cp's OEIS Frontend

A024998 a(n) = number of (s(0), s(1), ..., s(n)) such that every s(i) is an integer, s(0) = 0, |s(i) - s(i-1)| = 1 for i = 1,2; |s(i) - s(i-1)| <= 1 for i >= 3, s(n) = 1. Also a(n) = T(n,n-1), where T is the array defined in A024996.

%I A024998 #14 Feb 14 2014 13:20:33
%S A024998 1,0,3,6,19,52,150,428,1232,3552,10275,29790,86559,251980,734773,
%T A024998 2145822,6275145,18373296,53856153,158025186,464112297,1364247180,
%U A024998 4013353932,11815188000,34807249134,102606325136,302646363725,893175905778
%N A024998 a(n) = number of (s(0), s(1), ..., s(n)) such that every s(i) is an integer, s(0) = 0, |s(i) - s(i-1)| = 1 for i = 1,2; |s(i) - s(i-1)| <= 1 for i >= 3, s(n) = 1. Also a(n) = T(n,n-1), where T is the array defined in A024996.
%C A024998 Comments from _David Callan_, Jul 15 2004: "a(n+1) is the number of Motzkin (2n)-paths whose longest level segment has length n. (A level segment is a maximal sequence of contiguous flatsteps.) Example: with n=3, the paths counted by a(4) are FFFUDF, FFFUFD, FUDFFF, FUFFFD, UFDFFF, UFFFDF. Here is a bijection to the sequences (s(i)) above.
%C A024998 "Given such a Motzkin (2n)-path, delete the (unique) longest level segment to split the path into A,B. Form the path BUA (U can be recovered as the up step immediately following the rightmost of the lowest points on this path). This path will not start or end with F.
%C A024998 "Transfer the level segment (if any) following the first step to the end. Code the resulting path with 1 for U, 0 for F and -1 for D. Then take partial sums (including the empty sum) to get a sequence (s(i)). Example: UF^9UFFUDDDF -> U,UFFUDDDF -> UFFUDDDFUU -> UUDDDFUUFF -> 1,1,-1,-1,-1,0,1,1,0,0 -> (0,1,2,1,0,-1,-1,0,1,1,1)."
%H A024998 Vincenzo Librandi, <a href="/A024998/b024998.txt">Table of n, a(n) for n = 1..200</a>
%F A024998 a(n+1)=sum(j=1..n/2, C[n-1, 2j-1] C[2j+1, j]). This sum counts the Motzkin (2n)-paths above by number j of up steps and the sequences (s(i)) by number j of indices i for which s(i) - s(i-1) = -1. GF: (1-x)^2 ( (1-x)(1-2x-3x^2)^(-1/2) - 1 )/(2x). - _David Callan_, Jul 15 2004
%F A024998 Conjecture: (n+1)*a(n) +(-3*n+2)*a(n-1) +(-n-7)*a(n-2) +3*(n-4)*a(n-3)=0. - _R. J. Mathar_, Jun 23 2013
%F A024998 a(n) ~ 2 * 3^(n+1/2) / (9 * sqrt(Pi*n)). - _Vaclav Kotesovec_, Feb 13 2014
%t A024998 Rest[CoefficientList[Series[(1-x)^2*((1-x)*(1-2*x-3*x^2)^(-1/2)-1)/(2*x), {x, 0, 20}], x]] (* _Vaclav Kotesovec_, Feb 13 2014 *)
%K A024998 nonn
%O A024998 1,3
%A A024998 _Clark Kimberling_