A108674 a(n) = (n+1)^2 * (n+2)^2 * (2*n+3) / 12.

1, 15, 84, 300, 825, 1911, 3920, 7344, 12825, 21175, 33396, 50700, 74529, 106575, 148800, 203456, 273105, 360639, 469300, 602700, 764841, 960135, 1193424, 1470000, 1795625, 2176551, 2619540, 3131884, 3721425, 4396575, 5166336, 6040320, 7028769, 8142575

Offset: 0

Author

Emeric Deutsch, Jun 17 2005

Keywords

• nonn
• easy

Comments

Kekulé numbers for certain benzenoids.

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

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

Partial sums of A008354. - J. M. Bergot, Dec 19 2014

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

References

• S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 231, # 33).

Links

• Muniru A Asiru, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000537, A126274.

Cf. A000217, A000330, A008354 (first diffs.), A132458.

Programs

GAP

List([0..40],k->(k+1)^2*(k+2)^2*(2*k+3)/12); # Muniru A Asiru, Feb 18 2019

Maple

A108674 :=n->(n+1)^2*(n+2)^2*(2*n+3)/12;
seq(A108674(n),n=0..35);

Mathematica

Table[(n^2+4*n^3+5*n^4+2*n^5)/12, {n, 40}] (* Enrique Pérez Herrero, Feb 27 2013 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1,15,84,300,825,1911},40] (* Harvey P. Dale, Apr 04 2024 *)

PARI

a(n)=(n+1)^2*(n+2)^2*(2*n+3)/12 \\ Charles R Greathouse IV, Feb 27 2013

Formula

G.f.: (1+z)*(1+8*z+z^2)/(1-z)^6.

a(n) = Sum_{j=1..n+1} j^2 Sum_{i=1..n+1} i. - Alexander Adamchuk, Jun 25 2006

a(n) = A000330(n+1) * A000217(n+1). - Daniel Suteu, Nov 26 2020

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

From Amiram Eldar, May 29 2022: (Start)

Sum_{n>=0} 1/a(n) 192*log(2) - 132.

Sum_{n>=0} (-1)^n/a(n) = 2*Pi^2 - 48*Pi + 132. (End)