A004187 -id:A004187 - cp's OEIS frontend

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

0, 1, 10, 99, 980, 9701, 96030, 950599, 9409960, 93149001, 922080050, 9127651499, 90354434940, 894416697901, 8853812544070, 87643708742799, 867583274883920, 8588189040096401, 85014307126080090, 841554882220704499, 8330534515080964900, 82463790268588944501, 816307368170808480110

Offset: 0

N. J. A. Sloane

Indices of square numbers which are also generalized pentagonal numbers.
If t(n) denotes the n-th triangular number, t(A105038(n))=a(n)*a(n+1). - Robert Phillips (bobanne(AT)bellsouth.net), May 25 2008
The n-th term is a(n) = ((5+sqrt(24))^n - (5-sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, May 31 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 10'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
a(n) and b(n) (A001079) are the nonnegative proper solutions of the Pell equation b(n)^2 - 6*(2*a(n))^2 = +1. See the cross reference to A001079 below. - Wolfdieter Lang, Jun 26 2013
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,9}. - Milan Janjic, Jan 25 2015
For n > 1, this also gives the number of (n-1)-decimal-digit numbers which avoid a particular two-digit number with distinct digits. For example, there are a(5) = 9701 4-digit numbers which do not include "39" as a substring; see Wikipedia link. - Charles R Greathouse IV, Jan 14 2016
All possible solutions for y in Pell equation x^2 - 24*y^2 = 1. The values for x are given in A001079. - Herbert Kociemba, Jun 05 2022
Dickson on page 384 gives the Diophantine equation "(20)  24x^2 + 1 = y^2" and later states "... three consecutive sets (x_i, y_i) of solutions of (20) or 2x^2 + 1 = 3y^2 satisfy x_{n+1} = 10x_n - x_{n-1}, y_{n+1} = 10y_n - y_{n-1} with (x_1, y_1) = (0, 1) or (1, 1), (x_2, y_2) = (1, 5) or (11, 9), respectively." The first set of values (x_n, y_n) = (A001079(n-1), a(n-1)). - Michael Somos, Jun 19 2023

			a(2)=10 and (3(-8)^2-(-8))/2=10^2, a(3)=99 and (3(81)^2-(81))/2=99^2. - _Michael Somos_, Sep 05 2006
G.f. = x + 10*x^2 + 99*x^3 + 980*x^4 + 9701*x^5 + 96030*x^6 + ...

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 12.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
D. Fortin, B-spline Toeplitz inverse under corner perturbations, International Journal of Pure and Applied Mathematics, Volume 77, No. 1, 2012, 107-118. - From _N. J. A. Sloane_, Oct 22 2012
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=10, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=12.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Wikipedia, Curse of 39
Jianqiang Zhao, Finite Multiple zeta Values and Finite Euler Sums, arXiv:1507.04917 [math.NT], 2015.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A001079, A001109, A001353, A001906, A004187, A004254, A018913, A046173, A049310, A054320, A072256, A101950, A105138, A108741 (squares).
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), this sequence (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=5;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[ n eq 1 select 0 else n eq 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..20] ]; // Vincenzo Librandi, Aug 19 2011

A004189 := proc(n)
    option remember;
    if n <= 1 then
        n ;
    else
        10*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A004189(n),n=0..20) ; # R. J. Mathar, Apr 30 2017
seq( simplify(ChebyshevU(n-1, 5)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n-1,1,5], {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008; modified by G. C. Greubel, Jun 06 2019 *)
LinearRecurrence[{10, -1}, {0, 1}, 20] (* Jean-François Alcover, Nov 15 2017 *)
ChebyshevU[Range[21] -2, 5] (* G. C. Greubel, Dec 23 2019 *)

{a(n) = subst(poltchebi(n+1) - 5*poltchebi(n), 'x, 5) / 24}; /* Michael Somos, Sep 05 2006 */

a(n)=([9,1;8,1]^(n-1)*[1;1])[1,1] \\ Charles R Greathouse IV, Jan 14 2016

vector(21, n, n--; polchebyshev(n-1, 2, 5) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number1(n,10,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,5) for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = S(2*n-1, sqrt(12))/sqrt(12) = S(n-1, 10); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
A001079(n) = sqrt(24*(a(n)^2)+1), that is a(n) = sqrt((A001079(n)^2-1)/24).
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ( (5+2*sqrt(6))^n - (5-2*sqrt(6))^n )/(4*sqrt(6)).
G.f.: x/(1-10*x+x^2). (End)
a(-n) = -a(n). - Michael Somos, Sep 05 2006
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 9*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 10*a(n-1) - a(n-2). (End)
a(n+1) = Sum_{k=0..n} A101950(n,k)*9^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/2*(2 + sqrt(6)).
Product {n >= 2} (1 - 1/a(n)) = 1/5*(2 + sqrt(6)). (End)
a(n) = (A054320(n-1) + A072256(n))/2. - Richard R. Forberg, Nov 21 2013
a(2*n - 1) = A046173(n).
E.g.f.: exp(5*x)*sinh(2*sqrt(6)*x)/(2*sqrt(6)). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*8^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*12^k. - Peter Bala, Jul 18 2023

