A075796 -id:A075796 - cp's OEIS frontend

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799

Offset: 0

N. J. A. Sloane

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
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
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
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).
Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone and Diego 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.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
Nathan D. Cahill, John R. D'Errico and John P. Spence, Complex Factorizations of the Fibonacci and Lucas Numbers, Fibonacci Quarterly, 1(41):13-19, 2003.
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Paul Curtz, A269500, SeqFan list, March 2, 2016.
Murray Elder and Arkadius Kalka, Logspace computations for rigid Garside groups, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
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.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Tian-Xiao He and Louis W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Seong Ju Kim, Ryan Stees and Laura Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
Tanya Khovanova, Recursive Sequences.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
Serge Perrine, Some properties of the equation x^2=5y^2-4, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
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.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
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 sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019

a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

A002878 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]);
    else
        3*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)

a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011

for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015

a002878 = [1, 4]
for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
print(a002878) # Gennady Eremin, Feb 05 2022

[lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 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 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A001077 Numerators of continued fraction convergents to sqrt(5).

Original entry on oeis.org

1, 2, 9, 38, 161, 682, 2889, 12238, 51841, 219602, 930249, 3940598, 16692641, 70711162, 299537289, 1268860318, 5374978561, 22768774562, 96450076809, 408569081798, 1730726404001, 7331474697802, 31056625195209

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A001076(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = -1.
a(2*n) with b(2*n) := A001076(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,9) = A023039(n), n >= 0 and a(2*n+1) = 2*S(2*n, 2*sqrt(5)) = A075796(n+1), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
From Greg Dresden, May 21 2023: (Start)
For n >= 2, 8*a(n) is the number of ways to tile this T-shaped figure of length n-1 with four colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 8*a(6) = 23112 different tilings.
    _
   || _  
   |||_|||
   |_|
(End)

			1  2  9  38  161  (A001077)
-, -, -, --, ---, ...
0  1  4  17   72  (A001076)
1 + 2*x + 9*x^2 + 38*x^3 + 161*x^4 + 682*x^5 + 2889*x^6 + 12238*x^7 + ... - _Michael Somos_, Aug 11 2009

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).
V. Thébault, Les Récréations Mathématiques, Gauthier-Villars, Paris, 1952, p. 282.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Ex. 1, pp. 237-238.
Tanya Khovanova, Recursive Sequences
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
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A001076, A023039, A049629, A052924, A078343, A164581, A179237, A180148, A329723, A374439.

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001077:=(-1+2*z)/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
with(combinat): a:=n->fibonacci(n+1, 4)-2*fibonacci(n, 4): seq(a(n), n=0..30); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{4, 1}, {1, 2}, 30]
Join[{1},Numerator[Convergents[Sqrt[5],30]]] (* Harvey P. Dale, Mar 23 2016 *)
CoefficientList[Series[(1-2*x)/(1-4*x-x^2), {x, 0, 30}], x] (* G. C. Greubel, Dec 19 2017 *)
LucasL[3*Range[0,30]]/2 (* Rigoberto Florez, Apr 03 2019 *)
a[ n_] := LucasL[n, 4]/2; (* Michael Somos, Nov 02 2021 *)

{a(n) = fibonacci(3*n) / 2 + fibonacci(3*n - 1)}; /* Michael Somos, Aug 11 2009 */

a(n)=if(n<2,n+1,my(t=4);for(i=1,n-2,t=4+1/t);numerator(2+1/t)) \\ Charles R Greathouse IV, Dec 05 2011

x='x+O('x^30); Vec((1-2*x)/(1-4*x-x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, May 14 2009

