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Solution: Catalan numbers

                                          429
                                    132   429
                               42   132   297
                         14    42    90   165
                    5    14    28    48    75
              2     5     9    14    20    27
        1     2     3     4     5     6     7
  1     1     1     1     1     1     1     1

  This are the Narayana numbers. 
  On the diagonal are the Catalan numbers.

  Recursion for the Narayana numbers:

    N(x,y) = N(x-1,y) + N(x,y-1)

  An explicit formular can be found by
  the reflection principle of Andre (see [C1]).

The answer of: sequences@research.att.com

%I A0108 M1460 N0577
%S A0108 1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,
%T A0108 9694845,35357670,129644790,477638700,1767263190,6564120420,24466267020
%N A0108 Catalan numbers: $C(n)^=^C(2n,n)/(n+1)$.
%R A0108 AMM 72 973 65. RCI 101. C1 53. PLC 2 109 71. MAG 61 211 88.
%K A0108 core,nonn
%F A0108 roman {G.f.:}~~{ 1 ~-~ ( 1 ~-~ 4^x) sup {1/2} } over {2^x} .
%O A0108 0,3
%p A0108 f:=n->binomial(2*n,n)/(n+1);
%B A0108 A0108.
%A A0108 njas
%Z A0108 FLAG.

[AMM] = { American Mathematical Monthly}.
[C1]  = L. Comtet, { Advanced Combinatorics}, Reidel, Dordrecht, Holland, 1974.
[MAG] = { The Mathematical Gazette}.
[RCI] = J. Riordan, { Combinatorial Identities}, Wiley, NY, 1968.
[PLC] = Volume 2: { Proceedings of the Second Louisiana Conference on Combinatorics,
        Graph Theory and Computing}, edited R. C. Mullin et\ al., 1971.