A001090 a(n) = 8*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 8, 63, 496, 3905, 30744, 242047, 1905632, 15003009, 118118440, 929944511, 7321437648, 57641556673, 453811015736, 3572846569215, 28128961537984, 221458845734657, 1743541804339272, 13726875588979519, 108071462907496880, 850844827670995521, 6698687158460467288

Offset: 0

N. J. A. Sloane

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 15*y^2 = 1; the corresponding values of x are in A001091. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 02 2014]
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 8'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,...,7}. - Milan Janjic, Jan 25 2015
From Klaus Purath, Jul 25 2024: (Start)
For any three consecutive terms (x, y, z) y^2 - x*z = 1 always applies.
a(n) = (t(i+2n) - t(i))/(t(i+n+1) - t(i+n-1)) where (t) is any recurrence t(k) = 9t(k-1) - 9t(k-2) + t(k-3) or t(k) = 8t(k-1) - t(k-2) without regard to initial values.
In particular, if the recurrence (t) of the form (9,-9,1) has the initial values t(0) = 1, t(1) = 2, t(2) = 9, a(n) = t(n) - 1 applies. (End)

			G.f. = x + 8*x^2 + 63*x^3 + 496*x^4 + 3905*x^5 + 30744*x^6 + 242047*x^7 + ...

Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..100 from T. D. Noe)
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
H. Brocard, Notes élémentaires sur le problème de Peel, Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
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=8, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=10.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Shobhna Singh and Felix Flicker, Exact Solution to the Quantum and Classical Dimer Models on the Spectre Aperiodic Monotiling, arXiv:2309.14447 [cond-mat.str-el], 2023.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A001091, A001906, A004187, A004254, A049310, A070997.
Equals one-third A136325.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), this sequence (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=4;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

I:=[0,1]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017

A001090:=1/(1-8*z+z**2); # Simon Plouffe in his 1992 dissertation
seq( simplify(ChebyshevU(n-1, 4)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n-1, 1, 4], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
LinearRecurrence[{8,-1},{0,1},30] (* Harvey P. Dale, Aug 29 2012 *)
a[n_]:= ChebyshevU[n-1, 4]; (* Michael Somos, May 28 2014 *)
CoefficientList[Series[x/(1-8*x+x^2), {x,0,20}], x] (* G. C. Greubel, Dec 20 2017 *)

{a(n) = subst(poltchebi(n+1) - 4 * poltchebi(n), x, 4) / 15}; /* Michael Somos, Apr 05 2008 */

{a(n) = polchebyshev(n-1, 2, 4)}; /* Michael Somos, May 28 2014 */

my(x='x+O('x^30)); concat([0], Vec(x/(1-8*x-x^2))) \\ G. C. Greubel, Dec 20 2017

[lucas_number1(n,8,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,4) for n in (0..20)] # G. C. Greubel, Dec 23 2019

15*a(n)^2 - A001091(n)^2 = -1.
a(n) = sqrt((A001091(n)^2 - 1)/15).
a(n) = S(2*n-1, sqrt(10))/sqrt(10) = S(n-1, 8); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310, with S(-1, x) := 0.
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ((4+sqrt(15))^n - (4-sqrt(15))^n)/(2*sqrt(15)).
G.f.: x/(1-8*x+x^2). (End)
Limit_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
[A070997(n-1), a(n)] = [1,6; 1,7]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(-n) = -a(n). - Michael Somos, Apr 05 2008
a(n+1) = Sum_{k=0..n} A101950(n,k)*7^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = (1/3)*(3 + sqrt(15)).
Product_{n >= 2} (1 - 1/a(n)) = (1/8)*(3 + sqrt(15)).
(End)
a(n) = A136325(n)/3. - Greg Dresden, Sep 12 2019
E.g.f.: exp(4*x)*sinh(sqrt(15)*x)/sqrt(15). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*6^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*10^k. - Peter Bala, Jul 17 2023

More terms from Wolfdieter Lang, Aug 02 2000

A033890 a(n) = Fibonacci(4*n + 2).

