A049684 -id:A049684 - cp's OEIS frontend

A049683 a(n) = (Lucas(6*n) - 2)/16.

0, 1, 20, 361, 6480, 116281, 2086580, 37442161, 671872320, 12056259601, 216340800500, 3882078149401, 69661065888720, 1250017107847561, 22430646875367380, 402501626648765281, 7222598632802407680, 129604273763794572961, 2325654329115499905620

Offset: 0

Clark Kimberling

This is the r = 20 member of the r-family of sequences S_r(n), n >= 1, defined in A092184 where more information can be found.

Colin Barker, Table of n, a(n) for n = 0..750
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A000032, A004146, A023039, A049660, A049682, A049684, A092184.

List([0..30], n-> (Lucas(1,-1,6*n)[2] - 2)/16 ); # G. C. Greubel, Dec 14 2019

[(Lucas(6*n) -2)/16: n in [0..30]]; // G. C. Greubel, Dec 02 2017

with(combinat); seq( (5*fibonacci(3*n)^2 -2*(1-(-1)^n))/16, n=0..30); # G. C. Greubel, Dec 14 2019

LinearRecurrence[{19,-19,1}, {0,1,30}, 20] (* or *) Table[(LucasL[6*n] -2)/16, {n,0,30}] (* G. C. Greubel, Dec 02 2017 *)

concat(0, Vec(x*(1+x)/((1-x)*(1-18*x+x^2)) + O(x^30))) \\ Colin Barker, Jun 03 2016

vector(31, n, (5*fibonacci(3*n-3)^2 -2*(1+(-1)^n))/16 ) \\ G. C. Greubel, Dec 14 2019

[(lucas_number2(6*n,1,-1) -2)/16 for n in (0..30)] # G. C. Greubel, Dec 14 2019

a(n) = (-2 + (9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/16. - Ralf Stephan, Apr 14 2004
a(n) = (T(n, 9) - 1)/8 with Chebyshev's polynomials of the first kind evaluated at x = 9: T(n, 9) = A023039(n). Wolfdieter Lang, Oct 18 2004
G.f.: x*(1 + x)/((1 - x)*(1 - 18*x + x^2)) = x*(1 + x)/(1 - 19*x + 19*x^2 - x^3). (from the Stephan link, see A092184).
exp( Sum_{n >= 1} 16*a(n)*x^n/n ) = 1 + 2*Sum_{n >= 1} Fibonacci(6*n)*x^n. - Peter Bala, Jun 03 2016
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3) for n>2. - Colin Barker, Jun 03 2016

A217273 T(n,k)=Number of nXk arrays of occupancy after each element moves to some horizontal or vertical neighbor.

Original entry on oeis.org

0, 1, 1, 2, 9, 2, 4, 64, 64, 4, 8, 441, 1472, 441, 8, 16, 3025, 33856, 33856, 3025, 16, 32, 20736, 772800, 2572816, 772800, 20736, 32, 64, 142129, 17640000, 194881600, 194881600, 17640000, 142129, 64, 128, 974169, 402494400, 14755689729

Offset: 1

R. H. Hardin Sep 29 2012

Table starts
....0..........1...............2..................4.....................8
....1..........9..............64................441..................3025
....2.........64............1472..............33856................772800
....4........441...........33856............2572816.............194881600
....8.......3025..........772800..........194881600...........48909857500
...16......20736........17640000........14755689729........12276607432401
...32.....142129.......402494400......1117092569476......3080781263731464
...64.....974169......9183772224.....84567501131481....773143936621737444
..128....6677056....209543184576...6401965110822961.194021789111058265560
..256...45765225...4781079618624.484642857129680209
..512..313679521.109088244863424
.1024.2149991424

			Some solutions for n=3 k=4
..0..1..0..2....0..2..2..1....0..2..2..1....1..0..0..1....1..0..1..1
..1..1..2..2....1..0..1..1....1..2..1..0....1..1..1..1....3..1..2..0
..2..0..1..0....1..1..2..0....0..1..2..0....0..2..3..1....1..0..2..0

R. H. Hardin, Table of n, a(n) for n = 1..97

Column 2 is A049684

A103434 a(n) = Sum_{i=1..n} Fibonacci(2i)^2.

Original entry on oeis.org

0, 1, 10, 74, 515, 3540, 24276, 166405, 1140574, 7817630, 53582855, 367262376, 2517253800, 17253514249, 118257345970, 810547907570, 5555578007051, 38078498141820, 260993908985724, 1788878864758285, 12261158144322310

Offset: 0

Ralf Stephan, Feb 08 2005

Index entries for linear recurrences with constant coefficients, signature (9, -16, 9, -1).

