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A108647
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a(n) = (n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144.
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3
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1, 20, 150, 700, 2450, 7056, 17640, 39600, 81675, 157300, 286286, 496860, 828100, 1332800, 2080800, 3162816, 4694805, 6822900, 9728950, 13636700, 18818646, 25603600, 34385000, 45630000, 59889375, 77808276, 100137870, 127747900
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OFFSET
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0,2
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COMMENTS
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Kekulé numbers for certain benzenoids.
a(n-4), n>=4, is the number of ways to have n identical objects in m=4 of altogether n distinguishable boxes (n-4 boxes stay empty). - Wolfdieter Lang, Nov 13 2007
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REFERENCES
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S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, no. 23).
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LINKS
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Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Index to sequences with linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).
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FORMULA
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a(n)=C(n+4,4)*C(n+3,2)(n+1)/3. - Paul Barry, May 13 2006
G.f.: (1+12*x+18*x^2+4*x^3)/(1-x)^8.
a(n)= 4*C(n,4)^2/n, n >= 4. - Zerinvary Lajos, May 09 2008
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EXAMPLE
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a(2)=150 because n=6 identical balls can be put into m=4 of n=6 distinguishable boxes in binomial(6,4)*(4!/(3!*1!)+ 4!/(2!*2!)) = 15*(4 + 6) =150 ways. The m=4 part partitions of 6, namely (1^3,3) and (1^2,2^2) specify the filling of each of the 15 possible four box choices. - Wolfdieter Lang, Nov 13 2007
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MAPLE
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a:=(n+1)^2*(n+2)^2*(n+3)^2*(n+4)/144: seq(a(n), n=0..30);
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PROG
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(Mupad) 4*binomial(n, 4)^2/n $ n = 4..35; # Zerinvary Lajos, May 09 2008
(Haskell)
a108647 = flip a103371 3 . (+ 3) -- Reinhard Zumkeller, Apr 04 2014
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CROSSREFS
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Fourth column of triangle A103371.
Sequence in context: A100190 A189494 A022680 * A164605 A000492 A015866
Adjacent sequences: A108644 A108645 A108646 * A108648 A108649 A108650
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KEYWORD
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nonn,changed
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AUTHOR
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Emeric Deutsch, Jun 13 2005
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STATUS
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approved
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