Original entry on oeis.org

1, 8, 55, 377, 2584, 17711, 121393, 832040, 5702887, 39088169, 267914296, 1836311903, 12586269025, 86267571272, 591286729879, 4052739537881, 27777890035288, 190392490709135, 1304969544928657, 8944394323791464, 61305790721611591, 420196140727489673

Offset: 0

N. J. A. Sloane

(x,y) = (a(n), a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033888. - Floor van Lamoen, Dec 10 2001
This sequence consists of the odd-indexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The even-indexed terms of A001906 are in A033888. Limit_{n->infinity} a(n)/a(n-1) = phi^4 = (7 + 3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Indices of square numbers which are also 12-gonal. - Sture Sjöstedt, Jun 01 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with 3'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
If we let b(0) = 0 and, for n >= 1, b(n) = A033890(n-1), then the sequence b(n) will be F(4n-2) and the first difference is L(4n) or A056854. F(4n-2) is also the ratio of golden spiral length (rounded to the nearest integer) after n rotations. L(4n) is also the pitch length ratio. See illustration in links. - Kival Ngaokrajang, Nov 03 2013
The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
Solutions y of Pell equation x^2 - 5*y^2 = 4; corresponding x values are in A342710 (see A342709). - Bernard Schott, Mar 19 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Nathan D. Cahill and Darren A. Narayan, Fibonacci and Lucas Numbers as Tridiagonal Matrix Determinants, Fib. Quart. 42, no. 3, Aug. 2004, pp. 216-221. See p. 219.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Kival Ngaokrajang, Illustration of golden spiral length and pitch ratio
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 linear recurrences with constant coefficients, signature (7,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001906, A100047.

Cf. A342709, A342710.

[Fibonacci(4*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

A033890 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,8]);
    else
        7*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

Table[Fibonacci[4n + 2], {n, 0, 14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{7, -1}, {1, 8}, 50] (* G. C. Greubel, Jul 13 2017 *)
a[n_] := (GoldenRatio^(2 (1 + 2 n)) - GoldenRatio^(-2 (1 + 2 n)))/Sqrt[5]
Table[a[n] // FullSimplify, {n, 0, 21}] (* Gerry Martens, Aug 20 2025 *)

PARI
```
a(n)=fibonacci(4*n+2);
```

G.f.: (1+x)/(1-7*x+x^2).
a(n) = 7*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=8.
a(n) = S(n,7) + S(n-1,7) = S(2*n,sqrt(9) = 3), where S(n,x) = U(n,x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n,7) = A004187(n+1), S(n,3) = A001906(n+1).
a(n) = ((7+3*sqrt(5))^n - (7-3*sqrt(5))^n + 2*((7+3*sqrt(5))^(n-1) - ((7-3*sqrt(5))^(n-1)))) / (3*(2^n)*sqrt(5)). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -9). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-7)*(-1)^n, where L is defined as in A108299; see also A049685 for L(n,+7). - Reinhard Zumkeller, Jun 01 2005
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),7/2) + f(a(n-2),7/2). - Marcos Carreira, Dec 27 2006
a(n+1) = 8*a(n) - 8*a(n-1) + a(n-2); a(1)=1, a(2)=8, a(3)=55. - Sture Sjöstedt, May 27 2009
a(n) = A167816(4*n+2). - Reinhard Zumkeller, Nov 13 2009
a(n)=b such that (-1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2-sin(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n) = A000045(A016825(n)). - Michel Marcus, Mar 22 2015
a(n) = A001906(2*n+1). - R. J. Mathar, Apr 30 2017
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + 3*sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Apr 14 2025
From Peter Bala, Jun 08 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/9 [telescoping series: 3/(a(n) - 1/a(n)) = 1/Fibonacci(4*n+4) + 1/Fibonacci(4*n)].
Product_{n >= 1} (a(n) + 3)/(a(n) - 3) = 5/2 [telescoping product:
(a(n) + 3)/(a(n) - 3) = b(n)/b(n-1), where b(n) = (Lucas(4*n+4) - 3)/(Lucas(4*n+4) + 3)].
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(9/5) [telescoping product:
(a(n) + 1)/(a(n) - 1) = c(n)/c(n-1) for n >= 1, where c(n) = Fibonacci(2*n+2)/Lucas(2*n+2)]. (End)
From Gerry Martens, Aug 20 2025: (Start)
a(n) = ((3 + sqrt(5))^(1 + 2*n) - (3 - sqrt(5))^(1 + 2*n)) / (2^(1 + 2*n)*sqrt(5)).
a(n) = Sum_{k=0..2*n} binomial(2*n + k + 1, 2*k + 1). (End)

A056854 a(n) = Lucas(4*n).

