A004254 -id:A004254 - cp's OEIS frontend

A003690 Number of spanning trees in K_3 X P_n.

3, 75, 1728, 39675, 910803, 20908800, 479991603, 11018898075, 252954664128, 5806938376875, 133306628004003, 3060245505715200, 70252340003445603, 1612743574573533675, 37022849875187828928, 849912803554746531675, 19510971631883982399603

Offset: 1

Frans J. Faase

Column 3 of A173958. The sequence a(n)/3 is linear divisibility sequence of the fourth order; it is the case P1 = 25, P2 = 46, Q = 1 of the three parameter family of divisibility sequences found by Williams and Guy. - Peter Bala, Apr 27 2014

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

Vincenzo Librandi, 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. [From _Paul Raff_, Mar 06 2009]
P. Raff, Analysis of the Number of Spanning Trees of K_3 x P_n. Contains sequence, recurrence, generating function, and more. [From _Paul Raff_, Mar 06 2009] [broken link]
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for sequences related to trees
Index entries for linear recurrences with constant coefficients, signature (24,-24,1).

Cf. A100047, A173958 (column 3).

I:=[3,75,1728]; [n le 3 select I[n] else 24*Self(n-1)-24*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Apr 28 2014

CoefficientList[Series[3 (1 + x)/((1 - x) (1 - 23 x + x^2)), {x, 0, 20}], x] (* Vincenzo Librandi, Apr 28 2014 *)

Vec(3*x*(1+x)/((1-x)*(1-23*x+x^2)) + O(x^25)) \\ Colin Barker, Mar 06 2016

a(n) = (A090731(n)-2)/7.
a(n) = 24*a(n-1) - 24*a(n-2) + a(n-3), n>3.
G.f.: 3*x*(1+x)/((1-x)*(1-23*x+x^2)). - R. J. Mathar, Dec 16 2008
a(n) = 3*(A004254(n))^2. - R. K. Guy, seqfan list, Mar 28 2009, - R. J. Mathar, Jun 03 2009
From Peter Bala, Apr 27 2014: (Start)
Product {n >= 2} (1 - 3/a(n)) = 1/2 + sqrt(21)/10.
a(n) = (2/7)*( T(n,23/2) - 1), where T(n,x) is the Chebyshev polynomial of the first kind.
a(n) = 3 * the bottom left entry of the 2 X 2 matrix T(n,M), where M is the 2 X 2 matrix [0, -23/2; 1, 25/2].
a(n) = 3*U(n-1,5/2)^2, where U(n,x) is the Chebyshev polynomial of the second kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(n) = (-2+(2/(23+5*sqrt(21)))^n+(1/2*(23+5*sqrt(21)))^n)/7. - Colin Barker, Mar 06 2016

A078364 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 15, 224, 3345, 49951, 745920, 11138849, 166336815, 2483913376, 37092363825, 553901543999, 8271430796160, 123517560398401, 1844491975179855, 27543862067299424, 411313439034311505

Offset: 0

Wolfdieter Lang, Nov 29 2002

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 221*a^2 = +4 with companion sequence b(n)=A078365(n+1), n>=0.
This is the m=17 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..16 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362 and A007655. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 15's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,14}. - Milan Janjic, Jan 23 2015

Harvey P. Dale, Table of n, a(n) for n = 0..850
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=15, q=-1.
Tanya Khovanova, Recursive Sequences
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=17.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (15,-1).

a(n) = sqrt((A078365(n+1)^2 - 4)/221), n>=0, (Pell equation d=221, +4).

Cf. A077428, A078355 (Pell +4 equations).

LinearRecurrence[{15,-1},{1,15},30] (* Harvey P. Dale, Oct 16 2011 *)

[lucas_number1(n,15,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = 15*a(n-1) - a(n-2), n>= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(17))/sqrt(17) = S(n, 15); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (15+sqrt(221))/2 and am = (15-sqrt(221))/2.
G.f.: 1/(1 - 15*x + x^2). - Philippe Deléham, Nov 17 2008
a(n) = Sum_{k=0..n} A101950(n,k)*14^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/13*(13 + sqrt(221)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/30*(13 + sqrt(221)). - Peter Bala, Dec 23 2012
For n>=1, a(n) = U(n-1,15/2), where U(k,x) is Chebyshev polynomial of the second kind. - Milan Janjic, Jan 23 2015