Partial sums of A049684.
Bisection of A002571 and |A077916|.
Cf. A000045.

[(1/5)*(Fibonacci(4*n+2)-2*n-1): n in [0..50]]; // Vincenzo Librandi, Apr 20 2011

Accumulate[Fibonacci[Range[0,40,2]]^2] (* Harvey P. Dale, Nov 14 2013 *)
LinearRecurrence[{9, -16, 9, -1},{0, 1, 10, 74},21] (* Ray Chandler, Sep 23 2015 *)

G.f.: x(1+x) / ((1-7x+x^2)(1-x)^2).
a(n) = (1/5)*(Fibonacci(4n+2) - 2n - 1).
a(n) = Sum_{i=0..2n} (-1)^i*Fibonacci(i)*Fibonacci(i+1). - Rigoberto Florez, May 04 2019

A221507 T(n,k)=Number of nXk arrays of occupancy after each element moves to some horizontal, diagonal or antidiagonal neighbor, without consecutive moves in the same direction.

Original entry on oeis.org

0, 1, 0, 0, 9, 0, 1, 44, 64, 0, 0, 446, 664, 441, 0, 1, 3334, 31663, 12495, 3025, 0, 0, 27540, 730874, 2691211, 221614, 20736, 0, 1, 219070, 22664752, 223123842, 219689604, 4010052, 142129, 0, 0, 1766136, 617892584, 26917553016, 61691224718

Offset: 1

R. H. Hardin Jan 18 2013

Table starts
.0......1........0...........1...........0...........1.........0...........1
.0......9.......44.........446........3334.......27540....219070.....1766136
.0.....64......664.......31663......730874....22664752.617892584.17782943460
.0....441....12495.....2691211...223123842.26917553016
.0...3025...221614...219689604.61691224718
.0..20736..4010052.18054912942
.0.142129.72195950
.0.974169
.0

			Some solutions for n=3 k=4
..1..1..3..0....0..0..0..0....0..0..1..1....1..0..1..0....1..0..1..1
..1..1..1..1....1..3..2..2....0..4..1..0....1..2..0..3....0..3..0..1
..0..3..0..0....2..1..1..0....2..0..2..1....1..1..2..0....2..0..2..1

R. H. Hardin, Table of n, a(n) for n = 1..48

Column 2 is A049684

A181878 Coefficient array for square of Chebyshev S-polynomials.

Original entry on oeis.org

1, 1, 1, -2, 1, 4, -4, 1, 1, -6, 11, -6, 1, 9, -24, 22, -8, 1, 1, -12, 46, -62, 37, -10, 1, 16, -80, 148, -128, 56, -12, 1, 1, -20, 130, -314, 367, -230, 79, -14, 1, 25, -200, 610, -920, 771, -376, 106, -16, 1, 1, -30, 295, -1106, 2083, -2232, 1444, -574, 137, -18, 1, 36, -420, 1897, -4352, 5776, -4744, 2486, -832, 172, -20, 1, 1, -42, 581, -3108, 8518, -13672, 13820, -9142, 4013, -1158, 211, -22, 1

Offset: 0

Wolfdieter Lang, Dec 22 2010

For the coefficients of Chebyshev polynomials S(n,x) see A049310.
The row length sequence for this array is A109613 = {1,1,3,3,5,5,...}.
The row polynomials (in x^2) for even row numbers are
  S(2*k,x)^2 = Sum_{m=0..2*k} a(2*k,m)*x^(2*m), k >= 0.
For odd row numbers the row polynomials (in x^2) are
  (S(2*k+1,x)^2)/x^2 = Sum_{m=0..2*k} a(2*k+1,m)*x^(2*m), k >= 0.
The o.g.f. for the polynomials S(n,x)^2 is
  S(x,z):=((1+z)/(1-z))/(1 + (2-x^2)z +z^2). See the link for a proof. Therefore the coefficients constitute the Riordan array (1/(1-x^2),x/(1+x)^2) found as A158454.
The o.g.f. for S(2*k,sqrt(x))^2 is
  (1-2*(1-x)*z+z^2)/((1-z)*(1 - (2-4*x+x^2)*z + z^2)).
The o.g.f. for (S(2*k+1,sqrt(x))^2)/x is
  ((1+z)/(1-z))/(1 - (2-4*x+x^2)*z + z^2).
The row sums A011655(n+1) are the same as those for the triangle A158454.
The alternating row sums for even numbered rows (-1)^n*A007598(n+1) coincide with those of triangle A158454. For odd row numbers n=2k+1 these sums are A049684(k+1), k >= 0 (squares of even-indexed Fibonacci numbers).

