1,2

Number of lattice paths from (0,0) to the line x=n-1 that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1), and three types of steps H=(1,0) (left factors of 3-Motzkin steps). Example: a(3)=17 because we have UD, UU, 9 HH paths, 3 HU paths, and 3 UH paths. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jan 22 2004

Also a(n) = number of integer strings s(0), ..., s(n) counted by array U in A026386 that have s(n)=1; a(n) = U(2n-1, n-1).

The Hankel transform of [1,1,4,17,75,339,1558,...] is [1,3,8,21,55,144,377,...] (see A001906) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Apr 13 2007

Number of peaks in all skew Dyck paths of semilength n. A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down), and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps. Example: a(2)=4 because in the 3 (=A002212(2)) skew Dyck paths (UD)(UD), U(UD)D, and U(UD)L we have altogether 4 peaks (shown between parentheses). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 25 2007

Hankel transform of this sequence gives A000012 = [1,1,1,1,1,1,...] . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 24 2007

E. Deutsch, E. Munarini, and S. Rinaldi, Skew Dyck paths (in preparation).

J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.

G.f.: (1/2)/(5*x^2-x)*(1-5*x-(1-6*x+5*x^2)^(1/2)). E.g.f.: exp(3*x)*(BesselI(0, 2*x)+BesselI(1, 2*x)). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Oct 03 2003

G.f.= [(1-z)/sqrt(1-6z+5z^2)-1]/2 = z + 4z^2 + 17z^3 + ... - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jan 22 2004

a(n) = coefficient of t^n in (1+t)(1+3t+t^2)^(n-1). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jan 30 2004

a(n)=A026380(2n-2). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 18 2004

a(n)=[2(3n-2)a(n-1)-5(n-2)a(n-2)]/n for n>=2; a(0)=0, a(1)=1. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 18 2004

a(n+1) = sum(k=0, n, binomial(n, k)*sum(i=0, k, binomial(k+i, i))) - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 06 2004

a(n+1) = sum(k=0, n, binomial(n, k)*binomial(2*k+1, k+1)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 06 2004

a(n)=Sum(k*A126182(n-1,k-1),k=1..n). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 25 2007

Cf. A002212, A026375.

Cf. A026380.

Half the values of A026387. Bisection of A026380 and A026392.

Cf. A126182.

Sequence in context: A049027 A026751 A081568 this_sequence A117439 A081910 A026773

Adjacent sequences: A026375 A026376 A026377 this_sequence A026379 A026380 A026381

nonn

Clark Kimberling (ck6(AT)evansville.edu)