OFFSET
0,3
COMMENTS
Used when showing that the regular simplex is not "scisssors-dissectable" to a cube, thus answering Hilbert's third problem.
From Peter Bala, Apr 01 2018: (Start)
This sequence is (1/2) * the Lucas sequence V(n,2,9). The companion Lucas sequence U(n,2,9) is A127357.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
J. L. Dupont, Scissors Congruences, Group Homology and Characteristic Classes, World Scientific, 2001. See p. 4.
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (2,-9).
FORMULA
a(0) = 1, a(1) = 1; for n >= 2, a(n) = 2*a(n-1) - 9*a(n-2). - Warut Roonguthai, Oct 11 2005
a(n) = (1/2)*(1-2*i*2^(1/2))^n+(1/2)*(1+2*i*2^(1/2))^n, where i=sqrt(-1). - Vladeta Jovovic, Apr 19 2003
a(n) is the permanent of the matrix M^n, where M = [i, 2; 1, i]. - Simone Severini, Apr 27 2007
a(n) = Product_{i=1..n} (2 - tan((i-1/2)*Pi/(2*n))^2). - Gerry Martens, May 26 2011
G.f.: (1-x)/(1-2*x+9*x^2). - Colin Barker, Jun 21 2012
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(8*k+1)/(x*(8*k+9) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
E.g.f.: exp(x)*cos(2*sqrt(2)*x). - Vladimir Reshetnikov, Oct 15 2016
MAPLE
f:=proc(n) option remember; if n <= 1 then RETURN(1); fi; 2*f(n-1)-9*f(n-2); end;
MATHEMATICA
Table[ n/2 3^n GegenbauerC[ n, 1/3 ], {n, 24} ]
CoefficientList[Series[(1 - x)/(1 - 2 x + 9 x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 17 2013 *)
LinearRecurrence[{2, -9}, {1, 1}, 30] (* Harvey P. Dale, Jan 30 2016 *)
PROG
(PARI) {a(n)= if(n<0, 0, 3^(n-1)* subst(3* poltchebi(abs(n)), x, 1/3))} /* Michael Somos, Mar 14 2007 */
CROSSREFS
KEYWORD
sign,easy
AUTHOR
EXTENSIONS
Better description from Vladeta Jovovic, Apr 19 2003
Edited by N. J. A. Sloane, Feb 22 2007. Among other things, I changed the offset and the beginning of the sequence, so some of the formulas may need to be adjusted slightly.
STATUS
approved