OFFSET
0,2
COMMENTS
Conjecture: For k>1, the interlaced polynomials b(2*k-1) = a(k)/7 and b(2*k) = (a(k+1) - a(k)) / 7 are the Fibonacci numbers (A000045). - Avi Friedlich, May 25 2015
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Boothby, T.; Burkert, J.; Eichwald, M.; Ernst, D. C.; Green, R. M.; Macauley, M. On the cyclically fully commutative elements of Coxeter groups, J. Algebr. Comb. 36, No. 1, 123-148 (2012), Table 1 CFC Type H.
J. H. Conway and N. J. A. Sloane, Sphere Packings, Lattices and Groups, Springer, 2nd ed., 1993.
C. Series and D. Wright, Non-Euclidean geometry and Indra's pearls, Plus magazine, Jul 12 2011, (see Fig 1a)
A. Stakhov and A. S. Aranson, Hyperbolic Fibonacci and Lucas Functions, Applied Mathematics, 2(1); 2011. - Avi Friedlich, May 28 2015
Wikipedia, Uniform tiling 73
Index entries for linear recurrences with constant coefficients, signature (3,-1).
FORMULA
a(n+1) = 3*a(n) - a(n-1).
a(n) = 7*A001906(n), n>0.
G.f.: (1 + 4*x + x^2)/(1 - 3*x + x^2). - Colin Barker, Apr 14 2012
a(0)=1, and a(n) = F(2*n+3) + 2*L(2*n-1) for n>0, where F(n) is the n-th Fibonacci number and L(n) is the n-th Lucas number. - Rigoberto Florez, Jul 30 2019
EXAMPLE
G.f. = 1 + 7*x + 21*x^2 + 56*x^3 + 147*x^4 + 385*x^5 + 1008*x^6 + ...
MATHEMATICA
CoefficientList[Series[(1+4*x+x^2)/(1-3*x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 15 2012 *)
a[ n_] := Boole[n == 0] + 7 Fibonacci[2 n]; (* Michael Somos, Jun 07 2015 *)
Table[If[n == 0, 1, Fibonacci[2*n+3] + 2*LucasL[2*n-1]], {n, 0, 20}] (* Rigoberto Florez, Jul 30 2019 *)
LinearRecurrence[{3, -1}, {1, 7, 21}, 30] (* Harvey P. Dale, Oct 24 2020 *)
PROG
(PARI) {a(n) = (n==0) + 7 * fibonacci(2*n)}; /* Michael Somos, Jun 07 2015 */
(Magma) [1] cat [Fibonacci(2*n+3)+2*Lucas(2*n-1):n in [1..30]]; // Marius A. Burtea, Jul 31 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved