cp's OEIS Frontend

A198636 One half of total number of round trips, each of length 2n, on the graph P_6 (o-o-o-o-o-o).

%I A198636 #90 Feb 11 2018 10:49:09
%S A198636 3,5,13,38,117,370,1186,3827,12389,40169,130338,423065,1373466,
%T A198636 4459278,14478659,47011093,152642789,495626046,1609284589,5225309458,
%U A198636 16966465802,55089756851,178875298901,580804419201,1885860059450,6123349080945
%N A198636 One half of total number of round trips, each of length 2n, on the graph P_6 (o-o-o-o-o-o).
%C A198636 See the array and triangle A198632 for the general graph P_N case (there N is n and the length is l=2*k).
%H A198636 M. F. Hasler, <a href="/A198636/b198636.txt">Table of n, a(n) for n = 0..499</a>
%H A198636 S. Barbero, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Barbero/barbero15.html">Dickson Polynomials, Chebyshev Polynomials, and Some Conjectures of Jeffery</a>, Journal of Integer Sequences, 17 (2014), #14.3.8.
%H A198636 S. Barbero, U. Cerruti, N. Murru, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Barbero/barbero11.html">Identities Involving Zeros of Ramanujan and Shanks Cubic Polynomials</a>, J. Integer Seq., Vol. 16 (2013), Article 13.8.1, pp. 10-12.
%H A198636 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-6,1).
%F A198636 a(n) = w(6,2*n)/2, n>=0, with w(6,l) the total number of closed walks on the graph P_6 (the simple path with 6 points (vertices) and 5 lines (or edges)).
%F A198636 O.g.f. for w(6,l) (with zeros for odd l): y*(d/dy)S(6,y)/S(6,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See also A198632 for a rewritten form.
%F A198636 O.g.f.: (3-10*x+6*x^2)/(1-5*x+6*x^2-x^3). - _Colin Barker_, Jan 02 2012
%F A198636 Conjecture: a(n) = 2^(2*n)*(sum_{k=1,2,3} (cos(k*Pi/7))^(2*n)). - _L. Edson Jeffery_, Jan 21 2012 (in fact this conjecture was recently proved in [Barbero, et al.])
%F A198636 a(n) = 7*(binomial(2n-1,n-1) + sum_{k = 1..floor(n/7)} binomial(2n,n-7k)) - 2^(2n-1). - _M. Lawrence Glasser_, Feb 20 2013
%F A198636 Let r,s,t be the roots of x^3 + x^2 - 2x - 1; then apparently a(n) = r^(2n) + s^(2n) + t^(2n). - _James R. Buddenhagen_, Nov 03 2013 [This is equivalent to the conjecture by _L. Edson Jeffery_.]
%F A198636 a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3). - _M. F. Hasler_, Nov 05 2013
%F A198636 G.f.: F(x) = (sum_{r=0..2} ((3-r)*(-1)^r*binomial(6-r,r))*x^r)/(sum_{s=0..3} ((-1)^s*binomial(6-s,s))*x^s). - _L. Edson Jeffery_, Nov 23 2013
%e A198636 With the graph P_6 as 1-2-3-4-5-6:
%e A198636 n=0: a(0)=3 because w(6,0)=6, the number of vertices.
%e A198636 n=2: a(2)=5 because the 10 round trips of length 2 are 121, 212, 232, 323, 343, 434, 454, 545, 565 and 656.
%t A198636 Table[7 (Binomial[2 n - 1, n - 1] + Sum[Binomial[2 n, n - 7 k], {k, Floor[n/7]}]) - 2^(2 n - 1) - (7/2) Boole[n == 0], {n, 0, 25}] (* _Michael De Vlieger_, Jul 17 2017 *)
%o A198636 (PARI) vec_A198636(Nmax)=Vec((3-10*x+6*x^2)/(1-5*x+6*x^2-x^3)+O(x^Nmax)) \\ Indices will start at 1 in this vector. - _M. F. Hasler_, Nov 03 2013
%o A198636 (PARI) {a(n) = if( n<0, n=-n; polcoeff( (3 - 12*x + 5*x^2) / (1 - 6*x + 5*x^2 - x^3) + x * O(x^n), n), polcoeff( (3 - 10*x + 6*x^2) / (1 - 5*x + 6*x^2 -x^3) + x * O(x^n), n))}; /* _Michael Somos_, Jul 17 2017 */
%Y A198636 Cf. A198632, A198633, A005248, A198635.
%K A198636 nonn,easy,walk
%O A198636 0,1
%A A198636 _Wolfdieter Lang_, Nov 03 2011