A056854 -id:A056854 - cp's OEIS frontend

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

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)

A014448 Even Lucas numbers: L(3n).

Original entry on oeis.org

2, 4, 18, 76, 322, 1364, 5778, 24476, 103682, 439204, 1860498, 7881196, 33385282, 141422324, 599074578, 2537720636, 10749957122, 45537549124, 192900153618, 817138163596, 3461452808002, 14662949395604, 62113250390418

Offset: 0

Mohammad K. Azarian

This is the Lucas sequence V(4,-1). - Bruno Berselli, Jan 08 2013

			a(4) = L(3 * 4) = L(12) = 322. - _Indranil Ghosh_, Feb 05 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1591
Pooja Bhadouria, Deepika Jhala and Bijendra Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1 (2014), pp. 81-92; sequence L_{4,n}.
H. H. Ferns, Problem B-115, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 2 (1967), p. 202; Identities for F_{kn} and L{kn}, Solution to Problem B-115 by Stanley Rabinowitz, ibid., Vol. 6, No. 1 (1968), pp. 92-93.
Tanya Khovanova, Recursive Sequences.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
Roberto Tauraso, Some Congruences for Central Binomial Sums Involving Fibonacci and Lucas Numbers, JIS 19 (2016) # 16.5.4
Wikipedia, Lucas sequence: Specific names.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A000045, A001077.

Cf. Lucas(k*n): A005248 (k = 2), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(3*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

Table[LucasL[3*n], {n,0,100}] (* G. C. Greubel, Nov 07 2018 *)

PARI
```
polsym(x^2-4*x-1,100)
```

a(n)=sum(k=0,n,binomial(n,k)*(fibonacci(n+k-1)+fibonacci(n+k+1))) \\ Paul D. Hanna, Oct 19 2010

[lucas_number2(n,4,-1) for n in range(0, 23)] # Zerinvary Lajos, May 14 2009

G.f.: (2-4*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) +a(n-2) with n>1, a(0)=2, a(1)=4.
a(n) = (2+sqrt(5))^n + (2-sqrt(5))^n.
a(n) = 2*A001077(n).
a(n) = A000032(3*n).
a(n) = Sum_{k=0..n} C(n,k)*Lucas(n+k). - Paul D. Hanna, Oct 19 2010
a(n) = Fibonacci(6*n)/Fibonacci(3*n), n>0. - Gary Detlefs, Dec 26 2010
From Peter Bala, Mar 22 2015: (Start)
a(n) = ( Fibonacci(3*n + 2*k) - F(3*n - 2*k) )/Fibonacci(2*k) for nonzero integer k.
a(n) = ( Fibonacci(3*n + 2*k + 1) + F(3*n - 2*k - 1) )/Fibonacci(2*k + 1) for arbitrary integer k. (End)
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 20*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = L(n)*(L(n-1)*L(n+1) + 2*(-1)^n). - J. M. Bergot, Feb 05 2016
From Peter Bala Oct 14 2019: (Start)
Sum_{n >= 1} 1/( a(n) + (-1)^(n+1)*20/a(n) ) = 3/16.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + (-1)^(n+1)*20/a(n) ) = 1/16. (End)
a(n) = (15*Fibonacci(n)^2*Lucas(n) + Lucas(n)^3)/4 (Ferns, 1967). - Amiram Eldar, Feb 06 2022
E.g.f.: 2*exp(2*x)*cosh(sqrt(5)*x). - Stefano Spezia, Jan 18 2025

More terms from Erich Friedman

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)

A049684 a(n) = Fibonacci(2n)^2.

Original entry on oeis.org

0, 1, 9, 64, 441, 3025, 20736, 142129, 974169, 6677056, 45765225, 313679521, 2149991424, 14736260449, 101003831721, 692290561600, 4745030099481, 32522920134769, 222915410843904, 1527884955772561, 10472279279564025, 71778070001175616, 491974210728665289

Offset: 0

Clark Kimberling

This is the r=9 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.