Original entry on oeis.org

2, 7, 47, 322, 2207, 15127, 103682, 710647, 4870847, 33385282, 228826127, 1568397607, 10749957122, 73681302247, 505019158607, 3461452808002, 23725150497407, 162614600673847, 1114577054219522, 7639424778862807, 52361396397820127, 358890350005878082, 2459871053643326447

Offset: 0

Barry E. Williams, Aug 29 2000

a(n) and b(n) := A004187(n) are the nonnegative proper and improper solutions of the Pell equation a(n)^2 - 5*(3*b(n))^2 = +4. See the cross-reference to A004187 below. - Wolfdieter Lang, Jun 26 2013

Lucas numbers of the form n^2-2. - Michel Lagneau, Aug 11 2014

			Pell equation: n = 0, 2^2 - 45*0^2 = +4 (improper);  n = 1, 7^2 - 5*(3*1)^2 = +4; n=2, 47^2 - 5*(3*7)^2 = +4. - _Wolfdieter Lang_, Jun 26 2013

R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Dale Gerdemann, Combinatorial proofs of Zeckendorf family identities, Fib. Quart. 46/47 (2009) 249. [_N. J. A. Sloane_, Dec 05 2009]
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A004187, A005248.
Cf. quadrisection of A000032: this sequence (first), A056914 (second), A246453 (third, without 11), A288913 (fourth).
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[Lucas(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 7; a[n_] := 7a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{7,-1},{2,7},25] (* or *) LucasL[4*Range[0,25]] (* Harvey P. Dale, Aug 08 2011 *)

a(n)=if(n<0,0,polsym(1-7*x+x^2,n)[n+1])

a(n)=if(n<0,0,2*subst(poltchebi(n),x,7/2))

[lucas_number2(n,7,1) for n in range(27)] #Zerinvary Lajos, Jun 25 2008

a(n) = 7*a(n-1) - a(n-2) with a(0)=2, a(1)=7.
a(n) = A000032(4*n), where A000032 = Lucas numbers.
a(n) = 7*S(n-1, 7) - 2*S(n-2, 7) = S(n, 7) - S(n-2, 7) = 2*T(n, 7/2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 7) = A004187(n), n>=0. See A049310 and A053120.
a(n) = ((7+sqrt(45))/2)^n + ((7-sqrt(45))/2)^n.
G.f.: (2-7x)/(1-7x+x^2).
a(n) = A005248(2*n); bisection of A005248.
a(n) = Fibonacci(8*n)/Fibonacci(4*n), n>0. - Gary Detlefs, Dec 26 2010
a(n) = 2 + 5*Fibonacci(2*n)^2 = 2 + 5*A049684(n), n >= 0. This is in Koshy's book (reference under A065563) 15. on p. 88. Compare with the above Chebyshev T formula. - Wolfdieter Lang, Aug 27 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(7 - 3*sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.14242 42709 40138 85949 ... = 2 + 1/(7 + 1/(47 + 1/(322 + ...))).
Also F(-alpha) = 0.85670 72882 04563 14901 ... has the continued fraction representation 1 - 1/(7 - 1/(47 - 1/(322 - ...))) and the simple continued fraction expansion 1/(1 + 1/((7-2) + 1/(1 + 1/((47-2) + 1/(1 + 1/((322-2) + 1/(1 + ...))))))). Cf. A005248.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((7^2-4) + 1/(1 + 1/((47^2-4) + 1/(1 + 1/((322^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + (1/2)*F(alpha)/F(-alpha) = 1.16675297774947414828... has the simple continued fraction expansion 1 + 1/((7 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/(15127 - 2) + 1/(1 + ...))))). (End)
a(n) = Fibonacci(4*n+2) - Fibonacci(4*n-2), where Fibonacci(-2) = -1. - Bruno Berselli, May 25 2015
a(n) = sqrt(45*(A004187(n))^2+4).
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(4*n+4)/F(4) - F(4*n-4)/F(4) = A004187(n+1) - A004187(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^4 = [2, 3; 3, 5].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) (mod p^k) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
5*Sum_{n >= 1} 1/(a(n) - 9/a(n)) = 1: (9 = Lucas(4)+2 and 5 = Lucas(4)-2)
9*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4)*( theta_3((7-3*sqrt(5))/2)^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. Cf. A153415.
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4)*( 1 - theta_3((3*sqrt(5)-7)/2)^2 ).
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 7*x^2 + 48*x^3 + 329*x^4 + ... is the o.g.f. for A004187. (End)
E.g.f.: 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
a(2k+1)/7 is the numerator of the continued fraction [3*sqrt(5), 3*sqrt(5), ..., 3*sqrt(5)] with 2k copies of 3*sqrt(5), for k>0. - Greg Dresden and Tracy Z. Wu, Sep 10 2020
a(n) = Sum_{k>=1} Lucas(2*n*k)/(Lucas(2*n)^k). - Diego Rattaggi, Jan 20 2025

