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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. 1

%I #17 Oct 25 2022 08:51:41

%S 1,3,7,18,46,120,316,840,2248,6048,16336,44256,120160,326784,889792,

%T 2424960,6613120,18043392,49247488,134450688,367134208,1002645504,

%U 2738510848,7480215552,20433258496,55818559488,152486858752

%N 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 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 Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-floretion.html">Sequences Related to Floretions</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-4).

%F 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 (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 G.f.: ( 1-x-3*x^2 )*x / ( (2*x-1)*(2*x^2+2*x-1) ). - _R. J. Mathar_, Sep 11 2019

%F 4*a(n) = 2^n + 2*A026150(n). - _R. J. Mathar_, Oct 25 2022

%Y First differences of A038508.

%K easy,nonn

%O 1,2

%A _Herbert Kociemba_, Jun 02 2004

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