A078366 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 17, 288, 4879, 82655, 1400256, 23721697, 401868593, 6808044384, 115334885935, 1953885016511, 33100710394752, 560758191694273, 9499788548407889, 160935647131239840, 2726406212682669391, 46187969968474139807, 782469083251377707328, 13255786445304946884769

Offset: 0

Wolfdieter Lang, Nov 29 2002

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 285*a^2 = +4 with companion sequence b(n)=A078367(n+1), n >= 0.
This is the m=19 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..18 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362, A007655, A078364 and A077412. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 17's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,16}. - Milan Janjic, Jan 23 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..800
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=17, q=-1.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=19.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (17,-1).

Cf. A077428, A078355 (Pell +4 equations). Cf. A078367.

a:=[1,17,288];; for n in [4..20] do a[n]:=17*a[n-1]-a[n-2]; od; a; # G. C. Greubel, May 25 2019

I:=[1, 17, 288]; [n le 3 select I[n] else 17*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

CoefficientList[Series[1/(1 - 17 x + x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
LinearRecurrence[{17,-1},{1,17},20] (* Harvey P. Dale, Aug 02 2018 *)

my(x='x+O('x^20)); Vec(1/(1-17*x+x^2)) \\ G. C. Greubel, May 25 2019

[lucas_number1(n,17,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = 17*a(n-1) - a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(19))/sqrt(19) = S(n, 17), where S(n, x) = U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (17+sqrt(285))/2 and am = (17-sqrt(285))/2.
G.f.: 1/(1-17*x+x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*16^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = (1/15)*(15 + sqrt(285)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = (1/34)*(15 + sqrt(285)). - Peter Bala, Dec 23 2012
For n >= 1, a(n) = U(n-1,13/2), where U(k,x) represents Chebyshev polynomial of the second order. - Milan Janjic, Jan 23 2015
a(n) = sqrt((A078367(n+1)^2 - 4)/285), n>=0, (Pell equation d=285, +4).
E.g.f.: exp(17*x/2)*(sqrt(285)*cosh(sqrt(285)*x/2) + 17*sinh(sqrt(285)*x/2))/sqrt(285). - Stefano Spezia, Aug 19 2023

A097778 Chebyshev polynomials S(n,23) with Diophantine property.

Original entry on oeis.org

1, 23, 528, 12121, 278255, 6387744, 146639857, 3366328967, 77278926384, 1774048977865, 40725847564511, 934920445005888, 21462444387570913, 492701300469125111, 11310667466402306640, 259652650426783927609

Offset: 0

Wolfdieter Lang, Aug 31 2004

All positive integer solutions of Pell equation b(n)^2 - 525*a(n)^2 = +4 together with b(n)=A090731(n+1), n>=0. Note that D=525=21*5^2 is not squarefree.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 23's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,22}. - Milan Janjic, Jan 25 2015

			(x,y) = (23;1), (527;23), (12098;528), ... give the positive integer solutions to x^2 - 21*(5*y)^2 =+4.

Indranil Ghosh, Table of n, a(n) for n = 0..733
R. Flórez, R. A. Higuita, A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
Index entries for sequences relate d to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (23,-1).

A004254, A049310.

LinearRecurrence[{23,-1},{1,23},20] (* Harvey P. Dale, May 06 2016 *)

[lucas_number1(n,23,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = S(n, 23) = U(n, 23/2) = S(2*n+1, 5)/5 with S(n, x) = U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x). S(n, 5) = A004254(n+1).
a(n) = 23*a(n-1)-a(n-2), n >= 1; a(0)=1, a(1)=23; a(-1)=0.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap := (23+5*sqrt(21))/2 and am := (23-5*sqrt(21))/2.
G.f.: 1/(1-23*x+x^2).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*22^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/21*(21 + 5*sqrt(21)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/46*(21 + 5*sqrt(21)). - Peter Bala, Dec 23 2012

A207824 Triangle of coefficients of Chebyshev's S(n,x+5) polynomials (exponents of x in increasing order).