Apparently, this sequence consists of those nonnegative integers k for which x*(x^2-1)*y*(y^2-1) = k*(k^2-1) has a solution in nonnegative integers x, y. If k = a(n), x = A000045(2*n-1) and y = A000045(2*n+1) are a solution. See A374375 for numbers k*(k^2-1) that can be written as a product of two or more factors of the form x*(x^2-1). - Pontus von Brömssen, Jul 14 2024

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 27.
H. J. H. Tuenter, Fibonacci summation identities arising from Catalan's identity, Fib. Q., 60:4 (2022), 312-319.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059.
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.
Pridon Davlianidze, Problem B-1264, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; It's All About Catalan, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 1, k=2.
R. Stephan, Boring proof of a nonlinearity
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

First differences give A033890.
First differences of A103434.
Bisection of A007598 and A064841.
a(n) = A064170(n+2) - 1 = (1/5) A081070.
Cf. A000032, A000045, A001622, A001906, A056854, A065563, A092184, A374375.

Join[{a=0, b=1}, Table[c=7*b-1*a+2; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Fibonacci[Range[0, 40, 2]]^2 (* Harvey P. Dale, Mar 22 2012 *)
Table[Fibonacci[n - 1] Fibonacci[n + 1] - 1, {n, 0, 40, 2}] (* Bruno Berselli, Feb 12 2015 *)
LinearRecurrence[{8, -8, 1},{0, 1, 9},21] (* Ray Chandler, Sep 23 2015 *)

numlib::fibonacci(2*n)^2 $ n = 0..35; // Zerinvary Lajos, May 13 2008

PARI
```
a(n)=fibonacci(2*n)^2
```

[fibonacci(2*n)^2 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009

G.f.: (x+x^2) / ((1-x)*(1-7*x+x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) with n>2, a(0)=0, a(1)=1, a(2)=9.
a(n) = 7*a(n-1) - a(n-2) + 2 = A001906(n)^2.
a(n) = (A000032(4*n)-2)/5. [This is in Koshy's book (reference under A065563) on p. 88, attributed to Lucas 1876.] - Wolfdieter Lang, Aug 27 2012
a(n) = 1/5*(-2 + ( (7+sqrt(45))/2 )^n + ( (7-sqrt(45))/2 )^n). - Ralf Stephan, Apr 14 2004
a(n) = 2*(T(n, 7/2)-1)/5 with twice the Chebyshev polynomials of the first kind evaluated at x=7/2: 2*T(n, 7/2)= A056854(n). - Wolfdieter Lang, Oct 18 2004
a(n) = F(2*n-1)*F(2*n+1)-1, see A064170 - Bruno Berselli, Feb 12 2015
a(n) = Sum_{i=1..n} F(4*i-2) for n>0. - Bruno Berselli, Aug 25 2015
From Peter Bala, Nov 20 2019: (Start)
Sum_{n >= 1} 1/(a(n) + 1) = (sqrt(5) - 1)/2.
Sum_{n >= 1} 1/(a(n) + 4) = (3*sqrt(5) - 2)/16. More generally, it appears that
Sum_{n >= 1} 1/(a(n) + F(2*k+1)^2) = ((2*k+1)*F(2*k+1)*sqrt(5) - Lucas(2*k+1))/ (2*F(2*k+1)*F(4*k+2)) for k = 0,1,2,....
Sum_{n >= 2} 1/(a(n) - 1) = (8 - 3*sqrt(5))/9. (End)
E.g.f.: (1/5)*(-2*exp(x) + exp((16*x)/(1 + sqrt(5))^4) + exp((1/2)*(7 + 3*sqrt(5))*x)). - Stefano Spezia, Nov 23 2019
Product_{n>=2} (1 - 1/a(n)) = phi^2/3, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021
a(n) = A092521(n-1)+A092521(n). - R. J. Mathar, Nov 22 2024

Better description and more terms from Michael Somos

A087215 Lucas(6n): a(n) = 18a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.

