A003696 Number of spanning trees in P_4 X P_n.

1, 56, 2415, 100352, 4140081, 170537640, 7022359583, 289143013376, 11905151192865, 490179860527896, 20182531537581071, 830989874753525760, 34214941811800329425, 1408756312731277540744

Offset

1,2

Comments

Also number of domino tilings of the 7 X (2n-1) rectangle with upper left corner removed. - Alois P. Heinz, Apr 14 2011

A linear divisibility sequence of order 8; a(n) divides a(m) whenever n divides m. It is the product of a 2nd-order Lucas sequence and a 4th-order linear divisibility sequence. - Peter Bala, Apr 27 2014

References

F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.

Links

T. D. Noe, Table of n, a(n) for n = 1..200

F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Preliminary version of paper that appeared in Ars Combin. 49 (1998), 129-154.

F. Faase, Counting Hamiltonian cycles in product graphs

F. Faase, Results from the counting program

P. Raff, Spanning Trees in Grid Graphs, arXiv:0809.2551 [math.CO], 2008.

P. Raff, Analysis of the Number of Spanning Trees of P_4 x P_n. Contains sequence, recurrence, generating function, and more.

Roettger, E. L.; Williams, H. C.; Guy, R. K., Some extensions of the Lucas functions, Springer Proceedings in Mathematics & Statistics 43, 271-311 (2013), chapter 5.

Index to divisibility sequences

Index entries for sequences related to trees

Index entries for linear recurrences with constant coefficients, signature (56, -672, 2632, -4094, 2632, -672, 56, -1).

Formula

a(1) = 1,

a(2) = 56,

a(3) = 2415,

a(4) = 100352,

a(5) = 4140081,

a(6) = 170537640,

a(7) = 7022359583,

a(8) = 289143013376 and

a(n) = 56a(n-1) - 672a(n-2) + 2632a(n-3) - 4094a(n-4) + 2632a(n-5) - 672a(n-6) + 56a(n-7) - a(n-8).

G.f.: x(x^6-49x^4+112x^3-49x^2+1) / (x^8-56x^7 +672x^6-2632x^5 +4094x^4 -2632x^3 +672x^2-56x+1). - Paul Raff, Mar 06 2009

From Peter Bala, Apr 27 2014: (Start)

a(n) = Resultant( U(3,(x-4)/2),U(n-1,x/2) ), where U(n,x) denotes the Chebyshev polynomial of the second kind. The polynomial U(3,(x-4)/2) = x^3 - 12*x^2 + 46*x - 56 (see A159764) has zeros z_1 = 4, z_2 = 4 + sqrt(2) and z_3 = 4 - sqrt(2). Hence a(n) = U(n-1,2)*U(n-1,1/2*(4 + sqrt(2)))*U(n-1,1/2*(4 - sqrt(2))).

a(n) = A001353(n)*A161158(n-1). (End)

a(n) = (9/3968)*(A028469(n+3)-A028469(n-4)) - (497/3968)*(A028469(n+2)-A028469(n-3)) + (5687/3968)*(A028469(n+1)-A028469(n-2)) - (19983/3968)*(A028469(n)-A028469(n-1)), n>3. - Sergey Perepechko, May 02 2016

a(n) = -a(-n) for all n in Z. - Michael Somos, Oct 31 2022

Maple

seq(resultant(simplify(ChebyshevU(3, (x-4)*(1/2))), simplify(ChebyshevU(n-1, (1/2)*x)), x), n = 1 .. 14); # Peter Bala, Apr 27 2014

Mathematica

LinearRecurrence[{56, -672, 2632, -4094, 2632, -672, 56, -1}, {1, 56, 2415, 100352, 4140081, 170537640, 7022359583, 289143013376}, 20] (* Jean-François Alcover, Feb 28 2020 *)

Prog

(PARI) {a(n) = polresultant((x-4)*(x^2-8*x+14), polchebyshev(n-1, 2, x/2))}; /* Michael Somos, Oct 31 2022 */

Crossrefs

A row of A116469. - N. J. A. Sloane, May 27 2012

Bisection of A189004. - Alois P. Heinz, Sep 20 2012

A001353, A161158, A159764.

Sequence in context: A280803 A124101 A198948 * A199709 A205227 A328352

Adjacent sequences: A003693 A003694 A003695 * A003697 A003698 A003699

Keyword

nonn,easy

Author

Frans J. Faase

Extensions

Added recurrence from Faase's web page. - N. J. A. Sloane, Feb 03 2009

Status

approved