G.f.: (1-2*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) + a(n-2), a(0)=1, a(1)=2.
a(n) = ((2 + sqrt(5))^n + (2 - sqrt(5))^n)/2.
a(n) = A014448(n)/2.
Limit_{n->infinity} a(n)/a(n-1) = phi^3 = 2 + sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = ((-i)^n)*T(n, 2*i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
Binomial transform of A084057. - Paul Barry, May 10 2003
E.g.f.: exp(2x)cosh(sqrt(5)x). - Paul Barry, May 10 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)*5^k*2^(n-2k). - Paul Barry, Nov 15 2003
a(n) = 4*a(n-1) + a(n-2) when n > 2; a(1) = 1, a(2) = 2. - Alex Vinokur (alexvn(AT)barak-online.net), Oct 25 2004
a(n) = A001076(n+1) - 2*A001076(n) = A097924(n) - A015448(n+1); a(n+1) = A097924(n) + 2*A001076(n) = A097924(n) + 2(A048876(n) - A048875(n)). - Creighton Dement, Mar 19 2005
a(n) = F(3*n)/2 + F(3*n-1) where F() = Fibonacci numbers A000045. - Gerald McGarvey, Apr 28 2007
a(n) = A000032(3*n)/2.
For n >= 1: a(n) = (1/2)*Fibonacci(6*n)/Fibonacci(3*n) and a(n) = integer part of (2 + sqrt(5))^n. - Artur Jasinski, Nov 28 2011
a(n) = Sum_{k=0..n} A201730(n,k)*4^k. - Philippe Deléham, Dec 06 2011
a(n) = A001076(n) + A015448(n). - R. J. Mathar, Jul 06 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) is the (1,1)-entry of the matrix W^n with W=[2, sqrt(5); sqrt(5), 2]. - Carmine Suriano, Mar 21 2014
From Rigoberto Florez, Apr 03 2019: (Start)
a(n) = A099919(n) + A049651(n) if n > 0.
a(n) = 1 + Sum_{k=0..n-1} L(3*k + 1) if n >= 0, L(n) = n-th Lucas number (A000032). (End)
From Christopher Hohl, Aug 22 2021: (Start)
For n >= 2, a(2n-1) = A079962(6n-9) + A079962(6n-3).
For n >= 1, a(2n) = sqrt(20*A079962(6n-3)^2 + 1). (End)
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*4^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*4^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 4^n*Sum_{k=0..n} A374439(n, k)*(-1/4)^k. - Peter Luschny, Jul 26 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) + 5*(-1)^(n+1)/a(n)) = 3/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = b(n) - b(n+1), where b(n) = (1/4) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5*(-1)^(n+1)/a(n)) = 1/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = c(n) + c(n+1), where c(n) = (1/4) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A007805 a(n) = Fibonacci(6*n + 3)/2.

Original entry on oeis.org

1, 17, 305, 5473, 98209, 1762289, 31622993, 567451585, 10182505537, 182717648081, 3278735159921, 58834515230497, 1055742538989025, 18944531186571953, 339945818819306129, 6100080207560938369

Offset: 0

James A. Raymond, Clark Kimberling

Hypotenuse (z) of Pythagorean triples (x,y,z) with |2x-y|=1.
x(n) := 2*A049629(n) and y(n) := a(n), n >= 0, give all positive solutions of the Pell equation x^2 - 5*y^2 = -1. See the Gregory V. Richardson formula, where his x is the y here and A075796(n+1) = x(n). - Wolfdieter Lang, Jun 20 2013
Positive numbers n such that 5*n^2 - 1 is a square (A075796(n+1)^2). - Gregory V. Richardson, Oct 13 2002

G. C. Greubel, Table of n, a(n) for n = 0..795 (terms 0..100 from T. D. Noe)
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
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 (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045.
Row 18 of array A094954.
Row 2 of array A188647.
Cf. similar sequences listed in A238379.

a007805 = (`div` 2) . a000045 . (* 3) . (+ 1) . (* 2)
-- Reinhard Zumkeller, Mar 26 2013

I:=[1, 17]; [n le 2 select I[n] else 18*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

seq(combinat:-fibonacci(6*n+3)/2, n=0..30); # Robert Israel, Sep 10 2014

LinearRecurrence[{18, -1}, {1, 17}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
Table[Fibonacci[6n+3]/2, {n, 0, 20}] (* Harvey P. Dale, Dec 17 2011 *)
CoefficientList[Series[(1-x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