Original entry on oeis.org

2, 18, 322, 5778, 103682, 1860498, 33385282, 599074578, 10749957122, 192900153618, 3461452808002, 62113250390418, 1114577054219522, 20000273725560978, 358890350005878082, 6440026026380244498, 115561578124838522882, 2073668380220713167378

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

a(n+1)/a(n) converges to 9 + sqrt(80) = 17.9442719... a(0)/a(1) = 2/18; a(1)/a(2) = 18/322; a(2)/a(3) = 322/5778; a(3)/a(4) = 5778/103682; etc.
Lim_{n -> oo} a(n)/a(n+1) = 0.05572809000084... = 1/(9 + sqrt(80)) = 9 - sqrt(80).
From Peter Bala, Oct 13 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = (1/2)*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^6) = 1.0555459720... = 1 + 1/(18 + 1/(322 + 1/(5778 + ...))).
Also F(-Phi^6) = 0.9444348576... has the continued fraction representation 1 - 1/(18 - 1/(322 - 1/(5788 - ...))) and the simple continued fraction expansion 1/(1 + 1/((18 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/((5788 - 2) + 1/(1 + ...))))))).
F(Phi^6)*F(-Phi^6) = 0.9968944099... has the simple continued fraction expansion 1/(1 + 1/((18^2 - 4) + 1/(1 + 1/((322^2 - 4) + 1/(1 + 1/((5788^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^6)/F(-Phi^6) = 1.0588241282... has the simple continued fraction expansion 1 + 1/((18 - 2) + 1/(1 + 1/((5778 - 2) + 1/(1 + 1/(1860498 - 2) + 1/(1 + ...))))). (End)

			a(4) = 103682 = 18*a(3) - a(2) = 18*5778 - 322 = (9 + sqrt(80))^4 + (9 - sqrt(80))^4 = 103681.99999035512... + 0.00000964487... = 103682.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Colin Barker, Table of n, a(n) for n = 0..750
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence R_4.
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 linear recurrences with constant coefficients, signature (18,-1).

Cf. A074919.
Row 2 * 2 of array A188645.
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[ Lucas(6*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 18; a[n_] := 18a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
Table[LucasL[6n], {n, 0, 18}]  (* or *) CoefficientList[Series[2*(1 - 9*x)/(1 - 18*x + x^2), {x, 0, 17}], x] (* Indranil Ghosh, Mar 15 2017 *)

Vec(2*(1-9*x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016

a(n) = if(n<2, 17^n + 1, 18*a(n - 1) - a(n - 2));
for(n=0, 17, print1(a(n),", ")) \\ Indranil Ghosh, Mar 15 2017

a(n) = A000032(6*n).
a(n) = 18*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 18.
a(n) = (9 + sqrt(80))^n + (9 - sqrt(80))^n.
G.f.: 2*(1-9*x)/(1-18*x+x^2). - Philippe Deléham, Nov 17 2008
a(n) = 2*A023039(n). - R. J. Mathar, Oct 22 2010
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(6*n+6)/F(6) - F(6*n-6)/F(6) = A049660(n+1) - A049660(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^6 = [5, 8; 8, 13].
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).
16*Sum_{n >= 1} 1/(a(n) - 20/a(n)) = 1: (20 = Lucas(6) + 2 and 16 = Lucas(6) - 2)
20*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 16/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/16 - 20*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 20)).
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/20 + 16*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(9-4*sqrt(5)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(4*sqrt(5)-9))^2 )/4,
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 18*x^2 + 323*x^3 + ... is the o.g.f. for A049660. (End)
E.g.f.: 2*exp(9*x)*cosh(4*sqrt(5)*x). - Stefano Spezia, Oct 18 2019
a(n) = L(2n-1)^2 * F(2n+1) + L(2n+1)^2 * F(2n-1), where F(n) = A000045(n) and L(n) = A000032(n). - Diego Rattaggi, Nov 12 2020
From  Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^3 - 3*Lucas(2*n) = 2*T(3, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3(9 - sqrt(80))^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3(sqrt(80) - 9)^2), where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A153415 Decimal expansion of Sum_{n>=1} 1/A000032(2*n).

