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A098402
a(n) = (0^n + 4^n * binomial(2*n,n))/2.
4
1, 4, 48, 640, 8960, 129024, 1892352, 28114944, 421724160, 6372720640, 96865353728, 1479398129664, 22684104654848, 348986225459200, 5384358907084800, 83278084429578240, 1290810308658462720, 20045524793284362240, 311819274562201190400, 4857816066863765913600
OFFSET
0,2
COMMENTS
It seems that a(n) is the number of pairs of binary vectors of length 2*n which are orthogonal. (Define binary vectors here to have components of value +1 or -1. There are no pairs of binary vectors of odd length which are orthogonal.) For example, the a(1) = 4 pairs of binary vectors of length 2 are (-1,-1) and (1,-1), (-1,-1) and (-1,1), (1,-1) and (1,1), (-1,1) and (1,1). Tested up to and including a(8). - R. J. Mathar, Apr 15 2013
Tested up to and including a(210). - R. H. Hardin, May 08 2013
LINKS
FORMULA
G.f.: 8*x/( sqrt(1 - 16*x)*(1 - sqrt(1 - 16*x)) ).
a(n+1) = 4*A098400(n).
n*a(n) - 8*(2*n-1)*a(n-1) = 0. - R. J. Mathar, Nov 09 2012
a(n) ~ 16^n/(2*sqrt(Pi*n)). - Ilya Gutkovskiy, Aug 03 2016
a(n) = A055372(2*n,n). - Alois P. Heinz, Jan 21 2020
From Amiram Eldar, Jan 16 2024: (Start)
Sum_{n>=0} 1/a(n) = 17/15 + 32*arcsin(1/4)/(15*sqrt(15)).
Sum_{n>=0} (-1)^n/a(n) = 15/17 - 32*arcsinh(1/4)/(17*sqrt(17)). (End)
MATHEMATICA
Table[(Boole[n == 0] + 4^n Binomial[2 n, n])/2, {n, 0, 18}] (* or *)
CoefficientList[Series[8 x/(# (1 - #)) &@ Sqrt[1 - 16 x], {x, 0, 18}], x] (* Michael De Vlieger, Aug 03 2016 *)
PROG
(Magma) [(0^n + 4^n*(n+1)*Catalan(n))/2: n in [0..40]]; // G. C. Greubel, Dec 27 2023
(SageMath) [(4^n*binomial(2*n, n) + int(n==0))/2 for n in range(41)] # G. C. Greubel, Dec 27 2023
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Sep 06 2004
STATUS
approved