A000531 From area of cyclic polygon of 2n + 1 sides.

1, 7, 38, 187, 874, 3958, 17548, 76627, 330818, 1415650, 6015316, 25413342, 106853668, 447472972, 1867450648, 7770342787, 32248174258, 133530264682, 551793690628, 2276098026922, 9373521044908, 38546133661492

Offset: 1

Author

Simon Plouffe

Keywords

• nonn
• easy
• nice

Comments

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(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

Equals the Catalan sequence, A000108, convolved with A002457: (1, 6, 30, 140, ...). - Gary W. Adamson, May 14 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

References

• W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.

Links

• T. D. Noe, Table of n, a(n) for n = 1..100
• Arvind Ayyer, Towards a human proof of Gessel's conjecture, arXiv:0902.2329 [math.CO], 2009, JIS 12 (2009) 09.4.2
• Hacène Belbachir, Toufik Djellal, Jean-Gabriel Luque, On the self-convolution of generalized Fibonacci numbers, arXiv:1703.00323 [math.CO], 2017.
• F. Bowman, Cyclic pentagons, Math. Gaz. 36, (1952). 244-250. MR0051523.
• A. Burstein and S. Elizalde, Total occurrence statistics on restricted permutations, arXiv preprint arXiv:1305.3177 [math.CO], 2013.
• C. Homberger, Expected patterns in permutation classes, Electronic Journal of Combinatorics, 19(3) (2012), P43.
• Cheyne Homberger, Patterns in Permutations and Involutions: A Structural and Enumerative Approach, arXiv preprint 1410.2657 [math.CO], 2014.
• Yasuhiko Kamiyama, The Euler characteristic of the regular spherical polygon spaces, arXiv:1803.05559 [math.GT], 2018.
• A. F. Moebius, Über die Gleichungen, mittelst welcher aus den Seiten eines in einen Kreis zu beschreibenden Vielecks der Halbmesser des Kreises und die Fläche des Vielecks gefunden werden, Gesammelte Werke, vol. 1., pp. 407-438.
• D. P. Robbins, Areas of polygons inscribed in a circle, Amer. Math. Monthly, 102 (1995), 523-530.
• Eric Weisstein's World of Mathematics, Cyclic Polygon
• Y. Q. Zhao, Introduction to Probability with Applications

Crossrefs

Cf. A002457 (Banach's modified matchbox problem), A135404, A002457, A258431.

Programs

Maple

f := proc(n) sum((n-k)*binomial(2*n+1,k),k=0..n-1); end;

Mathematica

a[n_] := ((2n+1)!/n!^2-4^n)/2; Table[a[n], {n, 1, 22}] (* Jean-François Alcover, Dec 07 2011, after Pari *)

PARI

a(n)=if(n<1,0,((2*n+1)!/n!^2-4^n)/2)

Formula

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

a(n) = Sum_{k=0..n} A039599(n,k)*k^2, for n>=1. - Philippe Deléham, Jun 10 2007

a(n) = Sum_{k=0..n} A106566(n,k)*A001788(k). - Philippe Deléham, Oct 31 2008

(Conjecture) a(n)=2^(2*n)*sum_{k=1..n} cos(k*Pi/(2*n+1))^2*n. - L. Edson Jeffery, Jan 21 2012

Extensions

Moebius reference from Michael Somos

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