Original entry on oeis.org

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

Offset: 0

Eric W. Weisstein, Dec 25 2008

From Peter Bala, Oct 15 2019: (Start)
c = (1/4)*(theta_3( (3-sqrt(5))/2 )^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A056854.
Series acceleration formulas (L(n) = A000032(n)):
c = 1 - 5*Sum_{n >= 1} 1/( L(2*n)*(L(2*n)^2 - 5) ).
c = (1/6) + 15*Sum_{n >= 1} 1/( L(2*n)*(L(2*n)^2 + 5) ).
c = (11/16) - 10*Sum_{n >= 1} (L(2*n)^2 - 10)/( L(2*n)*(L(2*n)^2 - 5)*(L(2*n)^2 - 20) ). (End)
Compare with Sum_{n >= 1} 1/(L(2*n) - sqrt(5)) = phi and Sum_{n >= 1} 1/(L(2*n) + sqrt(5)) = 2 - phi, where phi = (sqrt(5) + 1)/2. - Peter Bala, Nov 23 2019
This constant is transcendental (Duverney et al., 1997). - Amiram Eldar, Oct 30 2020

			0.56617767581138455027...

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.

Daniel Duverney, Keiji Nishioka, Kumiko Nishioka and Iekata Shiokawa, Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers, Proceedings of the Japan Academy, Series A, Mathematical Sciences, Vol. 73, No. 7 (1997), pp. 140-142.
Index entries for transcendental numbers

Cf. A000032, A000122, A056854, A093540, A153416.

First[ RealDigits[ N[(EllipticTheta[3, 0, GoldenRatio^(-2)]^2 - 1)/4, 120], 10, 105]](* Jean-François Alcover, Jun 07 2012, after Eric W. Weisstein *)

th3(x)=1 + 2*suminf(n=1,x^n^2)
phi=(sqrt(5)+1)/2
(th3(phi^-2)^2-1)/4 \\ Charles R Greathouse IV, Jun 06 2016

A087265 Lucas numbers L(8*n).

Original entry on oeis.org

2, 47, 2207, 103682, 4870847, 228826127, 10749957122, 505019158607, 23725150497407, 1114577054219522, 52361396397820127, 2459871053643326447, 115561578124838522882, 5428934300813767249007, 255044350560122222180447

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 19 2003

a(n+1)/a(n) converges to (47+sqrt(2205))/2 = 46.9787137... a(0)/a(1)=2/47; a(1)/a(2)=47/2207; a(2)/a(3)=2207/103682; a(3)/a(4)=103682/4870847; etc. Lim_{n->infinity} a(n)/a(n+1) = 0.02128623625... = 2/(47+sqrt(2205)) = (47-sqrt(2205))/2.
a(n) = a(-n). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^8) = 1.0212763906... = 1 + 1/(47 + 1/(2207 + 1/(103682 + ...))).
Also F(-Phi^8) = 0.9787231991... has the continued fraction representation 1 - 1/(47 - 1/(2207 - 1/(103682 - ...))) and the simple continued fraction expansion 1/(1 + 1/((47 - 2) + 1/(1 + 1/((2207 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + ...))))))).
F(Phi^8)*F(-Phi^8) = 0.9995468962... has the simple continued fraction expansion 1/(1 + 1/((47^2 - 4) + 1/(1 + 1/((2207^2 - 4) + 1/(1 + 1/((103682^2 - 4) + 1/(1 + ...))))))).
1/2 + 1/2*F(Phi^8)/F(-Phi^8) = 1.0217391349... has the simple continued fraction expansion 1 + 1/((47 - 2) + 1/(1 + 1/((103682 - 2) + 1/(1 + 1/(228826127 - 2) + 1/(1 + ...))))). (End)

			a(4) = 4870847 = 47*a(3) - a(2) = 47*103682 - 2207=((47+sqrt(2205))/2)^4 + ( (47-sqrt(2205))/2)^4 =4870846.999999794696 + 0.000000205303 = 4870847.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..596
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 linear recurrences with constant coefficients, signature (47,-1).