Original entry on oeis.org

1, 5, 1, 24, 10, 1, 115, 73, 15, 1, 551, 470, 147, 20, 1, 2640, 2828, 1190, 246, 25, 1, 12649, 16310, 8631, 2400, 370, 30, 1, 60605, 91371, 58275, 20385, 4225, 519, 35, 1, 290376, 501150, 374115, 157800, 41140, 6790, 693, 40, 1

Offset: 0

Philippe Deléham, Feb 20 2012

Riordan array (1/(1-5*x+x^2), x/(1-5*x+x^2)).
Subtriangle of triangle given by (0, 5, -1/5, 1/5, 0, 0, 0, 0, 0, 0, 0, 0...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
Unsigned version of A123967 and A179900.
For 1<=k<=n, T(n,k) equals the number of (n-1)-length  words over {0,1,2,3,4,5} containing k-1 letters equal 5 and avoiding 01. -  Milan Janjic, Dec 20 2016

			Triangle begins :
  1
  5, 1
  24, 10, 1
  115, 73, 15, 1
  551, 470, 147, 20, 1
  2640, 2828, 1190, 246, 25, 1
  12649, 16310, 8631, 2400, 370, 30, 1
  ...
Triangle (0, 5, -1/5, 1/5, 0, 0, 0,...) DELTA (1, 0, 0, 0, ...) begins :
  1
  0, 1
  0, 5, 1
  0, 24, 10, 1
  0, 115, 73, 15, 1
  0, 551, 470, 147, 20, 1
  0, 2640, 2828, 1190, 246, 25, 1
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150)
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. Triangles of coefficients of Chebyshev's S(n,x+k) polynomials : A207824 (k = 5), A207823 (k = 4), A125662 (k = 3), A078812 (k = 2), A101950 (k = 1), A049310 (k = 0), A104562 (k = -1), A053122 (k = -2), A207815 (k = -3), A159764 (k = -4), A123967 (k = -5).

With[{n = 8}, DeleteCases[#, 0] & /@ CoefficientList[Series[1/(1 - 5 x + x^2 - y x), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)

row(n) = Vecrev(polchebyshev(n, 2, (x+5)/2)); \\ Michel Marcus, Apr 26 2018

Recurrence : T(n,k) = 5*T(n-1,k) + T(n-1,k-1) - T(n-2,k).
G.f.: 1/(1-5*x+x^2-y*x).
Diagonal sums are 5^n = A000351(n).
Row sums are A001109(n+1).
T(n,0) = A004254(n+1), T(n+1,n) = 5n+5 = A008587(n+1).

A107905 Decimal expansion of (5+sqrt(21))/2.

Original entry on oeis.org

4, 7, 9, 1, 2, 8, 7, 8, 4, 7, 4, 7, 7, 9, 2, 0, 0, 0, 3, 2, 9, 4, 0, 2, 3, 5, 9, 6, 8, 6, 4, 0, 0, 4, 2, 4, 4, 4, 9, 2, 2, 2, 8, 2, 8, 8, 3, 8, 3, 9, 8, 5, 9, 5, 1, 3, 0, 3, 6, 2, 1, 0, 6, 1, 9, 5, 3, 4, 3, 4, 2, 1, 2, 7, 7, 3, 8, 8, 5, 4, 4, 3, 3, 0, 2, 1, 8, 0, 7, 7, 9, 7, 4, 6, 7, 2, 2, 5, 1, 6

Offset: 1

Jonathan Vos Post, Jun 22 2007

			4.7912878474779200032940235968640042444922282883839859513036...
The zeros at 15, 16 and 17 digits after the decimal point allow for a good rational approximation. The continued fraction is [4,1,3,1,3,1,3,...] = 4 + 1/(1+ 1/(3+ 1/(1+ 1/(3+ 1/(1+ 1/(3+ 1(/1+ ...

D. Mumford et al., Indra's Pearls, Cambridge 2002; see p. 317. [From N. J. A. Sloane, Nov 22 2009]

Daniel Starodubtsev, Table of n, a(n) for n = 1..10000
Emma Y. Jin and Christian M. Reidys, Asymptotic Enumeration of RNA Structures with Pseudoknots, arXiv:0706.3137 [q-bio.BM], 2007, Theorem 5, p. 15.
Index entries for algebraic numbers, degree 2.