a(n)=fibonacci(6*n+3)/2 \\ Edward Jiang, Sep 09 2014

x='x+O('x^30); Vec((1-x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

G.f.: (1-x)/(1-18*x+x^2).
a(n) = 18*a(n-1) - a(n-2), n > 1, a(0)=1, a(1)=17.
a(n) = A134495(n)/2 = A001076(2n+1).
a(n+1) = 9*a(n) + 4*sqrt(5*a(n)^2-1). - Richard Choulet, Aug 30 2007, Dec 28 2007
a(n) = ((2+sqrt(5))^(2*n+1) - (2-sqrt(5))^(2*n+1))/(2*sqrt(5)). - Dean Hickerson, Dec 09 2002
a(n) ~ (1/10)*sqrt(5)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = 8*phi + 5 = 9 + 4*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) = q(n, 16). - Benoit Cloitre, Dec 06 2002
a(n) = 19*a(n-1)- 19*a(n-2) + a(n-3); f(x) = (sqrt(5)/10)*((2+sqrt(5))*(9+4*sqrt(5))^(x-1) - (2-sqrt(5))*(9-4*sqrt(5))^(x-1)). - Antonio Alberto Olivares, May 15 2008
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3). - Antonio Alberto Olivares, Jun 19 2008
a(n) = b(n+1) - b(n), n >= 0, with b(n) := F(6*n)/F(6) = A049660(n). First differences. See the o.g.f.s. - Wolfdieter Lang, 2012
a(n) = S(n,18) - S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/4^2. Compare with A001519 and A097843. - Peter Bala, Nov 29 2013
a(n) = 9*a(n-1) + 8*A049629(n-1), n >= 1, a(0) = 1. This is just the rewritten Chebyshev S(n, 18) recurrence. - Wolfdieter Lang, Aug 26 2014
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Fibonacci(6*n + 6 - 2*k) - Fibonacci(6*n + 2*k) )/( Fibonacci(6 - 2*k) - Fibonacci(2*k) ), for k an arbitrary integer.
a(n) = ( Fibonacci(6*n + 6 - 2*k - 1) + Fibonacci(6*n + 2*k + 1) )/( Fibonacci(6 - 2*k - 1) + Fibonacci(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n)) n>=1 = [1, 0, 17, 0, 305, 0, 5473, 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 = -20, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = sqrt(2 + (9-4*sqrt(5))^(1+2*n) + (9+4*sqrt(5))^(1+2*n))/(2*sqrt(5)). - Gerry Martens, Jun 04 2015

Better description and more terms from Michael Somos

A023039 a(n) = 18*a(n-1) - a(n-2).

Original entry on oeis.org

1, 9, 161, 2889, 51841, 930249, 16692641, 299537289, 5374978561, 96450076809, 1730726404001, 31056625195209, 557288527109761, 10000136862780489, 179445175002939041, 3220013013190122249, 57780789062419261441

Offset: 0

David W. Wilson

The primitive Heronian triangle 3*a(n) +- 2, 4*a(n) has the latter side cut into 2*a(n) +- 3 by the corresponding altitude and has area 10*a(n)*A060645(n). - Lekraj Beedassy, Jun 25 2002
Chebyshev polynomials T(n,x) evaluated at x=9.
{a(n)} gives all (unsigned, integer) solutions of Pell equation a(n)^2 - 80*b(n)^2 = +1 with b(n) = A049660(n), n >= 0.
{a(n)} gives all possible solutions for x in Pell equation x^2 - D*y^2 = 1 for D=5, D=20 and D=80. The corresponding values for y are A060645 (D=5), A207832 (D=20) and A049660 (D=80). - Herbert Kociemba, Jun 05 2022
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=sqrt(5). - Benoit Cloitre, Feb 24 2004
For all terms x of the sequence, 5*x^2 - 5 is a square, A004292(n)^2.
The a(n) are the x-values in the nonnegative integer solutions of x^2 - 5y^2 = 1, see A060645(n) for the corresponding y-values. - Sture Sjöstedt, Nov 29 2011
Rightmost digits alternate repeatedly: 1 and 9 in fact, a(2) = 18*9 - 1 == 1 (mod 10); a(3) = 18*1 - 9 == 9 (mod 10) therefore a(2n) == 1 (mod 10), a(2n+1) == 9 (mod 10). - Carmine Suriano, Oct 03 2013

			G.f. = 1 + 9*x + 161*x^2 + 2889*x^3 + 51841*x4 + 930249*x^5 + 16692641*x^6 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..750 (terms 0..200 from Vincenzo Librandi)
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A001077, A115032.
Row 2 of array A188645.
Row 4 of A322790.

I:=[1, 9]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 13 2012

a := n -> hypergeom([n, -n], [1/2], -4):
seq(simplify(a(n)), n=0..16); # Peter Luschny, Jul 26 2020

LinearRecurrence[{18, -1}, {1, 9}, 50] (* Sture Sjöstedt, Nov 29 2011 *)
CoefficientList[Series[(1-9*x)/(1-18*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

{a(n) = fibonacci(6*n) / 2 + fibonacci(6*n - 1)}; /* Michael Somos, Aug 11 2009 */