Cf. A000032. Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

a(n) = A000032(8n).

[ Lucas(8*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a:= n-> (Matrix([[2,47]]). Matrix([[47,1],[ -1,0]])^(n))[1,1]:
seq(a(n), n=0..14);  # Alois P. Heinz, Aug 07 2008

LucasL[8*Range[0,20]] (* or *) LinearRecurrence[{47,-1},{2,47},20] (* Harvey P. Dale, Oct 23 2017 *)

a(n) = 47*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 47.
a(n) = ((47+sqrt(2205))/2)^n + ((47-sqrt(2205))/2)^n
(a(n))^2 = a(2n)+2.
G.f.: (2-47*x)/(1-47*x+x^2). - Alois P. Heinz, Aug 07 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(8*n+8)/F(8) - F(8*n-8)/F(8) = A049668(n+1) - A049668(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^8 = [13, 21; 21, 34].
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).
45*Sum_{n >= 1} 1/(a(n) - 49/a(n)) = 1: (49 = Lucas(8) + 2 and 45 = Lucas(8) - 2)
49*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 45/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 47*x^2 + 2208*x^3 + ... is the o.g.f. for A049668. (End)
E.g.f.: 2*exp(47*x/2)*cosh(21*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^4 - 4*Lucas(2*n)^2 + 2 = 2*T(4, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind; more generally, for k >= 0, Lucas(2*k*n) = 2*T(k, Lucas(2*n)/2).
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (47 - sqrt(2205))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(2205) - 47)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

Terms a(22)-a(27) from John W. Layman, Jun 14 2004

A056914 a(n) = Lucas(4*n+1).

Original entry on oeis.org

1, 11, 76, 521, 3571, 24476, 167761, 1149851, 7881196, 54018521, 370248451, 2537720636, 17393796001, 119218851371, 817138163596, 5600748293801, 38388099893011, 263115950957276, 1803423556807921, 12360848946698171

Offset: 0

Barry E. Williams, Jul 11 2000

a(n) = (t(i+4n+1) - t(i))/(t(i+2n+1) - t(i+2n)), where (t) is any sequence of the form t(n+2) = 4t(n+1) - 4t(n) + t(n-1) or of the form t(n+1) = 3t(n) - t(n-1) without regard to initial values as long as t(i+2n+1) - t(i+2n) != 0. - Klaus Purath, Jun 24 2024

V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers, A Publication of the Fibonacci Association, Houghton Mifflin Co., 1969, pp. 27-29.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. (A056914) = sqrt{5*(A033889)^2 - 4}.

Cf. quadrisection of A000032: A056854 (first), this sequence (second), A246453 (third, without 11), A288913 (fourth).

List([0..30], n-> Lucas(1,-1,4*n+1)[2] ); # G. C. Greubel, Jan 16 2020

[Lucas(4*n+1): n in [0..30]]; // G. C. Greubel, Dec 24 2017

with(combinat); seq(fibonacci(4*n+2)+fibonacci(4*n), n = 0..30); # G. C. Greubel, Jan 16 2020

