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A025172 Let phi = arccos(1/3), the dihedral angle of the regular tetrahedron. Then cos(n*phi) = a(n)/3^n.

%I A025172 #61 Feb 20 2025 06:41:39
%S A025172 1,1,-7,-23,17,241,329,-1511,-5983,1633,57113,99529,-314959,-1525679,
%T A025172 -216727,13297657,28545857,-62587199,-382087111,-200889431,3037005137,
%U A025172 7882015153,-11569015927,-94076168231,-84031193119,678623127841,2113526993753
%N A025172 Let phi = arccos(1/3), the dihedral angle of the regular tetrahedron. Then cos(n*phi) = a(n)/3^n.
%C A025172 Used when showing that the regular simplex is not "scisssors-dissectible" to a cube, thus answering Hilbert's third problem.
%C A025172 From _Peter Bala_, Apr 01 2018: (Start)
%C A025172 This sequence is (1/2) * the Lucas sequence V(n,2,9). The companion Lucas sequence U(n,2,9) is A127357.
%C A025172 Define a binary operation o on rational numbers by x o y = (x + y)/(1 - 2*x*y). This is a commutative and associative operation with identity 0. Then 2 o 2 o ... o 2 (n terms) = 2*A127357(n-1)/A025172(n). Cf. A088137 and A087455. (End)
%H A025172 Vincenzo Librandi, <a href="/A025172/b025172.txt">Table of n, a(n) for n = 0..1000</a>
%H A025172 J. L. Dupont, <a href="http://www.worldscientific.com/worldscibooks/10.1142/4598">Scissors Congruences, Group Homology and Characteristic Classes</a>, World Scientific, 2001. See p. 4.
%H A025172 Wikipedia, <a href="https://en.wikipedia.org/wiki/Lucas_sequence">Lucas sequence</a>
%H A025172 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-9).
%F A025172 a(0) = 1, a(1) = 1; for n >= 2, a(n) = 2*a(n-1) - 9*a(n-2). - _Warut Roonguthai_, Oct 11 2005
%F A025172 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
%F A025172 a(n) is the permanent of the matrix M^n, where M = [i, 2; 1, i]. - _Simone Severini_, Apr 27 2007
%F A025172 a(n) = Product_{i=1..n} (2 - tan((i-1/2)*Pi/(2*n))^2). - _Gerry Martens_, May 26 2011
%F A025172 G.f.: (1-x)/(1-2*x+9*x^2). - _Colin Barker_, Jun 21 2012
%F A025172 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
%F A025172 E.g.f.: exp(x)*cos(2*sqrt(2)*x). - _Vladimir Reshetnikov_, Oct 15 2016
%F A025172 a(n) = A127357(n)-A127357(n-1). - _R. J. Mathar_, Apr 07 2022
%p A025172 f:=proc(n) option remember; if n <= 1 then RETURN(1); fi; 2*f(n-1)-9*f(n-2); end;
%t A025172 Table[ n/2 3^n GegenbauerC[ n, 1/3 ], {n, 24} ]
%t A025172 CoefficientList[Series[(1 - x)/(1 - 2 x + 9 x^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, Oct 17 2013 *)
%t A025172 LinearRecurrence[{2,-9},{1,1},30] (* _Harvey P. Dale_, Jan 30 2016 *)
%o A025172 (PARI) {a(n)= if(n<0, 0, 3^(n-1)* subst(3* poltchebi(abs(n)), x, 1/3))} /* _Michael Somos_, Mar 14 2007 */
%Y A025172 Cf. A088137, A087455, A127357.
%K A025172 sign,easy
%O A025172 0,3
%A A025172 _Wouter Meeussen_
%E A025172 Better description from _Vladeta Jovovic_, Apr 19 2003
%E A025172 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.