			The irregular triangle a(n,m) begins:
  n\m  0    1    2      3     4      5     6    7   8   9  10 ...
  0:   1
  1:   1
  2:   1   -2    1
  3:   4   -4    1
  4:   1   -6   11     -6     1
  5:   9  -24   22     -8     1
  6:   1  -12   46    -62    37    -10     1
  7:  16  -80  148   -128    56    -12     1
  8:   1  -20  130   -314   367   -230    79  -14   1
  9:  25 -200  610   -920   771   -376   106  -16   1
  10:  1  -30  295  -1106  2083  -2232  1444 -574 137 -18   1
  ... Reformatted and extended by _Wolfdieter Lang_, Nov 24 2012

Wolfdieter Lang, First ten rows with more details and proofs.

Cf. A158454, A129818.

Join[{{1}, {1}}, CoefficientList[Table[ChebyshevU[n, Sqrt[x]/2]^2, {n, 2, 10}], x]] // Flatten (* Eric W. Weisstein, Apr 04 2018 *)
Join[{{1}, {1}}, CoefficientList[ChebyshevU[Range[2, 10], Sqrt[x]/2]^2, x]]  // Flatten (* Eric W. Weisstein, Apr 04 2018 *)

a(2*k,m) = (-1)^m*Sum_{j=0..k} binomial(2*k+m-1-2*j, 2*m-1), k >= 0.
a(2*k+1,m) = (-1)^m*Sum_{j=0..k} binomial(2*k+1+m-2*j, 2*m+1), k >= 0.
This derives from the formula for the entries of the Riordan array A158454.
For the o.g.f.s see the comment.

Corrected by Wolfdieter Lang, Jan 21 2011

A081070 Lucas(4n)-2, or 5*Fibonacci(2n)^2.

Original entry on oeis.org

0, 5, 45, 320, 2205, 15125, 103680, 710645, 4870845, 33385280, 228826125, 1568397605, 10749957120, 73681302245, 505019158605, 3461452808000, 23725150497405, 162614600673845, 1114577054219520, 7639424778862805

Offset: 0

R. K. Guy, Mar 04 2003

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Hang Gu and Robert M. Ziff, Crossing on hyperbolic lattices, arXiv:1111.5626 [cond-mat.dis-nn], 2011-2012 (see Eq. 4).
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A049684.

[Lucas(4*n)-2: n in [0..30]]; // Vincenzo Librandi, Apr 21 2011

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,luc(4*n)-2) od: # James Sellers, Mar 05 2003

LinearRecurrence[{8, -8, 1}, {0, 5, 45}, 20] (* Jean-François Alcover, Nov 24 2017 *)

a(n) = 5*fibonacci(2*n)^2; \\ Michel Marcus, Nov 24 2017

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = 5*A049684(n).
G.f.: 5*x*(x+1)/((1-x)*(x^2-7*x+1)). - Colin Barker, Jun 24 2012

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).

Original entry on oeis.org

1, 1, 1, 5, 1, 16, 1, 45, 1, 121, 1, 320, 1, 841, 1, 2205, 1, 5776, 1, 15125, 1, 39601, 1, 103680, 1, 271441, 1, 710645, 1, 1860496, 1, 4870845, 1, 12752041, 1, 33385280, 1, 87403801, 1, 228826125, 1, 599074576, 1, 1568397605, 1, 4106118241, 1, 10749957120, 1, 28143753121

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 3. See also A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(3 - sqrt(5)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(3 - sqrt(5))) = 1.53879 34992 88095 08323 ... = 1 + 1/(1 + 1/(1 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + 1/(45 + ...))))))).
F((1/2*(3 - sqrt(5)))^2) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(3 - sqrt(5)))^3) = 1.05883 42773 67371 19975 ... = 1 + 1/(16 + 1/(1 + 1/(320 + 1/(1 + 1/(5776 + 1/(1 + 1/(103680 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4,0,1).

Cf. A001906, A002878, A004146, A049684, A081070, A081071, A174500 (N = 4), A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

a(2*n-1) = (1/2*(3 + sqrt(5)))^n + (1/2*(3 - sqrt(5)))^n - 2 = A004146(n); a(2*n) = 1.
a(4*n+1) = A081071(n) = A002878(n)^2;
a(4*n-1) = A081070(n) = 5*A049684(n) = 5*(A001906(n))^2.
a(n) = 4*a(n-2)-4*a(n-4)+a(n-6). G.f.: -(x^4+x^3-3*x^2+x+1) / ((x-1)*(x+1)*(x^2-x-1)*(x^2+x-1)). - Colin Barker, Jan 20 2013

More terms from Michel Marcus, Feb 21 2025

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A003755 Number of spanning trees in S_4 X P_n.

