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Triangle read by rows: T(n,k) is the number of walks (each step +-1) of length 2n which have a cumulative value of 0 last at step 2k.
6

%I #59 Aug 18 2024 23:03:51

%S 1,2,2,6,4,6,20,12,12,20,70,40,36,40,70,252,140,120,120,140,252,924,

%T 504,420,400,420,504,924,3432,1848,1512,1400,1400,1512,1848,3432,

%U 12870,6864,5544,5040,4900,5040,5544,6864,12870,48620,25740,20592,18480

%N Triangle read by rows: T(n,k) is the number of walks (each step +-1) of length 2n which have a cumulative value of 0 last at step 2k.

%C Since the triangle is symmetric, the probability that a one-dimensional random walk returns to the origin at all in the steps m through to 2m is 1/2 (for m odd).

%C Diagonal sums are A106183. - _Paul Barry_, Apr 24 2005

%D A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 79, ex. 3f.

%H G. C. Greubel, <a href="/A067804/b067804.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H B. C. Carlson, <a href="https://doi.org/10.1090%2Fs0025-5718-07-02049-2">Power series for inverse Jacobian elliptic functions</a>, Math. Comp., 77 (2008), 1615-1621, see p. 1617, equation (2.20).

%H C. M. Grinstead and J. L. Snell, <a href="http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html">Introduction to Probability</a> p. 482.

%H R. P. Kelisky, <a href="http://www.jstor.org/stable/2310630">Inverse elliptic functions and Legendre polynomials</a>, Amer. Math. Monthly 66 (1959), pp. 480-483. MR0103993 (21 #2755).

%H Michael Z. Spivey, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.121.06.537">A Combinatorial Proof for the Alternating Convolution of the Central Binomial Coefficients</a>, The American Mathematical Monthly 121.6 (2014): 537-540. [Suggested by _Roger L. Bagula_, Jun 21 2014]

%F T(n, k) = C(2k, k)*C(2n-2k, n-k) = C(2n, n)*C(n, k)^2/C(2n, 2k) = A000984(k)*A000984(n-k) = A000984(n)*A008459(n, k)/A007318(2n, 2k).

%F Row sums are 4^n = A000302(n).

%F G.f.: A(x,y) = 1/sqrt((1-4*x)*(1-4*x*y)). - _Vladeta Jovovic_, Dec 12 2003

%F Sum{k>=0} T(n, k)*(-3)^k = (-4)^n * A002426(n). Sum_{k>=0} T(n, k)/(2*k+1) = 2^(4*n)/((2*n+1)*C(2*n, n)). - _Philippe Deléham_, Dec 31 2003

%F O.g.f.: A(x,y) = 1 + x*d/dx(log(B(x,y))), where B(x,y) is the o.g.f. of A120406. - _Peter Bala_, Jul 17 2015

%e Triangle begins:

%e 1;

%e 2, 2;

%e 6, 4, 6;

%e 20, 12, 12, 20;

%e 70, 40, 36, 40, 70;

%e 252, 140, 120, 120, 140, 252;

%e ...

%e For a walk of length 4 (=2*2), 6 are only ever 0 at step 0, 4 are zero at step 2 but not step 4 and 6 are 0 at step 4.

%e For n=3,k=2, T(3,2)=12 since there are 12 monotonic paths from (0,0) to (2,2) and then on to (3,3). Using E for eastward steps and N for northward steps, the 12 paths are given by EENNNE, ENENNE, ENNENE, NNEENE, NENENE, NEENNE, EENNEN, ENENEN, ENNEEN, NNEEEN, NENEEN, NEENEN. - _Dennis P. Walsh_, Mar 23 2012

%t Table[Table[Binomial[2k,k]Binomial[2(n-k),n-k],{k,0,n}],{n,0,10}]//Grid (* _Geoffrey Critzer_, Jun 30 2013 *)

%t T[ n_, k_] := SeriesCoefficient[ D[ InverseJacobiSN[2 x, m] / 2, x], {x, 0, 2 n}, {m, 0, k}]; (* _Michael Somos_, May 06 2017 *)

%o (PARI) T(n, k) = binomial(2*k, k) * binomial(2*n-2*k, n-k) /* _Michael Somos_, Aug 20 2005 */

%o (Magma) /* As triangle */ [[Binomial(2*k, k)*Binomial(2*n-2*k, n-k): k in [0..n]]: n in [0.. 15]]; // _Vincenzo Librandi_, Jul 19 2015

%Y Columns include A000984, A028329. Central diagonal is A002894.

%Y Cf. A002426, A120406.

%K nonn,tabl,easy

%O 0,2

%A _Henry Bottomley_, Feb 07 2002