OFFSET
0,2
COMMENTS
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).
Diagonal sums are A106183. - Paul Barry, Apr 24 2005
REFERENCES
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 79, ex. 3f.
LINKS
G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
B. C. Carlson, Power series for inverse Jacobian elliptic functions, Math. Comp., 77 (2008), 1615-1621, see p. 1617, equation (2.20).
C. M. Grinstead and J. L. Snell, Introduction to Probability p. 482.
R. P. Kelisky, Inverse elliptic functions and Legendre polynomials, Amer. Math. Monthly 66 (1959), pp. 480-483. MR0103993 (21 #2755).
Michael Z. Spivey, A Combinatorial Proof for the Alternating Convolution of the Central Binomial Coefficients, The American Mathematical Monthly 121.6 (2014): 537-540. [Suggested by Roger L. Bagula, Jun 21 2014]
FORMULA
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).
Row sums are 4^n = A000302(n).
G.f.: A(x,y) = 1/sqrt((1-4*x)*(1-4*x*y)). - Vladeta Jovovic, Dec 12 2003
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
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
EXAMPLE
Triangle begins:
1;
2, 2;
6, 4, 6;
20, 12, 12, 20;
70, 40, 36, 40, 70;
252, 140, 120, 120, 140, 252;
...
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.
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
MATHEMATICA
Table[Table[Binomial[2k, k]Binomial[2(n-k), n-k], {k, 0, n}], {n, 0, 10}]//Grid (* Geoffrey Critzer, Jun 30 2013 *)
T[ n_, k_] := SeriesCoefficient[ D[ InverseJacobiSN[2 x, m] / 2, x], {x, 0, 2 n}, {m, 0, k}]; (* Michael Somos, May 06 2017 *)
PROG
(PARI) T(n, k) = binomial(2*k, k) * binomial(2*n-2*k, n-k) /* Michael Somos, Aug 20 2005 */
(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
CROSSREFS
KEYWORD
AUTHOR
Henry Bottomley, Feb 07 2002
STATUS
approved