More terms from James Sellers, Aug 31 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A049685 a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).

Original entry on oeis.org

1, 6, 41, 281, 1926, 13201, 90481, 620166, 4250681, 29134601, 199691526, 1368706081, 9381251041, 64300051206, 440719107401, 3020733700601, 20704416796806, 141910183877041, 972666870342481, 6666757908520326, 45694638489299801, 313195711516578281

Offset: 0

Clark Kimberling

In general, Sum_{k=0..n} binomial(2*n-k,k)j^(n-k) = (-1)^n*U(2n, I*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,7), where L is defined as in A108299; see also A033890 for L(n,-7). - Reinhard Zumkeller, Jun 01 2005
Take 7 numbers consisting of 5 ones together with any two successive terms from this sequence. This set has the property that the sum of their squares is 7 times their product. (R. K. Guy, Oct 12 2005.) See also A111216.
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
From Wolfdieter Lang, Feb 09 2021: (Start)
All positive solutions of the Diophantine equation x^2 + y^2 - 7*x*y = -5 are given by [x(n) = S(n, 7) - S(n-1, 7), y(n) = x(n-1)], for all integer numbers n, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-n, x) = -S(n-2, x), for n >= 2. x(n) = a(n), for n >= 0.
This indefinite binary quadratic form has discriminant D = +45. There is only this family representing -5 properly with x and y positive, and there are no improper solutions.
All proper and improper solutions of the generalized Pell equation X^2 - 45*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n) from the preceding comment, by X(n) = x(n) + x(n-1) = S(n-1, 7) - S(n-2, 7) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 7), for all integer numbers n. For positive integers X(n) = A056854(n) and Y(n) = A004187(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper family of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer numbers n.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

			a(3) = L(4*3 + 2)/3 = 843/3 = 281. - _Indranil Ghosh_, Feb 06 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1193
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 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 linear recurrences with constant coefficients, signature (7,-1).

Row 7 of array A094954. First differences of A004187.
Cf. A002310, A002320, A049310, A049685, A056854.
Cf. similar sequences listed in A238379.

[Lucas(4*n+2)/3: n in [0..30]]; // G. C. Greubel, Dec 17 2017

Table[LucasL[4*n+2]/3, {n,0,50}] (* or *) LinearRecurrence[{7,-1}, {1,6}, 50] (* G. C. Greubel, Dec 17 2017 *)

a(n)=(fibonacci(4*n+1)+fibonacci(4*n+3))/3 \\ Charles R Greathouse IV, Jun 16 2014

[lucas_number1(n,7,1)-lucas_number1(n-1,7,1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i); then q(n, 5)=a(n); a(n) = 7a(n-1) - a(n-2). - Benoit Cloitre, Nov 10 2002
From Ralf Stephan, May 29 2004: (Start)
a(n+2) = 7a(n+1) - a(n).
G.f.: (1-x)/(1-7x+x^2).
a(n)*a(n+3) = 35 + a(n+1)*a(n+2). (End)
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*5^k. - Paul Barry, Aug 30 2004
If another "1" is inserted at the beginning of the sequence, then A002310, A002320 and A049685 begin with 1, 2; 1, 3; and 1, 1; respectively and satisfy a(n+1) = (a(n)^2+5)/a(n-1). - Graeme McRae, Jan 30 2005
a(n) = (-1)^n*U(2n, i*sqrt(5)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
[a(n), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = S(n, 7) - S(n-1, 7) with Chebyshev S polynomials S(n-1, 7) = A004187(n), for n >= 0. - Wolfdieter Lang, Feb 09 2021
E.g.f.: exp(7*x/2)*(3*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/3. - Stefano Spezia, Apr 14 2025
From Peter Bala, May 04 2025: (Start)
a(n) = sqrt(2/9) * sqrt(1 - T(2*n+1, -7/2)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).
The aerated sequence [b(n)]n>=1 = [1, 0, 6, 0, 41, 0, 281, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -9, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/5 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A290903(n-1) - 1/A290903(n).) (End)

A033888 a(n) = Fibonacci(4*n).

