0,2

Expected number of matches remaining in Banach's modified matchbox problem (counted when last match is drawn from one of the two boxes), multiplied by 4^(n-1). - Michael Steyer (msteyer(AT)osram.de), Apr 13 2001

Sum_{n>=0} 1/a(n) = 2*Pi/3^(3/2) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 07 2009]

Hankel transform is (-1)^n*A014480(n). [From Paul Barry (pbarry(AT)wit.ie), Apr 26 2009]

Convolved with A000108: (1, 1, 1, 5, 14, 42,...) = A000531: (1, 7, 38, 187, 874,...). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 14 2009]

Convolution of A000302 and A000984 . [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), May 18 2009]

1.)Number of permutations of three distinct letters (ABC) 0 to n times >>("-". ABC, AABBCC, AAABBBCCC, etc...)and one after the other to resemble motif: AAA (3-0), AAAAAB (5-1), AAAAAAABB (7-2), AAAAAAAAABBB (9-3), AAAAAAAAAAABBBB (11-4), AAAAAAAAAAAAABBBBB (13-5), etc...,>>1,6,30,140,630,2772,12012,etc. fixed points. Example:motif: AAA (or BBB, or CCC) CAB CBA ACB BCA ABC BAC 6 * 1 fixed point etc... 2.) A005408*A000984 example: 7*20 =140 9*70 =630 11*252 =2772 13*924 =12012 15*3432 =51480 etc... [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 15 2009]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.

R. Chapman, Moments of Dyck paths, Discrete Math., 204 (1999), 113-117.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.

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

C. Jordan, Calculus of Finite Differences. Budapest, 1939, p. 449.

M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.

C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.

A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.

H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135.

J. Ser, Les Calculs Formels des S\'{e}ries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.

T. R. Van Oppolzer, Lehrbuch zur Bahnbestimmung der Kometen und Planeten, Vol. 2, Engelmann, Leipzig, 1880, p. 21.

T. D. Noe, Table of n, a(n) for n=0..200

Eric Weisstein's World of Mathematics, Central Beta Function

Y. Q. Zhao, Introduction to Probability with Applications

G.f.: (1-4x)^(-3/2). a(n-1)=binomial(2n, n)*n/2 = binomial(2n-1, n)*n.

a(n-1)=4^(n-1)*sum(binomial(n-1+i, i)*(n-i)/2^(n-1+i), i=0..n-1).

a(n) ~ 2*pi^(-1/2)*n^(1/2)*2^(2*n)*{1 + 3/8*n^-1 + ...} - Joe Keane (jgk(AT)jgk.org), Nov 21 2001

(2*n+2)!/(2*n!*(n+1)!) = (n+n+1)!/(n!*n!) = 1/beta(n+1, n+1) in A061928.

Sum(i * binomial(n, i)^2, i=0.. n) = n*binomial(2*n, n)/2 - Yong Kong (ykong(AT)curagen.com), Dec 26 2000

a(n) ~ 2*pi^(-1/2)*n^(1/2)*2^(2*n) - Joe Keane (jgk(AT)jgk.org), Jun 07 2002

a(n) = 1/Integral_{x=0..1} x^n (1-x)^n dx. - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 10 2003

E.g.f.: exp(2*x)*((1+4*x)*BesselI(0, 2*x)+4*x*BesselI(1, 2*x)). - Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 22 2003

a(n)=sum(i+j+k=n, binomial(2i, i)binomial(2j, j)binomial(2k, k)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 09 2003

Equals (2*n+1)*A000984(n) - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 12 2006

a(n-1)=Sum_{k, 0<=k<=n}A039599(n,k)*A000217(k), for n>=1 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Jun 10 2007

Sum of (n+1)-th row terms of triangle A132818. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 02 2007

a(n)=sum{k=0..n, C(2k,k)*4^(n-k)}. [From Paul Barry (pbarry(AT)wit.ie), Apr 26 2009]

A002457 := n-> (n+1)*binomial(2*(n+1), (n+1))/2;

a:=n->sum(abs(binomial(-n, -2*n)), j=1..n): seq(a(n), n=1..23); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 03 2007

a:=n->abs(sum((binomial(-n, n-2)), j=2..n)): seq(a(n), n=2..24); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 03 2007

restart:a:= proc(n) option remember; if n=0 then 1 else add((binomial (2*n, n))/2, j=0..n-1) fi end: seq (a(n), n=1..23); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 29 2009]

a[n_]:=(2*n+1)!/n!^2; [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 13 2008]

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

Cf. A033876. Also a(n)=f(n, n-3) where f is given in A034261.

Denominator of central elements of Leibniz's Harmonic Triangle A003506.

Cf. A000531 (Banach's original match problem). Equals A002011/4.

a(n) = A005430(n+1)/2 = A002011(n)/4.

Cf. A000984, A001803, A132818.

A000531 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 14 2009]

Sequence in context: A125316 A092439 A082149 this_sequence A137400 A026749 A003279

Adjacent sequences: A002454 A002455 A002456 this_sequence A002458 A002459 A002460

nonn,easy,nice,new

N. J. A. Sloane (njas(AT)research.att.com).