A094827 Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 9 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = 1, s(2n+1) = 4.

1, 4, 14, 48, 165, 571, 1988, 6953, 24396, 85786, 302104, 1064945, 3756519, 13256712, 46796545, 165225380, 583440086, 2060408640, 7276716445, 25700060995, 90770326604, 320598127113, 1132355884236, 3999522488002

Offset: 1

Author

Herbert Kociemba, Jun 13 2004

Keywords

• nonn
• easy

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))^(2n+1) counts (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < m and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = j, s(2n+1) = k.

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..1825
• 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).

Programs

Mathematica

LinearRecurrence[{7,-15,10,-1},{1,4,14,48},30] (* Harvey P. Dale, Jul 09 2020 *)

Formula

a(n) = (2/9)*Sum_{r=1..8} sin(r*Pi/9)*sin(4*r*Pi/9)*(2*cos(r*Pi/9))^(2*n+1).

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

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

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