Original entry on oeis.org

0, 3, 21, 144, 987, 6765, 46368, 317811, 2178309, 14930352, 102334155, 701408733, 4807526976, 32951280099, 225851433717, 1548008755920, 10610209857723, 72723460248141, 498454011879264, 3416454622906707, 23416728348467685, 160500643816367088, 1100087778366101931, 7540113804746346429

Offset: 0

N. J. A. Sloane

(x,y)=(a(n),a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033890. - Floor van Lamoen, Dec 10 2001
Sequence A033888 provides half of the solutions to the equation 5*x^2 + 4 is a square. The other half are found in A033890. Lim_{n->infinity} a(n)/a(n-1) = phi^4 = (7+3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002
Fibonacci numbers divisible by 3. - Reinhard Zumkeller, Aug 20 2011

			G.f. = 3*x + 21*x^2 + 144*x^3 + 987*x^4 + 6765*x^5 + 46368*x^6 + ...

Nathaniel Johnston, Table of n, a(n) for n = 0..300
Piero Filipponi and Marco Bucci, On the Integrity of Certain Fibonacci Sums, The Fibonacci Quarterly, Vol. 32, No. 3 (1994), pp. 245-252.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A033890, A337928, A337929.

Fourth column of array A102310.

[ Fibonacci(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 15 2011

A033888:=n->combinat[fibonacci](4*n): seq(A033888(n), n=0..30); # Wesley Ivan Hurt, Apr 26 2017

Table[Fibonacci[4*n], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
Table[Mod[Fibonacci[(4 n + 2)] , Fibonacci[(4 n + 1)]], {n, 1, 10}] (* Artur Jasinski, Nov 15 2011 (corrected by Iain Fox, Dec 18 2017) *)

numlib::fibonacci(n*4) $ n = 0..30; // Zerinvary Lajos, May 08 2008

a(n)=fibonacci(4*n) \\ Charles R Greathouse IV, Feb 03 2014

first(n) = Vec(3*x/(1 - 7*x + x^2) + O(x^n), -n) \\ Iain Fox, Dec 18 2017

a(n) = fibonacci(4*n + 2) % fibonacci(4*n + 1) \\ Iain Fox, Dec 18 2017

[lucas_number1(n,3,1)*lucas_number2(n,3,1) for n in range(0,21)] # Zerinvary Lajos, Jun 28 2008

[fibonacci(4*n) for n in range(0, 20)] # Zerinvary Lajos, May 15 2009

a(n) = 7*a(n-1) - a(n-2).
a(n) = ((7+3*sqrt(5))^(n-1) - (7-3*sqrt(5))^(n-1)) / ((2^(n-1))*sqrt(5)). - Gregory V. Richardson, Oct 13 2002
a(n) = Sum_{k=0..n} F(3*n-k)*binomial(n, k). - Benoit Cloitre, Jun 07 2004
a(n) = Lucas(2*n) * Lucas(n) * Fibonacci(n). - Ralf Stephan, Sep 25 2004
G.f.: 3*x/(1-7*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 3*A004187(n). - R. J. Mathar, Sep 03 2010
a(n) = Fibonacci[(4*n + 2)] modulo Fibonacci[(4*n + 1)]. - Artur Jasinski, Nov 15 2011 (corrected by Iain Fox, Dec 18 2017)
a(n) = (A337929(n) + A337928(n)) / 2. - Flávio V. Fernandes, Feb 06 2021
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Feb 07 2021
a(n) = Sum_{k>=0} Fibonacci(2*n*k)/Lucas(2*n)^k (Filipponi and Bucci, 1994). - Amiram Eldar, Jan 17 2022
From Peter Bala, May 22 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/3 (telescoping series: 3/(a(n) - 1/a(n)) = 1/A033890(n) + 1/A033890(n-1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) (telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (5 - 4/Fibonacci(2*n+1)^2)/(5 - 4/Fibonacci(2*n-1)^2) from which we get Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5 - 4/Fibonacci(2*n+1)^2). (End)

