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A067298
Generalized Catalan triangle, based on C(2,2; n) := A064340(n).
5
1, 1, 2, 4, 5, 9, 28, 32, 36, 64, 256, 284, 300, 328, 584, 2704, 2960, 3072, 3184, 3440, 6144, 31168, 33872, 34896, 35680, 36704, 39408, 70576, 380608, 411776, 422592, 429760, 436928, 447744, 478912
OFFSET
0,3
COMMENTS
For corresponding Catalan triangle with C(1,1; n) := A000108(n) see A028364.
Identity for each row n>=1: a(n,m)+a(n,n-(m+1))= a(n,n) = A067297(n) for m=0..floor((n-1)/2.). E.g., a(2k+1,k)= A067297(2*k+1)/2.
The columns (without leading zeros) give for m=0..3: A064340, A067299, 3*A067300, 8*A067301. The main diagonal gives A067297. The row sums give A067302.
FORMULA
a(n, m)= sum(C(2, 2; j)C(2, 2; n-j), j=0..m) if n>=m>=0 else 0.
G.f. for column m (without leading zeros): (c(m, x)*c(2, 2; x)-c2(m-1, x))/x^m, with c(2, 2; x)= (1-3*x*c(4*x))/(1-2*x*c(4*x))^2 (g.f. for C(2, 2; n)), c(x) g.f. for Catalan numbers A000108, c(m, x) := sum(C(2, 2; n)*x^n, n=0..m) and c2(m, x) := sum(A067297(n)*x^n, n=0..m) for m=0, 1, 2, ...
EXAMPLE
{1}; {1,2}; {4,5,9}; {28,32,36,64}; ...
CROSSREFS
Sequence in context: A073151 A279786 A255515 * A077389 A122991 A245512
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, Feb 05 2002
STATUS
approved