0,3

a(n) has the same number of factors of 2 as does A000108(n) (Catalan numbers). "Given a sequence of integers b = (b_0,b_1,b_2,...) one gives a Dyck path P of length 2n the weight wt(P) = b_{h_1} b_{h_2} ... b_{h_n}, where h_i is the height of the i-th ascent of P. The corresponding weighted Catalan number is C_n^b = sum_P wt(P), where the sum is over all Dyck paths of length 2n. So in particular, the ordinary Catalan numbers C_n correspond to b_i = 1 for all i >= 0. Let xi(n) stand for the base two exponent of n, i.e., the largest power of 2 dividing n. We give a condition on b which implies that xi(C_n^b) = xi(C_n). In the special case b_i=(2i+1)^2, this settles a conjecture of Postnikov about the number of plane Morse links." - Alexander Postnikov, Bruce Sagan.

Alexander Postnikov, Bruce Sagan, What power of two divides a weighted Catalan number?

O.g.f.: A(x) = 1/(1-x/(1-3^2*x/(1-5^2*x/(1-.../(1 - (2*n-1)^2*x/(1-... )))))) (continued fraction).

(PARI) {a(n)=local(CF=1/(1-(2*n+1)^2*x+x*O(x^n))); if(n==0, CF=1, for(i=1, n, CF=1/(1-(2*(n-i)+1)^2*x*CF))); polcoeff(CF, n)}

Sequence in context: A119297 A048345 A064343 this_sequence A113082 A046747 A006426

Adjacent sequences: A127820 A127821 A127822 this_sequence A127824 A127825 A127826

nonn

Paul D. Hanna (pauldhanna(AT)juno.com), Jan 30 2007