N a(n) = (n+1)(n+2)^2*(n+3)^2*(n+4)(2n+5)/720.

%C Kekulé numbers for certain benzenoids.

%C Partial sums of A114244. First differences of A006857. - _Peter Bala_, Sep 21 2007

%D S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 167-169, Table 10.5/II/2 and p. 105, eq. (ii) K(Ob(2,4,n))).

%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 G.f.: (1+x)(1 + 5x + x^2)/(1-x)^8.

%F a(n-2) = (1/6) * Sum_{1 <= x_1, x_2 <= n} (x_1)^2*x_2*(det V(x_1,x_2))^2 = 1/6*sum {1 <= i,j <= n} i^2*j*(i-j)^2, where V(x_1,x_2} is the Vandermonde matrix of order 2. - _Peter Bala_, Sep 21 2007

%F a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8). - _Harvey P. Dale_, Aug 21 2013

%F From _Amiram Eldar_, May 29 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 3550 - 5120*log(2).

%F Sum_{n>=0} (-1)^n/a(n) = 3430 - 1280*Pi + 60*Pi^2. (End)

%p a:=n->(n+1)*(n+2)^2*(n+3)^2*(n+4)*(2*n+5)/720: seq(a(n),n=0..30);

%t Table[((n+1)(n+2)^2 (n+3)^2 (n+4)(2n+5))/720,{n,0,30}] (* or *) LinearRecurrence[ {8,-28,56,-70,56,-28,8,-1},{1,14,90,385,1274,3528,8568,18810},30] (* _Harvey P. Dale_, Aug 21 2013 *)

%o (PARI) a(n)=(n+1)*(n+2)^2*(n+3)^2*(n+4)*(2*n+5)/720 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A005585, A006542, A107891.

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Nov 18 2005

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)