OFFSET
0,3
COMMENTS
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.
LINKS
Yibo Gao, Andrew Gu, Arithmetic of weighted Catalan numbers, arXiv:1908.03914 [math.CO], 2019.
Alexander Postnikov, Bruce Sagan, What power of two divides a weighted Catalan number?, arXiv:math/0601339 [math.CO], 2006.
Sarah Shader, Weighted Catalan Numbers and Their Divisibility Properties, Research Science Institute, MIT, 2014.
FORMULA
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).
G.f.: 1/Q(0), where Q(k) = 1 - (2*k+1)^2*x/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 17 2013
a(n) = the upper left term in M^n where M is an infinite square production matrix; M[i,j] = A016754(i-1) = (2*i-1)^2, i >= 1 and 1 <= j <= i+1, and M[i,j] = 0, i >= 1 and j >= i+2, see the examples. - Gary W. Adamson, Jul 18 2011
G.f.: Q(0), where Q(k) = 1 - x*(4*k+1)^2/( x*(4*k+1)^2 - 1/(1 - x*(4*k+3)^2/( x*(4*k+3)^2 - 1/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Oct 09 2013
a(n) ~ 2^(6*n + 1) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n + 1/2)). - Vaclav Kotesovec, Aug 26 2017
EXAMPLE
From Gary W. Adamson, Jul 18 2011: (Start)
The first few rows of matrix M are:
1, 1, 0, 0, 0, ...
9, 9, 9, 0, 0, ...
25, 25, 25, 25, 0, ...
49, 49, 49, 49, 49, ...
(End)
MAPLE
nmax:=13: M := Matrix(1..nmax+1, 1..nmax+1): for i from 1 to nmax do for j from 1 to i+1 do M[i, j]:= (2*i-1)^2 od: od: for n from 0 to nmax do B:=M^n: a(n):=B[1, 1] od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jul 21 2011
MATHEMATICA
nmax = 20; CoefficientList[Series[1/Fold[(1 - #2/#1) &, 1, Reverse[(2*Range[nmax + 1]-1)^2*x]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 26 2017 *)
PROG
(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)}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 30 2007
STATUS
approved