A094828 Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 9 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 5.

1, 5, 20, 75, 274, 988, 3536, 12597, 44745, 158632, 561683, 1987154, 7026408, 24835744, 87763945, 310088381, 1095490524, 3869911659, 13670143618, 48287147300, 170561502896, 602454835293, 2127962632993, 7516243783216

Offset

2,2

Comments

In general, a(n) = (2/m)*Sum_{r=1..m-1} sin(r*j*Pi/m)*sin(r*k*Pi/m)*(2*cos(r*Pi/m))^(2*n) counts (s(0), s(1), ..., s(2n)) such that 0 < s(i) < m and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = j, s(2n) = k.

Links

Michael De Vlieger, Table of n, a(n) for n = 2..1826

László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.

Index entries for linear recurrences with constant coefficients, signature (7,-15,10,-1).

Formula

a(n) = (2/9)*Sum_{r=1..8} sin(r*Pi/9)*sin(5*r*Pi/9)*(2*cos(r*Pi/9))^(2*n).

a(n) = 7*a(n-1) - 15*a(n-2) + 10*a(n-3) - a(n-4).

G.f.: -x^2*(-1+2*x) / ( (x-1)*(x^3-9*x^2+6*x-1) ).

a(n) = A094256(n-1) - 2*A094256(n-2). - R. J. Mathar, Nov 14 2019

3*a(n) = A094829(n+2) -5*A094829(n+1)+7*A094829(n)-1. - R. J. Mathar, Nov 14 2019

Mathematica

LinearRecurrence[{7, -15, 10, -1}, {1, 5, 20, 75}, 30] (* Harvey P. Dale, Apr 27 2020 *)

Crossrefs

Sequence in context: A248326 A022633 A092490 * A030191 A093131 A224422

Adjacent sequences: A094825 A094826 A094827 * A094829 A094830 A094831

Keyword

nonn,easy

Author

Herbert Kociemba, Jun 13 2004

Status

approved