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%I #30 Jul 25 2021 19:28:59
%S 1,0,1,2,1,6,7,12,31,40,91,170,281,602,1051,1988,3907,7044,13735,
%T 25962,48643,94094,177145,338184,647791,1228812,2356927,4500678,
%U 8595913,16486966,31521543,60419872,115870879,222045160,426275647,818054654
%N Number of weighted lattice paths of weight n having no (1,0)-steps of weight 1.
%C L_n is the set of lattice paths of weight n that start at (0,0) end on the horizontal axis and whose steps are of the following four kinds: an (1,0)-step with weight 1; an (1,0)-step with weight 2; a (1,1)-step with weight 2; a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.
%D M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
%D E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177.
%F a(n) = A182882(n,0).
%F G.f.: 1/sqrt(1-2*z^2+z^4-4*z^3).
%F It appears that a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k,k)*binomial(k,n-2*k): this gives correct values for a(0) through a(35). If true, then sequence equals antidiagonal sums of triangle A105868. - _Peter Bala_, Mar 06 2013
%F D-finite n*a(n) = (2*n - 2)*a(n-2) + (4*n - 6)*a(n-3) - (n - 2)*a(n-4), follows easily by differentiating the o.g.f. Maple's sumrecursion command verifies that Sum_{k = 0..floor(n/2)} binomial(n-k,k)*binomial(k,n-2*k) satisfies the same recurrence with the same initial conditions thus proving the above conjecture. - _Peter Bala_, Feb 07 2017
%F a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*hypergeom([-k, k-n, k-n], [1, -n], -1). - _Peter Luschny_, Feb 13 2018
%e a(3)=2. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and u=(1,1), d=(1,-1), the five paths of weight 3 are ud, du, hH, Hh, and hhh; two of them, namely ud and du, contain no h steps.
%p G:=1/sqrt(1-2*z^2+z^4-4*z^3): Gser:=series(G,z=0,40): seq(coeff(Gser,z,n), n=0..35);
%p # Alternatively (after Bala):
%p seq(add(binomial(n-k,k)*binomial(k,n-2*k), k=ceil(n/3)...floor(n/2)),n=0..35); # _Peter Luschny_, Feb 07 2017
%p # With natural summation bound:
%p a := n -> add((-1)^k*binomial(n,k)*hypergeom([-k,k-n,k-n], [1,-n], -1), k=0..n): seq(simplify(a(n)), n=0..35); # _Peter Luschny_, Feb 13 2018
%t CoefficientList[Series[1/Sqrt[1-2x^2+x^4-4x^3],{x,0,40}],x] (* _Harvey P. Dale_, Oct 16 2011 *)
%Y Cf. A182882. A105868.
%K nonn,easy
%O 0,4
%A _Emeric Deutsch_, Dec 11 2010
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