1,2

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 (msteyer(AT)osram.de), 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 ths 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

F. Bowman, Cyclic pentagons, Math. Gaz. 36, (1952). 244-250. MR0051523

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

A. F. Moebius, Ueber die Gleichungen, mittelst welcher aus den Seiten eines in einen Kreis zu beschreibenden Vielecks der Halbmesser des Kreises und die Fl\"ache des Vielecks gefunden werden [Crelle's Journal 1828 Band 3 p. 5-34], Gesammelte Werke, vol. 1., pp. 407-438.

D. P. Robbins, Areas of polygons inscribed in a circle, Amer. Math. Monthly, 102 (1995), 523-530.

T. D. Noe, Table of n, a(n) for n=1..100

A. F. Moebius, Ueber die Gleichungen, mittelst welcher aus den Seiten eines in einen Kreis zu beschreibenden Vielecks der Halbmesser des Kreises und die Fl\"ache des Vielecks gefunden werden, Gesammelte Werke, vol. 1., pp. 407-438.

D. P. Robbins, Areas of Polygons Inscribed in a Circle, Discrete & Computational Geometry 12, 223-236, 1994.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Y. Q. Zhao, Introduction to Probability with Applications

((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 (vladeta(AT)Eunet.yu), Sep 22 2003

a(n-1) = 4^n*sum(binomial(2*k+1, k)*4^(-k), k=0..n) = (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 (wolfdieter.lang(AT)physik.uni-karlsruhe.de) ]

a(n)=Sum_{k, 0<=k<=n}A039599(n,k)*k^2, for n>=1 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Jun 10 2007

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

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 (callan(AT)stat.wisc.edu), Dec 29 2006

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

Cf. A002457 (Banach's modified matchbox problem)

Adjacent sequences: A000528 A000529 A000530 this_sequence A000532 A000533 A000534

Sequence in context: A003352 A034858 A114290 this_sequence A099453 A026763 A037696

nonn,easy,nice

Simon Plouffe (plouffe(AT)math.uqam.ca)

Moebius reference from Michael Somos.