Equals 1+A090458. - R. J. Mathar, Aug 24 2008

Cf. A004253, A004254.

RealDigits[(5+Sqrt[21])/2,10,120][[1]] (* Harvey P. Dale, May 02 2011 *)

(sqrt(21)+5)/2 \\ Charles R Greathouse IV, Feb 11 2025

(4.791287...)^n = A090458 * A004254(n) + A004253(n). - Gary W. Adamson, Sep 11 2023
Equals lim_{n->oo} S(n, 5)/S(n-1, 5), with the S-Chebyshev polynomial (see A049310) S(n, 5) = A004254(n+1). - Wolfdieter Lang, Nov 15 2023
c^k = A004254(k)*c - A004254(k-1) for k >= 1, where c is the present constant. - Andrea Pinos, Jul 19 2024
Minimal polynomial: x^2 - 5*x + 1. - Stefano Spezia, Jul 02 2025

A136211 Denominators in continued fraction [0; 1, 3, 1, 3, 1, 3, ...].

Original entry on oeis.org

1, 4, 5, 19, 24, 91, 115, 436, 551, 2089, 2640, 10009, 12649, 47956, 60605, 229771, 290376, 1100899, 1391275, 5274724, 6665999, 25272721, 31938720, 121088881, 153027601, 580171684, 733199285, 2779769539, 3512968824

Offset: 1

Gary W. Adamson, Dec 21 2007

A136210(n)/A136211(n) tends to 0.791287847... = [0; 1, 3, 1, 3, 1, 3, ...] = (sqrt(21) - 3)/2 = the inradius of a right triangle with hypotenuse 3, legs 2 and sqrt(21).
The number 0.791287847... = (sqrt(21) - 3)/2 arises in finding a number which is 5 less than its square; the result is: 2.791287847... because (2.791287847...)^2 = 7.791287847... In general the quadratic equation for finding such numbers is x^2 - x = N, so x = (1 + sqrt(1 + 4N))/2. - Alexander R. Povolotsky, Dec 23 2007
Prepending a 1 to the sequence gives [1, 1, 4, 5, 19, 24, ...]. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) with the parameters R = 3 and Q = -1. It is a strong divisibility sequence, that is, GCD(a(n),a(m)) = a(GCD(n,m)) for all natural numbers n and m. - Peter Bala, May 14 2014

			a(4) = 19 = 3*a(3) + a(2) = 3*5 + 4.
a(5) = 24 = a(4) + a(3) = 19 + 5.
T^3 = [19, 72; 24, 91], where the bottom row [24, 91] = [a(5), a(6)].

G. C. Greubel, Table of n, a(n) for n = 1..1000
P. Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6
J. L. Ramirez, F. Sirvent, A q-Analogue of the Bi-Periodic Fibonacci Sequence, J. Int. Seq. 19 (2016) # 16.4.6, t_n at a=1, b=3.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0, 5, 0, -1).

Cf. A136210.

Denominator[NestList[(3/(3+#))&,0,60]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
a[n_] := FromContinuedFraction[ Join[{0}, 3 - 2*Array[Mod[#, 2]&, n]]] // Denominator; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, May 15 2014 *)

x='x + O('x^25); Vec(x*(1+4*x-x^3)/(1-5*x^2+x^4)) \\ G. C. Greubel, Feb 18 2017

a(1) = 1, a(2) = 4, then for n>2, a(2n) = 3*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Let T = the 2 X 2 matrix [1, 3; 1, 4]. Then T^n = [A136210(2n-1), A136210(2n); a(2n-1), a(2n)].
From R. J. Mathar, May 18 2008: (Start)
O.g.f.: x*(1+4*x-x^3)/(1-5*x^2+x^4).
a(2*n) = A004253(n+1).
a(2*n+1) = A004254(n). (End)
a(n)*a(n+1) = A099025(n). - R. K. Guy, May 18 2008
{-a(n) + 5 a(n + 2) - a(n + 4), a(0) = 1, a(1) = 4, a(2) = 5, a(3) = 19}. - Robert Israel, May 14 2008

A139400 Number of spanning trees in the graph P_6 x P_n.

