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%I #82 Apr 04 2024 12:06:11
%S 1,15,84,300,825,1911,3920,7344,12825,21175,33396,50700,74529,106575,
%T 148800,203456,273105,360639,469300,602700,764841,960135,1193424,
%U 1470000,1795625,2176551,2619540,3131884,3721425,4396575,5166336,6040320,7028769,8142575
%N a(n) = (n+1)^2 * (n+2)^2 * (2*n+3) / 12.
%C Kekulé numbers for certain benzenoids.
%C This is the case P(3,n) of the family of sequences defined in A132458. - Ottavio D'Antona (dantona(AT)dico.unimi.it), Oct 31 2007
%C Using the triangular numbers 0, 1, 3, ..., create a sequence of advancing sums of k-tuples with k=n*(n+1)/2 of the odd numbers: 0, 1, 15, 84, 300, 825, 1911, 3920, ... . This begins 0, then 1, then 3+5+7=15, then 9+11+13+15+17+19=84, then 21+23+...+39=300 and so on. - _J. M. Bergot_, Dec 08 2014
%C Partial sums of A008354. - _J. M. Bergot_, Dec 19 2014
%C Coefficients in the terminating series identity 1 - 15*n/(n + 4) + 84*n*(n - 1)/((n + 4)*(n + 5)) - 300*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 2,3,4,.... Cf. A000330. - _Peter Bala_, Feb 12 2019
%D S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 231, # 33).
%H Muniru A Asiru, <a href="/A108674/b108674.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F G.f.: (1+z)*(1+8*z+z^2)/(1-z)^6.
%F a(n) = Sum_{j=1..n+1} j^2 Sum_{i=1..n+1} i. - _Alexander Adamchuk_, Jun 25 2006
%F a(n) = A000330(n+1) * A000217(n+1). - _Daniel Suteu_, Nov 26 2020
%F E.g.f.: exp(x)*(12 + 168*x + 330*x^2 + 184*x^3 + 35*x^4 + 2*x^5)/12. - _Stefano Spezia_, Mar 02 2022
%F From _Amiram Eldar_, May 29 2022: (Start)
%F Sum_{n>=0} 1/a(n) 192*log(2) - 132.
%F Sum_{n>=0} (-1)^n/a(n) = 2*Pi^2 - 48*Pi + 132. (End)
%p A108674 :=n->(n+1)^2*(n+2)^2*(2*n+3)/12;
%p seq(A108674(n),n=0..35);
%t Table[(n^2+4*n^3+5*n^4+2*n^5)/12, {n, 40}] (* _Enrique Pérez Herrero_, Feb 27 2013 *)
%t LinearRecurrence[{6,-15,20,-15,6,-1},{1,15,84,300,825,1911},40] (* _Harvey P. Dale_, Apr 04 2024 *)
%o (PARI) a(n)=(n+1)^2*(n+2)^2*(2*n+3)/12 \\ _Charles R Greathouse IV_, Feb 27 2013
%o (GAP) List([0..40],k->(k+1)^2*(k+2)^2*(2*k+3)/12); # _Muniru A Asiru_, Feb 18 2019
%Y Cf. A000537, A126274.
%Y Cf. A000217, A000330, A008354 (first diffs.), A132458.
%K nonn,easy
%O 0,2
%A _Emeric Deutsch_, Jun 17 2005
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