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A000531
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From area of cyclic polygon of 2n + 1 sides.
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18
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1, 7, 38, 187, 874, 3958, 17548, 76627, 330818, 1415650, 6015316, 25413342, 106853668, 447472972, 1867450648, 7770342787, 32248174258, 133530264682, 551793690628, 2276098026922, 9373521044908, 38546133661492
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OFFSET
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1,2
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COMMENTS
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Expected number of matches remaining in Banach's original matchbox problem (counted when empty box is chosen), multiplied by 2^(2*n-1). - Michael Steyer, Apr 13 2001
A conjectured definition: Let 0 < a_1 < a_2 <...<a_{2n} < 1. Then how many ways are there in which one can add or subtract all the a_i to get an odd number. For example, take n = 2. Then the options are a_1+a_2+a_3+a_4 = 1 or 3; one can change the sign of any of the a_i's and get 1; or -a_1-a_2+a_3+a_4 = 1. That's a total of 7, which is the 2nd number of this sequence. One of the definitions of the sequence (which was how I came across it) is the degree of the equation giving the area of a cyclic polygon in terms of the sides. I conjectured that for any set of side lengths there is a unique way of fitting them together for any possible winding number and any possible subset of sides which go round the circle in a retrograde manner. - Simon Norton (simon(AT)dpmms.cam.ac.uk), May 14 2001
a(n) = total weight of upsteps in all Dyck n-paths (A000108) when each upstep is weighted with its position in the path. For example, the Dyck path UDUUDUDD has upsteps in positions 1,3,4,6 and contributes 1+3+4+6=14 to the weight for Dyck 4-paths. The summand (n-k)*binomial(2*n+1, k) in the Maple formula below is the total weight of upsteps terminating at height n-k, 0<=k<=n-1. - David Callan, Dec 29 2006
Catalan transform of binomial transform of squares. - Philippe Deléham, Oct 31 2008
a(n) is also the number of walks of length 2n in the quarter plane starting and ending at the origin using steps {(1,1),(1,0),(-1,0), (-1,-1)} (which appear in Gessel's conjecture) in which the steps (1,0) and (-1,0) appear exactly once each. - Arvind Ayyer, Mar 02 2009
Total number of occurrences of the pattern 213 (or 132) in all skew-indecomposable (n+2)-permutations avoiding the pattern 123. For example, a(1) = 1, since there is one occurrence of the pattern 213 in the set {213, 132}. - Cheyne Homberger, Mar 13 2013
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REFERENCES
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W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
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LINKS
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FORMULA
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a(n) = ((2n+1)!/((n!)^2)-4^n)/2. - Simon Norton (simon(AT)dpmms.cam.ac.uk), May 14 2001
na(n) = (8n-2)a(n-1) - (16n-8)a(n-2), n>1. - Michael Somos, Apr 18 2003
E.g.f.: 1/2*((1+4*x)*exp(2*x)*BesselI(0, 2*x) + 4*x*exp(2*x)*BesselI(1, 2*x) - exp(4*x)). - Vladeta Jovovic, Sep 22 2003
a(n-1) = 4^n*sum_{k=0..n} binomial(2*k+1, k)*4^(-k) = (2*n+1)*(2*n+3)*C(n) - 2^(2*n+1) (C(n) = Catalan); g.f.: x*c(x)/(1-4*x)^(3/2), c(x): g.f. of Catalan numbers A000108. - Wolfdieter Lang
(Conjecture) a(n)=2^(2*n)*sum_{k=1..n} cos(k*Pi/(2*n+1))^2*n. - L. Edson Jeffery, Jan 21 2012
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MAPLE
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f := proc(n) sum((n-k)*binomial(2*n+1, k), k=0..n-1); end;
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MATHEMATICA
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a[n_] := ((2n+1)!/n!^2-4^n)/2; Table[a[n], {n, 1, 22}] (* Jean-François Alcover, Dec 07 2011, after Pari *)
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PROG
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(PARI) a(n)=if(n<1, 0, ((2*n+1)!/n!^2-4^n)/2)
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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