Original entry on oeis.org

1, 780, 380160, 170537640, 74795194705, 32565539635200, 14143261515284447, 6136973985625588560, 2662079368040434932480, 1154617875754582889149500, 500769437567956298239402223, 217185579535490113365186969600

Offset: 1

Paul Raff, Jun 09 2008; corrected recurrence Feb 03 2009

Also number of domino tilings of the 11 X (2n-1) rectangle with upper left corner removed. - Alois P. Heinz, Apr 14 2011

A linear divisibility sequence of order 32; a(n) divides a(m) whenever n divides m. It is the product of four linear divisibility sequences - three Lucas sequences of order 2 and one linear divisibility sequence of order 4. - Peter Bala, Apr 27 2014

			a(2) = 780, as can be verified from the seventh entry of A001353, which corresponds to the number of spanning trees of the same graph.

Paul Raff, Table of n, a(n) for n = 1..208
Paul Raff, Spanning Trees in Grid Graphs, arXiv:0809.2551 [math.CO], 2008.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (780, -194881, 22377420, -1419219792, 55284715980, -1410775106597, 24574215822780, -300429297446885, 2629946465331120, -16741727755133760, 78475174345180080, -273689714665707178, 716370537293731320, -1417056251105102122, 2129255507292156360, -2437932520099475424, 2129255507292156360, -1417056251105102122, 716370537293731320, -273689714665707178, 78475174345180080, -16741727755133760, 2629946465331120, -300429297446885, 24574215822780, -1410775106597, 55284715980, -1419219792, 22377420, -194881, 780, -1).

Row m=6 of A116469.

Bisection of A210724 (odd part). A001353, A001906, A004254, A159764, A161498.

seq(resultant(simplify(ChebyshevU(5, (x-4)*(1/2))), simplify(ChebyshevU(n-1, (1/2)*x)), x), n = 1 .. 12); # Peter Bala, Apr 27 2014

Array[Resultant[ChebyshevU[5, x/2-2], ChebyshevU[#-1, x/2], x] &, 20] (* Paolo Xausa, Mar 17 2024, after Peter Bala *)

a(n) = 780 a(n-1) - 194881 a(n-2) + 22377420 a(n-3) - 1419219792 a(n-4) + 55284715980 a(n-5) - 1410775106597 a(n-6) + 24574215822780 a(n-7) - 300429297446885 a(n-8) + 2629946465331120 a(n-9) - 16741727755133760 a(n-10)
+ 78475174345180080 a(n-11) - 273689714665707178 a(n-12) + 716370537293731320 a(n-13) - 1417056251105102122 a(n-14) + 2129255507292156360 a(n-15) - 2437932520099475424 a(n-16) + 2129255507292156360 a(n-17)
- 1417056251105102122 a(n-18) + 716370537293731320 a(n-19) - 273689714665707178 a(n-20) + 78475174345180080 a(n-21) - 16741727755133760 a(n-22) + 2629946465331120 a(n-23) - 300429297446885 a(n-24) + 24574215822780 a(n-25) - 1410775106597 a(n-26) + 55284715980 a(n-27) - 1419219792 a(n-28) + 22377420 a(n-29) - 194881 a(n-30) + 780 a(n-31) - a(n-32).
From Peter Bala, Apr 27 2014: (Start)
a(n) = Resultant( U(5,(x-4)/2), U(n-1,x/2) ), where U(n,x) denotes the Chebyshev polynomial of the second kind. The polynomial U(5,(x-4)/2) = x^5 - 20*x^4 + 156*x^3 - 592*x^2 + 1091*x - 780 (see A159764) has zeros z_1 = 3, z_2 = 4, z_3 = 5, z_4 = 4 + sqrt(3) and z_5 = 4 - sqrt(3). Hence a(n) = U(n-1,3/2)*U(n-1,2)*U(n-1,5/2)*U(n-1,1/2*(4 + sqrt(3)))*U(n-1,1/2*(4 - sqrt(3))).
a(n) = A001906(n)*A001353(n)*A004254(n)*A161498(n). (End)

A002310 a(n) = 5*a(n-1) - a(n-2), with a(0) = 1 and a(1) = 2.