Original entry on oeis.org

1, 54, 2240, 89964, 3596725, 143700480, 5740732439, 229334969304, 9161621922880, 365994298083150, 14620972301965259, 584087869159280640, 23333512405041243469, 932141942728566562746, 37237797134599264280000, 1487599121840339002010544, 59427552583207598523644161

Offset: 1

Frans J. Faase

nonn

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

Alois P. Heinz, 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 S_4 x P_n. Contains sequence, recurrence, generating function, and more.
Index entries for sequences related to trees
Index entries for linear recurrences with constant coefficients, signature (48, -336, 582, -336, 48, -1).

a:= n-> (Matrix([[1, 0, -1, -54, -2240, -89964]]). Matrix(6, (i,j)-> if (i=j-1) then 1 elif j=1 then [48,-336,582,-336,48,-1][i] else 0 fi)^(n-1))[1,1]: seq(a(n), n=1..14);  # Alois P. Heinz, Aug 01 2008

LinearRecurrence[{48, -336, 582, -336, 48, -1}, {1, 54, 2240, 89964, 3596725, 143700480}, 17] (* Jean-François Alcover, Aug 06 2018 *)

a(1) = 1,
a(2) = 54,
a(3) = 2240,
a(4) = 89964,
a(5) = 3596725,
a(6) = 143700480 and
a(n) = 48a(n-1) - 336a(n-2) + 582a(n-3) - 336a(n-4) + 48a(n-5) - a(n-6).
G.f.: x*(x^4+6*x^3-16*x^2+6*x+1)/ ((x^2-6*x+1)*(x^4-42*x^3+83*x^2-42*x+1)). - Paul Raff, Mar 06 2009
a(n) = A001109(n)*A049684(n). [R. Guy, seqfan list, Mar 28 2009] - R. J. Mathar, Jun 03 2009

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

A092136 Number of spanning trees in S_5 x P_n.

Original entry on oeis.org

1, 189, 24576, 3046869, 375175625, 46151368704, 5676383392121, 698151521972709, 85867005969063936, 10560944392853518125, 1298910307853115410641, 159755407182415993503744, 19648616177810537712940081, 2416620034547514872344613709

Offset: 1

Ralf Stephan, Mar 28 2004

nonn

P. Raff, Table of n, a(n) for n = 1..200
F. Faase, Counting Hamiltonian cycles in product graphs
P. Raff, Spanning Trees in Grid Graphs, arXiv:0809.2551 [math.CO], 2008.
P. Raff, Analysis of the Number of Spanning Trees of S_5 x P_n. Contains sequence, recurrence, generating function, and more.
P. Raff, Analysis of the Number of Spanning Trees of Grid Graphs.
Index entries for linear recurrences with constant coefficients, signature (144, -2640, 6930, -2640, 144, -1).

Cf. A001906, A004187, A049684.

LinearRecurrence[{7, -1}, {0, 1}, 13] LinearRecurrence[{3, -1}, {0, 1}, 13]^3 // Rest (* Jean-François Alcover, Oct 30 2018, after R. J. Mathar *)
LinearRecurrence[{144, -2640, 6930, -2640, 144, -1},{1, 189, 24576, 3046869, 375175625, 46151368704}, 14] (* Ray Chandler, Feb 28 2024 *)

a(n) = 144*a(n-1) - 2640*a(n-2) + 6930*a(n-3) - 2640*a(n-4) + 144*a(n-5) - a(n-6). [Modified by Paul Raff, Oct 30 2009]
G.f.: -x*(x^4 + 45*x^3 - 45*x-1)/(x^6 - 144*x^5 + 2640*x^4 - 6930*x^3 + 2640*x^2 - 144*x + 1). - Paul Raff (paul(AT)myraff.com), Mar 07 2009
a(n) = A004187(n)*(A001906(n))^3 = A004187(n)*A001906(n)*A049684(n). [See R. Guy, seqfan list, Mar 28 2009] - R. J. Mathar, Jun 03 2009

cp's OEIS Frontend

A049683 a(n) = (Lucas(6*n) - 2)/16.

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A217273 T(n,k)=Number of nXk arrays of occupancy after each element moves to some horizontal or vertical neighbor.

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A103434 a(n) = Sum_{i=1..n} Fibonacci(2i)^2.

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A221507 T(n,k)=Number of nXk arrays of occupancy after each element moves to some horizontal, diagonal or antidiagonal neighbor, without consecutive moves in the same direction.

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A181878 Coefficient array for square of Chebyshev S-polynomials.

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A081070 Lucas(4n)-2, or 5*Fibonacci(2n)^2.

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A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = 1/2*(3 - sqrt(5)).

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A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

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A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = 1/2*(3 - sqrt(5)).