LucasL[4*Range[0,30]+1] (* or *) LinearRecurrence[{7,-1}, {1,11}, 30] (* G. C. Greubel, Dec 24 2017 *)

my(x='x+O('x^30)); Vec((1+4*x)/(1-7*x+x^2)) \\ G. C. Greubel, Dec 24 2017

[lucas_number2(4*n+1,1,-1) for n in (0..30)] # G. C. Greubel, Jan 16 2020

a(n) = 7*a(n-1) - a(n-2), with a(0)=1, a(1)=11.
a(n) = (11*(((7+3*sqrt(5))/2)^n - ((7-3*sqrt(5))/2)^n) - (((7+3*sqrt(5))/2)^(n-1) - ((7-3*sqrt(5))/2)^(n-1)))/3*sqrt(5).
G.f.: (1+4*x)/(1-7*x+x^2). - Philippe Deléham, Nov 02 2008

A065705 a(n) = Lucas(10*n).

Original entry on oeis.org

2, 123, 15127, 1860498, 228826127, 28143753123, 3461452808002, 425730551631123, 52361396397820127, 6440026026380244498, 792070839848372253127, 97418273275323406890123, 11981655542024930675232002, 1473646213395791149646646123, 181246502592140286475862241127

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003

Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 2) + 1/(1 + ...))))). (End)

			a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..477
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 linear recurrences with constant coefficients, signature (123,-1).

Cf. A000032: a(n) = A000032(10*n).

Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(10*n): n in [0..90]]; // Vincenzo Librandi, Apr 14 2011

LucasL[10*Range[0, 20]] (* Paolo Xausa, Mar 04 2024 *)

a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
a(n)^2 = a(2*n) + 2.
G.f.: (2 - 123*x)/(1 - 123*x + x^2). -  Philippe Deléham, Nov 18 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
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).
121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x)). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^5 - 5*Lucas(2*n)^3 + 5*Lucas(2*n) = 2*T(5, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (123 - sqrt(15125))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(15125) - 123)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A089775 Lucas numbers L(12n).

Original entry on oeis.org

2, 322, 103682, 33385282, 10749957122, 3461452808002, 1114577054219522, 358890350005878082, 115561578124838522882, 37210469265847998489922, 11981655542024930675232002, 3858055874062761829426214722, 1242282009792667284144565908482, 400010949097364802732720796316482

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 09 2004

a(n+1)/a(n) converges to (322 + sqrt(103680))/2 = 321.996894379... a(0)/a(1) = 2/322; a(1)/a(2) = 322/103682; a(2)/a(3) = 103682/33385282; a(3)/a(4) = 33385282/10749957122; etc. Lim_{n -> inf} a(n)/a(n+1) = 0.00310562... = 2/(322 + sqrt(103680)) = (322 - sqrt(103680))/2.

			a(4) = 10749957122 = 322*a(3) - a(2) = 322*33385282 - 103682 = ((322 + sqrt(103680))/2)^4 + ((322 - sqrt(103680))/2)^4.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.

Nathaniel Johnston, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (322, -1).

Cf. A000032, A060964.
a(n) = A000032(12n).
Row 9 * 2 of array A188644
Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11).

[ Lucas(12*n) : n in [0..70]]; // Vincenzo Librandi, Apr 15 2011

Table[LucasL[12n], {n, 0, 13}] (* Indranil Ghosh, Mar 15 2017 *)

Vec((2 - 322*x)/(1 - 322*x + x^2) + O(x^14)) \\ Indranil Ghosh, Mar 15 2017

a(n) = 322*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 322
a(n) = ((322 + sqrt(103680))/2)^n + ((322 - sqrt(103680))/2)^n.
(a(n))^2 = a(2n) + 2.
G.f.: (2-322*x)/(1-322*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^6 - 6*Lucas(2*n)^4 + 9*Lucas(2*n)^2 - 2 = 2*T(6, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
320*Sum_{n >= 1} 1/(a(n) - 324/a(n)) = 1: (324 = Lucas(12) + 2 and 320 = Lucas(12) - 2)
324*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 320/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (322 - sqrt(103680))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(103680) - 322)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. (End)

a(11) - a(13) from Vincenzo Librandi, Apr 15 2011

cp's OEIS Frontend

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