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A108646
a(n) = (n+1)*(n+2)^2*(n+3)*(11*n^3 + 58*n^2 + 101*n + 60)/720.
4
1, 23, 194, 985, 3668, 11074, 28728, 66438, 140415, 276001, 511082, 900263, 1519882, 2473940, 3901024, 5982300, 8950653, 13101051, 18802210, 26509637, 36780128, 50287798, 67841720, 90405250, 119117115, 155314341, 200557098
OFFSET
0,2
COMMENTS
Kekulé numbers for certain benzenoids.
REFERENCES
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, no. 22).
FORMULA
From Chai Wah Wu, Jun 12 2016: (Start)
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) for n > 7.
G.f.: (1 + 15*x + 38*x^2 + 21*x^3 + 2*x^4)/(1 - x)^8. (End)
E.g.f.: (1/6!)*(720 + 15840*x + 53640*x^2 + 56520*x^3 + 24030*x^4 + 4548*x^5 + 377*x^6 + 11*x^7)*exp(x). - G. C. Greubel, Oct 19 2023
MAPLE
a:=(n+1)*(n+2)^2*(n+3)*(11*n^3+58*n^2+101*n+60)/720: seq(a(n), n=0..30);
MATHEMATICA
Table[(n+2)*(n+3)!*(11*n^3+58*n^2+101*n+60)/(6!*n!), {n, 0, 40}] (* G. C. Greubel, Oct 19 2023 *)
PROG
(Python)
A108646_list, m = [], [77, -85, 28, -1, 1, 1, 1, 1]
for _ in range(10001):
A108646_list.append(m[-1])
for i in range(7):
m[i+1] += m[i] # Chai Wah Wu, Jun 12 2016
(Magma) [(n+2)*(11*n^3+58*n^2+101*n+60)*Binomial(n+3, 3)/120: n in [0..40]]; // G. C. Greubel, Oct 19 2023
(SageMath) [(n+2)*(11*n^3+58*n^2+101*n+60)*binomial(n+3, 3)/120 for n in range(41)] # G. C. Greubel, Oct 19 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Emeric Deutsch, Jun 13 2005
STATUS
approved