1,1

Row sums yield A027307. T(n,n) = A108424(n).

Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

G.f.: tzA(z)A(tz)+tzA(z)A^2(tz), where A=1+zA^2+zA^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).

T(n, k) = A108424(k)*A027307(n-k) (there are explicit formulas in A108424 and A027307).

T(2,1)=4 because we have u(d)ud, u(d)Udd, Ud(d)ud and Ud(d)Udd, the d step of the first return being shown between parentheses.

Triangle begins:

2;

4,6;

20,12,34;

132,60,68,238;

...

a:=n->sum(2^(i+1)*binomial(2*n, i)*binomial(n, i+1), i=0..n-1)/n: b:=proc(n) if n=1 then 2 else (n*2^n*binomial(2*n, n)/((2*n-1)*(n+1)))*sum(binomial(n-1, j)^2/2^j/binomial(n+j+1, j), j=0..n-1): fi end: T:=proc(n, k) if k=n then b(n) else b(k)*a(n-k) fi end:for n from 1 to 9 do seq(T(n, k), k=1..n) od; > # yields sequence in triangular form

Cf. A027307, A108424, A108435.

Sequence in context: A185185 A234019 A279188 * A245766 A193774 A241210

Adjacent sequences: A108436 A108437 A108438 * A108440 A108441 A108442

nonn,tabl

Emeric Deutsch, Jun 05 2005

approved