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A054849
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a(n) = 2^(n-5)*binomial(n,5). Number of 5D hypercubes in an n-dimensional hypercube.
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21
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1, 12, 84, 448, 2016, 8064, 29568, 101376, 329472, 1025024, 3075072, 8945664, 25346048, 70189056, 190513152, 508035072, 1333592064, 3451650048, 8820883456, 22284337152, 55710842880, 137950658560, 338606161920
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OFFSET
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5,2
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COMMENTS
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With 5 leading zeros, binomial transform of binomial(n,5). - Paul Barry, Apr 10 2003
If X_1,X_2,...,X_n is a partition of a 2n-set X into 2-blocks then, for n>4, a(n) is equal to the number of (n+5)-subsets of X intersecting each X_i (i=1,2,...,n). - Milan Janjic, Jul 21 2007
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LINKS
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FORMULA
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O.g.f.: x^5/(1-2*x)^6.
E.g.f.: exp(2*x)*(x^5/5!) (with 5 leading zeros). (End)
a(n) = Sum_{i=5..n} binomial(i,5)*binomial(n,i). Example: for n=8, a(8) = 1*56 + 6*28 + 21*8 + 56*1 = 448. - Bruno Berselli, Mar 23 2018
Sum_{n>=5} 1/a(n) = 10*log(2) - 35/6.
Sum_{n>=5} (-1)^(n+1)/a(n) = 810*log(3/2) - 655/2. (End)
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MAPLE
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MATHEMATICA
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Table[2^(n-5)*Binomial[n, 5], {n, 5, 30}] (* G. C. Greubel, Aug 27 2019 *)
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PROG
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(Sage) [lucas_number2(n, 2, 0)*binomial(n, 5)/32 for n in range(5, 28)] # Zerinvary Lajos, Mar 10 2009
(PARI) vector(25, n, 2^(n-1)*binomial(n+4, 5)) \\ G. C. Greubel, Aug 27 2019
(Magma) [2^(n-5)*Binomial(n, 5): n in [5..30]]; // G. C. Greubel, Aug 27 2019
(GAP) List([5..30], n-> 2^(n-5)*Binomial(n, 5)); # G. C. Greubel, Aug 27 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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