A028412 Rectangular array of numbers Fibonacci(m(n+1))/Fibonacci(m), m >= 1, n >= 0, read by downward antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 4, 8, 3, 1, 7, 17, 21, 5, 1, 11, 48, 72, 55, 8, 1, 18, 122, 329, 305, 144, 13, 1, 29, 323, 1353, 2255, 1292, 377, 21, 1, 47, 842, 5796, 15005, 15456, 5473, 987, 34, 1, 76, 2208, 24447, 104005, 166408, 105937, 23184, 2584, 55, 1, 123, 5777

Offset: 0

N. J. A. Sloane

Every integer-valued quotient of two Fibonacci numbers is in this array. - Clark Kimberling, Aug 28 2008

Not only does 5 divide row 5, but 50 divides (-5 + row 5), as in A214984. - Clark Kimberling, Nov 02 2012

			   1   1    1      1       1        1
   1   3    4      7      11       18
   2   8   17     48     122      323
   3  21   72    329    1353     5796
   5  55  305   2255   15005   104005
   8 144 1292  15456  166408  1866294
  13 377 5473 105937 1845493 33489287
  ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 142.

Clark Kimberling, Table of n, a(n) for n = 0..1829
I. Strazdins, Lucas factors and a Fibonomial generating function, in Applications of Fibonacci numbers, Vol. 7 (Graz, 1996), 401-404, Kluwer Acad. Publ., Dordrecht, 1998.

Columns include A000045, A001906, A001076, A004187, A049666, A049660, A049667, A049668, A049669, A049670.
Rows include (essentially) A000032, A047946, A083564, A103226.
Main diagonal is A051294.
Transpose is A214978.

max = 11; col[m_] := CoefficientList[ Series[ 1/(1 - LucasL[m]*x + (-1)^m*x^2), {x, 0, max}], x]; t = Transpose[ Table[ col[m], {m, 1, max}]] ; Flatten[ Table[ t[[n - m + 1, m]], {n, 1, max }, {m, n, 1, -1}]] (* Jean-François Alcover, Feb 21 2012, after Paul D. Hanna *)
f[n_] := Fibonacci[n]; t[m_, n_] := f[m*n]/f[n]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]] (* array *)
t = Flatten[Table[t[k, n + 1 - k], {n, 1, 120}, {k, 1, n}]] (* sequence *) (* Clark Kimberling, Nov 02 2012 *)

{T(n,m)=polcoeff(1/(1 - Lucas(m)*x + (-1)^m*x^2 +x*O(x^n)),n)}

T(n, m) = Sum_{i_1>=0} Sum_{i_2>=0} ... Sum_{i_m>=0} C(n-i_m, i_1)*C(n-i_1, i_2)*C(n-i_2, i_3)*...*C(n-i_{m-1}, i_m).

G.f. for column m >= 1: 1/(1 - Lucas(m)*x + (-1)^m*x^2), where Lucas(m) = A000204(m). - Paul D. Hanna, Jan 28 2012

More terms from Erich Friedman, Jun 03 2001
Edited by Ralf Stephan, Feb 03 2005
Better description from Clark Kimberling, Aug 28 2008

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

Original entry on oeis.org

1, 8, 56, 385, 2640, 18096, 124033, 850136, 5826920, 39938305, 273741216, 1876250208, 12860010241, 88143821480, 604146740120, 4140883359361, 28382036775408, 194533374068496, 1333351581704065, 9138927697859960

Offset: 1

K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, Apr 06 2004

a(n) such that 9*(T(a(n)-1) + T(a(n+1)-1)) = 7*(T(a(n) + a(n+1) - 1)), where T(i) denotes the i-th triangular number.
Partial sums of Chebyshev sequence S(n,7) = U(n,7/2) = A004187(n+1). - Wolfdieter Lang, Aug 31 2004
From Klaus Purath, Aug 06 2025: (Start)
Numbers k such that both 3*k + 1 and 15*k + 1 are perfect squares. Also the sum of two consecutive terms is a square.
Take any recurrence (r) of the form (3,-1) with initial value 0 followed by an arbitrary positive integer i. Then the product of two consecutive terms of r divided by 3*i^2 gives the current sequence. (End)

			G.f. = x + 8*x^2 + 56*x^3 + 385*x^4 + 2640*x^5 + 18096*x^6 + ... - _Michael Somos_, Jan 23 2025

Michael De Vlieger, Table of n, a(n) for n = 1..1197
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A212336 for more sequences with g.f. of the type 1/(1 - k*x + k*x^2 - x^3).

Cf. A000032, A004187, A058038, A081079.

