A094806 Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 5.

1, 5, 20, 74, 264, 924, 3200, 11016, 37792, 129392, 442496, 1512224, 5165952, 17643456, 60250112, 205729920, 702452224, 2398414592, 8188884992, 27958972928, 95458646016, 325917686784, 1112755552256, 3799191029760, 12971261403136, 44286680330240

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))^(2n) 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

Harvey P. Dale, Table of n, a(n) for n = 2..1000

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 (6,-10,4).

Formula

a(n) = (1/4)*Sum_{k=1..7} sin(Pi*k/8)*sin(5*Pi*k/8)*(2*cos(Pi*k/8))^(2n).

a(n) = 6*a(n-1) - 10*a(n-2) + 4*a(n-3).

G.f.: x^2*(x-1) / ( (2*x-1)*(2*x^2-4*x+1) ).

a(n) = (-2^n+(-(2-sqrt(2))^n+(2+sqrt(2))^n)/sqrt(2))/4. - Colin Barker, Apr 27 2016

4*a(n) = 2*A007070(n-1) - 2^n.- R. J. Mathar, Nov 14 2019

Mathematica

f[n_] := FullSimplify[ TrigToExp[(1/4)Sum[ Sin[Pi*k/8]Sin[5Pi*k/8](2Cos[Pi*k/8])^(2n), {k, 1, 7}]]]; Table[ f[n], {n, 2, 25}] (* Robert G. Wilson v, Jun 18 2004 *)
LinearRecurrence[{6, -10, 4}, {1, 5, 20}, 30] (* Harvey P. Dale, Mar 04 2015 *)

Crossrefs

Sequence in context: A316222 A392266 A273718 * A289596 A026639 A248326

Adjacent sequences: A094803 A094804 A094805 * A094807 A094808 A094809

Keyword

nonn,easy

Author

Herbert Kociemba, Jun 11 2004

Status

approved