login
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.
1

%I #11 Aug 23 2024 10:48:48

%S 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 140,252,924,429,264,252,280,350,504,924,3432,1430,858,792,840,980,

%U 1260,1848,3432,12870,4862,2860,2574,2640,2940,3528,4620,6864,12870,48620

%N 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 Sum of entries in row n is (1/2)binomial(2n+2,n+1) = A001700(n).

%C T(n,0) = binomial(2n,n)/(n+1) = A000108(n) (the Catalan numbers).

%C T(n,n) = binomial(2n,n) = A000984(n) (the central binomial coefficients).

%F T(n,k) = binomial(2k,k)*binomial(2n-2k,n-k)/(n-k+1) (0 <= k <= n).

%F G.f. = G(t,x) = (1-sqrt(1-4x))/(2x*sqrt(1-4tx)).

%e Triangle starts:

%e 1;

%e 1, 2;

%e 2, 2, 6;

%e 5, 4, 6, 20;

%e 14, 10, 12, 20, 70;

%p 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 T[n_, k_] /; 0 <= k <= n := Binomial[2k, k]*Binomial[2n - 2k, n-k]/(n-k+1);

%t Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Aug 23 2024 *)

%Y Cf. A000108, A000984, A001700.

%K nonn,tabl

%O 0,3

%A _Emeric Deutsch_, Nov 22 2008