A092521:= func< n | (Lucas(4*n+2) -3)/15 >; // G. C. Greubel, Jun 12 2025

a[1] = 1; a[2] = 8; a[3] = 56; a[n_] := a[n] = 8 a[n - 1] - 8 a[n - 2] + a[n - 3]; Table[ a[n], {n, 20}] (* Robert G. Wilson v, Apr 08 2004 *)
Table[(LucasL[4n+2]-3)/15, {n, 1, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
LinearRecurrence[{8,-8,1},{1,8,56},30] (* Harvey P. Dale, Dec 27 2015 *)

Vec(x/((1-x)*(1-7*x+x^2)) + O(x^100)) \\ Altug Alkan, Oct 29 2015

def A092521(n): return (lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

G.f.: x/(1 - 8*x + 8*x^2 - x^3) = x/((1 - x)*(1 - 7*x + x^2)).
a(n) = 7*a(n-1) - a(n-2) + 1, n>=2, a(0):=0, a(1)=1.
a(n) = (S(n, 7)-S(n-1, 7) -1)/5, n>=1, with S(n, 7) = U(n, 7/2) = A004187(n+1).
a(n) = A058038(n)/3.
a(n) = (1/3)*Sum_{k=0..n} Fibonacci(4*k). - Gary Detlefs, Dec 07 2010
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 23 2025
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = A081079(n)/15.
E.g.f.: (1/15)*( exp(7*x/2)*( 3*cosh(p*x) + sqrt(5)*sinh(p*x) ) - 3*exp(x) ), where p = 3*sqrt(5)/2. (End)

Edited and extended by Robert G. Wilson v, Apr 08 2004

A018913 a(n) = 9*a(n - 1) - a(n - 2) for n>1, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 9, 80, 711, 6319, 56160, 499121, 4435929, 39424240, 350382231, 3114015839, 27675760320, 245967827041, 2186034683049, 19428344320400, 172669064200551, 1534593233484559, 13638670037160480, 121213437100959761

Offset: 0

R. K. Guy

Define the sequence L(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0. This is L(1,9).
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 9'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,...,8}. - Milan Janjic, Jan 25 2015
Not to be confused with the Pisot L(1,9) sequence, which is A001019. - R. J. Mathar, Feb 13 2016
Lim_{n->oo} a(n+1)/a(n) = (9 + sqrt(77))/2 = A092290 + 1 = 8.887482... - Wolfdieter Lang, Nov 16 2023

			G.f. = x + 9*x^2 + 80*x^3 + 711*x^4 + 6319*x^5 + 56160*x^6 + 499121*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 12.
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
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=9, q=-1.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
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=11.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).
Index entries for Pisot sequences

Cf. A000027, A001906, A001353, A004254, A001109, A004187, A001090.
Cf. A056918(n)=sqrt{77*(a(n))^2 +4}, that is, a(n)=sqrt((A056918(n)^2 - 4)/77).
Cf. A092290 + 1.

I:=[0, 1]; [n le 2 select I[n] else 9*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

CoefficientList[Series[x/(1 - 9*x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 23 2012 *)

concat(0, Vec(x/(1-9*x+x^2) + O(x^30))) \\ Michel Marcus, Sep 06 2017

[lucas_number1(n,9,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

G.f.: x/(1-9*x+x^2).
a(n) = S(2*n-1, sqrt(11))/sqrt(11) = S(n-1, 9); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
a(n) = (((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n)/sqrt(77). - Barry E. Williams, Aug 21 2000
a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*8^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/7*(7 + sqrt(77)).
Product {n >= 2} (1 - 1/a(n)) = 1/18*(7 + sqrt(77)). (End)
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*7^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*11^k. - Peter Bala, Jul 17 2023
E.g.f.: 2*exp(9*x/2)*sinh(sqrt(77)*x/2)/sqrt(77). - Stefano Spezia, Feb 23 2025
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = sqrt(11/7) [telescoping product: ((a(2*n) + 1)/(a(2*n) - 1))^2 = (11 - 4/(a(n+1) - a(n))^2)/(11 - 4/(a(n) - a(n-1))^2), leading to 11 - 7*Product_{k = 1..n} ((a(2*k) + 1)/(a(2*k) - 1))^2 = 4/A070998(n)^2]. - Peter Bala, May 18 2025

G.f. adapted to the offset by Vincenzo Librandi, Dec 23 2012

cp's OEIS Frontend

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

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A001090 a(n) = 8*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

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A033890 a(n) = Fibonacci(4*n + 2).

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A056854 a(n) = Lucas(4*n).

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A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

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A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.