A094832 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) = 3, s(2n+1) = 4.

1, 3, 10, 34, 117, 406, 1417, 4965, 17443, 61390, 216318, 762841, 2691574, 9500193, 33539833, 118428835, 418214706, 1476968554, 5216307805, 18423344550, 65070265609, 229827800509, 811757757123, 2867166603766, 10127007608998

Offset: 0

Author

Herbert Kociemba, Jun 13 2004

Keywords

• nonn

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 = 0..1824
• Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
• Index entries for linear recurrences with constant coefficients, signature (6,-9,1).

Crossrefs

Cf. A094833.

Programs

Mathematica

LinearRecurrence[{6,-9,1},{1,3,10},30] (* Harvey P. Dale, May 18 2011 *)

Formula

a(n+1) = 3*a(n) + A094833(n-1). - Philippe Deléham, Mar 18 2007

a(n) = (2/9)*Sum_{r=1..8} sin(r*Pi/3)*sin(4*r*Pi/9)*(2*cos(r*Pi/9))^(2n).

a(n) = 6a(n-1) - 9a(n-2) + a(n-3).

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

a(n) = A094833(n+2) - 3*A094833(n+1). - Philippe Deléham, Mar 18 2007