cp's OEIS Frontend

A094297 Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2.

%I A094297 #17 Oct 25 2022 08:51:41
%S A094297 1,3,7,18,46,120,316,840,2248,6048,16336,44256,120160,326784,889792,
%T A094297 2424960,6613120,18043392,49247488,134450688,367134208,1002645504,
%U A094297 2738510848,7480215552,20433258496,55818559488,152486858752
%N A094297 Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2.
%C A094297 In general, a(n,m,j,k) = (2/m)*Sum_{r=1..m-1) sin(j*r*Pi/m)*sin(k*r*Pi/m)*(1+2*cos(Pi*r/m))^n is the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < m and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = j, s(n) = k.
%H A094297 Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-floretion.html">Sequences Related to Floretions</a>
%H A094297 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-4).
%F A094297 a(n) = (1/3)*Sum_{k=1..5} sin(Pi*k/3)^2*(1+2*cos(Pi*k/6))^n or a(n) = (2^n + (1-sqrt(3))^n + (1 + sqrt(3))^n)/4.
%F A094297 (a(n)) seems to be given by tesseq(- 2'i + 2'j + 2'k - 2i' + 2j' + 2k' - 2'ii' + 2'jj' - 'kk' - 2.5'ik' - 1.5'jk' - 2.5'ki' - 1.5'kj' - e) (disregarding signs) - _Creighton Dement_, Nov 17 2004
%F A094297 G.f.: ( 1-x-3*x^2 )*x / ( (2*x-1)*(2*x^2+2*x-1) ). - _R. J. Mathar_, Sep 11 2019
%F A094297 4*a(n) = 2^n + 2*A026150(n). - _R. J. Mathar_, Oct 25 2022
%Y A094297 First differences of A038508.
%K A094297 easy,nonn
%O A094297 1,2
%A A094297 _Herbert Kociemba_, Jun 02 2004