

A000531


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


18



1, 7, 38, 187, 874, 3958, 17548, 76627, 330818, 1415650, 6015316, 25413342, 106853668, 447472972, 1867450648, 7770342787, 32248174258, 133530264682, 551793690628, 2276098026922, 9373521044908, 38546133661492
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

Expected number of matches remaining in Banach's original matchbox problem (counted when empty box is chosen), multiplied by 2^(2*n1).  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_1a_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 npaths (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 4paths. The summand (nk)*binomial(2*n+1, k) in the Maple formula below is the total weight of upsteps terminating at height nk, 0<=k<=n1.  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 skewindecomposable (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, JeanGabriel Luque, On the selfconvolution of generalized Fibonacci numbers, arXiv:1703.00323 [math.CO], 2017.
F. Bowman, Cyclic pentagons, Math. Gaz. 36, (1952). 244250. 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. 407438.
D. P. Robbins, Areas of polygons inscribed in a circle, Amer. Math. Monthly, 102 (1995), 523530.
Eric Weisstein's World of Mathematics, Cyclic Polygon
Y. Q. Zhao, Introduction to Probability with Applications


FORMULA

a(n) = ((2n+1)!/((n!)^2)4^n)/2.  Simon Norton (simon(AT)dpmms.cam.ac.uk), May 14 2001
na(n) = (8n2)a(n1)  (16n8)a(n2), 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(n1) = 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)/(14*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


MAPLE

f := proc(n) sum((nk)*binomial(2*n+1, k), k=0..n1); end;


MATHEMATICA

a[n_] := ((2n+1)!/n!^24^n)/2; Table[a[n], {n, 1, 22}] (* JeanFrançois Alcover, Dec 07 2011, after Pari *)


PROG

(PARI) a(n)=if(n<1, 0, ((2*n+1)!/n!^24^n)/2)


CROSSREFS

Cf. A002457 (Banach's modified matchbox problem), A135404, A002457, A258431.
Sequence in context: A249021 A114290 A277912 * A296769 A241524 A291822
Adjacent sequences: A000528 A000529 A000530 * A000532 A000533 A000534


KEYWORD

nonn,easy,nice


AUTHOR

Simon Plouffe


EXTENSIONS

Moebius reference from Michael Somos


STATUS

approved



