

A198636


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


6



3, 5, 13, 38, 117, 370, 1186, 3827, 12389, 40169, 130338, 423065, 1373466, 4459278, 14478659, 47011093, 152642789, 495626046, 1609284589, 5225309458, 16966465802, 55089756851, 178875298901, 580804419201, 1885860059450, 6123349080945
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OFFSET

0,1


COMMENTS

See the array and triangle A198632 for the general graph P_N case (there N is n and the length is l=2*k).


LINKS

M. F. Hasler, Table of n, a(n) for n = 0..499
S. Barbero, Dickson Polynomials, Chebyshev Polynomials, and Some Conjectures of Jeffery, Journal of Integer Sequences, 17 (2014), #14.3.8.
S. Barbero, U. Cerruti, N. Murru, Identities Involving Zeros of Ramanujan and Shanks Cubic Polynomials, J. Integer Seq., Vol. 16 (2013), Article 13.8.1, pp. 1012.
Index entries for linear recurrences with constant coefficients, signature (5,6,1).


FORMULA

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)).
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 Spolynomials (coefficients A049310). See also A198632 for a rewritten form.
O.g.f.: (310*x+6*x^2)/(15*x+6*x^2x^3).  Colin Barker, Jan 02 2012
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.])
a(n) = 7*(binomial(2n1,n1) + sum_{k = 1..floor(n/7)} binomial(2n,n7k))  2^(2n1).  M. Lawrence Glasser, Feb 20 2013
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.]
a(n) = 5*a(n1)  6*a(n2) + a(n3).  M. F. Hasler, Nov 05 2013
G.f.: F(x) = (sum_{r=0..2} ((3r)*(1)^r*binomial(6r,r))*x^r)/(sum_{s=0..3} ((1)^s*binomial(6s,s))*x^s).  L. Edson Jeffery, Nov 23 2013


EXAMPLE

With the graph P_6 as 123456:
n=0: a(0)=3 because w(6,0)=6, the number of vertices.
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.


MATHEMATICA

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 *)


PROG

(PARI) vec_A198636(Nmax)=Vec((310*x+6*x^2)/(15*x+6*x^2x^3)+O(x^Nmax)) \\ Indices will start at 1 in this vector.  M. F. Hasler, Nov 03 2013
(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 */


CROSSREFS

Cf. A198632, A198633, A005248, A198635.
Sequence in context: A175257 A190423 A278024 * A172023 A271667 A188583
Adjacent sequences: A198633 A198634 A198635 * A198637 A198638 A198639


KEYWORD

nonn,easy,walk


AUTHOR

Wolfdieter Lang, Nov 03 2011


STATUS

approved



