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A145890 Triangle read by rows: T(n,k) = B(k)C(n-k), where B(j) is the central binomial coefficient binomial(2j,j) (A000984) and C(j) is the Catalan number binomial(2j,j)/(j+1) (A000108); 0 <= k <= n.

%I A145890 #11 Aug 23 2024 10:48:48
%S A145890 1,1,2,2,2,6,5,4,6,20,14,10,12,20,70,42,28,30,40,70,252,132,84,84,100,
%T A145890 140,252,924,429,264,252,280,350,504,924,3432,1430,858,792,840,980,
%U A145890 1260,1848,3432,12870,4862,2860,2574,2640,2940,3528,4620,6864,12870,48620
%N A145890 Triangle read by rows: T(n,k) = B(k)C(n-k), where B(j) is the central binomial coefficient binomial(2j,j) (A000984) and C(j) is the Catalan number binomial(2j,j)/(j+1) (A000108); 0 <= k <= n.
%C A145890 Sum of entries in row n is (1/2)binomial(2n+2,n+1) = A001700(n).
%C A145890 T(n,0) = binomial(2n,n)/(n+1) = A000108(n) (the Catalan numbers).
%C A145890 T(n,n) = binomial(2n,n) = A000984(n) (the central binomial coefficients).
%F A145890 T(n,k) = binomial(2k,k)*binomial(2n-2k,n-k)/(n-k+1) (0 <= k <= n).
%F A145890 G.f. = G(t,x) = (1-sqrt(1-4x))/(2x*sqrt(1-4tx)).
%e A145890 Triangle starts:
%e A145890    1;
%e A145890    1,  2;
%e A145890    2,  2,  6;
%e A145890    5,  4,  6, 20;
%e A145890   14, 10, 12, 20, 70;
%p A145890 b:=proc(n) options operator, arrow: binomial(2*n,n) end proc: c:=proc(n) options operator, arrow: binomial(2*n, n)/(n+1) end proc: T:=proc(n,k) if k <= n then b(k)*c(n-k) else 0 end if end proc: for n from 0 to 8 do seq(T(n,k), k =0..n) end do; # yields sequence in triangular form
%t A145890 T[n_, k_] /; 0 <= k <= n := Binomial[2k, k]*Binomial[2n - 2k, n-k]/(n-k+1);
%t A145890 Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Aug 23 2024 *)
%Y A145890 Cf. A000108, A000984, A001700.
%K A145890 nonn,tabl
%O A145890 0,3
%A A145890 _Emeric Deutsch_, Nov 22 2008