Original entry on oeis.org

1, 2, 9, 43, 206, 987, 4729, 22658, 108561, 520147, 2492174, 11940723, 57211441, 274116482, 1313370969, 6292738363, 30150320846, 144458865867, 692144008489, 3316261176578, 15889161874401, 76129548195427, 364758579102734, 1747663347318243, 8373558157488481, 40120127440124162

Offset: 0

Joe Keane (jgk(AT)jgk.org)

Together with A002320 these are the two sequences satisfying ( a(n)^2+a(n-1)^2 )/(1 - a(n)a(n-1)) is an integer, in both cases this integer is -5. - Floor van Lamoen, Oct 26 2001

Limit_{n->oo} a(n+1)/a(n) = (5 + sqrt(21))/2 = A107905. - Wolfdieter Lang, Nov 17 2023

From a posting to Netnews group sci.math by ksbrown(AT)seanet.com (K. S. Brown) on Aug 15 1996.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 44, 56.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
Tanya Khovanova, Recursive Sequences
MathPages, N = (x^2 + y^2)/(1+xy) is a Square
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A002310, A002320, A004254, A049310, A049685, A054477, A107905.

a002310 n = a002310_list !! n
a002310_list = 1 : 2 :
   (zipWith (-) (map (* 5) (tail a002310_list)) a002310_list)
-- Reinhard Zumkeller, Oct 16 2011

LinearRecurrence[{5, -1}, {1, 2}, 25] (* T. D. Noe, Feb 22 2014 *)

Sequences A002310, A002320 and A049685 have this in common: each one satisfies a(n+1) = (a(n)^2+5)/a(n-1). - Graeme McRae, Jan 30 2005
G.f.: (1-3x)/(1-5x+x^2). - Philippe Deléham, Nov 16 2008
a(n) = S(n, 5) - 3*S(n-1, 5), for n >= 0, with the S-Chebyshev polynomial (see A049310) S(n, 5) = A004254(n+1). - Wolfdieter Lang, Nov 17 2023
E.g.f.: exp(5*x/2)*(21*cosh(sqrt(21)*x/2) - sqrt(21)*sinh(sqrt(21)*x/2))/21. - Stefano Spezia, Jul 07 2025

A110311 Expansion of 1/((1+x+x^2)(1+5x+x^2)).

Original entry on oeis.org

1, -6, 29, -138, 660, -3162, 15151, -72594, 347819, -1666500, 7984680, -38256900, 183299821, -878242206, 4207911209, -20161313838, 96598657980, -462831976062, 2217561222331, -10624974135594, 50907309455639, -243911573142600, 1168650556257360, -5599341208144200

Offset: 0

Creighton Dement, Jul 19 2005

In reference to the program code, A004254(n+1) = 1ibaseiseq[A*B](n).

Superseeker finds: a(n) + a(n+1) + a(n+2) = (-1)^n*A004254(n+3).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-6,-7,-6,-1).

Cf. A004253, A004254, A110307, A110308, A110309, A110310.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1+x+x^2)*(1+5*x+x^2)) )); // G. C. Greubel, Jan 02 2023

seriestolist(series(1/((x^2+5*x+1)*(x^2+x+1)), x=0,25));

LinearRecurrence[{-6,-7,-6,-1}, {1,-6,29,-138}, 40] (* G. C. Greubel, Jan 02 2023 *)

Vec(1/((1+x+x^2)*(1+5*x+x^2)) + O(x^25)) \\ Colin Barker, May 14 2019

def U(n,x): return chebyshev_U(n,x)
def A110311(n): return (1/4)*(5*U(n, -5/2) + U(n-1, -5/2) - U(n, -1/2) - U(n-1, -1/2))
[A110311(n) for n in range(41)] # G. C. Greubel, Jan 02 2023

a(n+2) = - 5*a(n+1) - a(n) + ((-1)^n)*A109265(n+1)/2.
a(n) = -6*a(n-1) - 7*a(n-2) - 6*a(n-3) - a(n-4) for n>3. - Colin Barker, May 14 2019
a(n) = (1/4)*(5*U(n, -5/2) + U(n-1, -5/2) - U(n, -1/2) - U(n-1, -1/2)), where U(n, x) = ChebyshevU(n, x). - G. C. Greubel, Jan 02 2023

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A110311 Expansion of 1/((1+x+x^2)(1+5x+x^2)).