OFFSET
5,2
COMMENTS
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
LINKS
G. C. Greubel, Table of n, a(n) for n = 5..1000
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Index entries for linear recurrences with constant coefficients, signature (12, -60, 160, -240, 192, -64).
FORMULA
a(n) = 2*a(n-1) + A003472(n-1).
From Paul Barry, Apr 10 2003: (Start)
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
From Amiram Eldar, Jan 06 2022: (Start)
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)
MAPLE
seq(binomial(n+5, 5)*2^n, n=0..22); # Zerinvary Lajos, Jun 13 2008
MATHEMATICA
Table[2^(n-5)*Binomial[n, 5], {n, 5, 30}] (* G. C. Greubel, Aug 27 2019 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Henry Bottomley, Apr 14 2000
EXTENSIONS
More terms from James A. Sellers, Apr 15 2000
STATUS
approved