x='x+O('x^30); Vec((1-9*x)/(1-18*x+x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) ~ (1/2)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->infinity} a(n)/a(n-1) = phi^6 = 9 + 4*sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 9) = (S(n, 18) - S(n-2, 18))/2, with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 18)=A049660(n+1).
a(n) = sqrt(80*A049660(n)^2 + 1) (cf. Richardson comment).
a(n) = ((9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/2.
G.f.: (1 - 9*x)/(1 - 18*x + x^2).
a(n) = cosh(2*n*arcsinh(2)). - Herbert Kociemba, Apr 24 2008
a(n) = A001077(2*n). - Michael Somos, Aug 11 2009
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = 2*A167808(6*n+1) - A167808(6*n+3).
Limit_{k->infinity} a(n+k)/a(k) = a(n) + A060645(n)*sqrt(5).
Limit_{n->infinity} a(n)/A060645(n) = sqrt(5).
(End)
a(n) = (1/2)*A087215(n) = (1/2)*(sqrt(5) + 2)^(2*n) + (1/2)*(sqrt(5) - 2)^(2*n).
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1/8. Compare with A005248, A002878 and A075796. - Peter Bala, Nov 29 2013
a(n) = 2*A115032(n-1) - 1 = S(n, 18) - 9*S(n-1, 18), with A115032(-1) = 1, and see the above formula with S(n, 18) using its recurrence. - Wolfdieter Lang, Aug 22 2014
a(n) = A128052(3n). - A.H.M. Smeets, Oct 02 2017
a(n) = A049660(n+1) - 9*A049660(n). - R. J. Mathar, May 24 2018
a(n) = hypergeom([n, -n], [1/2], -4). - Peter Luschny, Jul 26 2020
a(n) = L(6*n)/2 for L(n) the Lucas sequence A000032(n). - Greg Dresden, Dec 07 2021
a(n) = cosh(6*n*arccsch(2)). - Peter Luschny, May 25 2022

Chebyshev and Pell comments from Wolfdieter Lang, Nov 08 2002

Sture Sjöstedt's comment corrected and reformulated by Wolfdieter Lang, Aug 24 2014

A103134 a(n) = Fibonacci(6n+4).

Original entry on oeis.org

3, 55, 987, 17711, 317811, 5702887, 102334155, 1836311903, 32951280099, 591286729879, 10610209857723, 190392490709135, 3416454622906707, 61305790721611591, 1100087778366101931, 19740274219868223167, 354224848179261915075, 6356306993006846248183

Offset: 0

Creighton Dement, Jan 24 2005

Gives those numbers which are Fibonacci numbers in A103135.
Generally, for any sequence where a(0)= Fibonacci(p), a(1) = F(p+q) and Lucas(q)*a(1) +- a(0) = F(p+2q), then a(n) = L(q)*a(n-1) +- a(n-2) generates the following Fibonacci sequence: a(n) = F(q(n)+p). So for this sequence, a(n) = 18*a(n-1) - a(n-2) = F(6n+4): q=6, because 18 is the 6th Lucas number (L(0) = 2, L(1)=1); F(4)=3, F(10)=55 and F(16)=987 (F(0)=0 and F(1)=1). See Lucas sequence A000032. This is a special case where a(0) and a(1) are increasing Fibonacci numbers and Lucas(m)*a(1) +- a(0) is another Fibonacci. - Bob Selcoe, Jul 08 2013
a(n) = x + y where x and y are solutions to x^2 = 5*y^2 - 1. (See related sequences with formula below.) - Richard R. Forberg, Sep 05 2013

Colin Barker, Table of n, a(n) for n = 0..750
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Subsequence of A033887.
Cf. A000032, A000045, A001906, A001519, A015448, A014445, A033888, A033889, A033890, A033891, A049310, A049660, A102312, A099100, A134490, A134491, A134492, A134493, A134494, A134495, A103134, A134497, A134498, A134499, A134500, A134501, A134502, A134503, A134504.
Cf. A103135.

[Fibonacci(6*n +4): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Table[Fibonacci[6n+4], {n, 0, 30}]
LinearRecurrence[{18,-1},{3,55},20] (* Harvey P. Dale, Mar 29 2023 *)
Table[ChebyshevU[3*n+1, 3/2], {n, 0, 20}] (* Vaclav Kotesovec, May 27 2023 *)

a(n)=fibonacci(6*n+4) \\ Charles R Greathouse IV, Feb 05 2013

G.f.: (x+3)/(x^2-18*x+1).
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=3, a(1)=55. - Philippe Deléham, Nov 17 2008
a(n) = A007805(n) + A075796(n), as follows from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((15-7*sqrt(5)+(9+4*sqrt(5))^(2*n)*(15+7*sqrt(5))))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
a(n) = S(3*n+1, 3) = 3*S(n,18) + S(n-1,18), with the Chebyshev S polynomials (A049310), S(-1, x) = 0, and S(n, 18) = A049660(n+1). - Wolfdieter Lang, May 08 2023

Edited by N. J. A. Sloane, Aug 10 2010

A167808 Numerator of x(n), where x(n) = x(n-1) + x(n-2) with x(0)=0, x(1)=1/2.

Original entry on oeis.org

0, 1, 1, 1, 3, 5, 4, 13, 21, 17, 55, 89, 72, 233, 377, 305, 987, 1597, 1292, 4181, 6765, 5473, 17711, 28657, 23184, 75025, 121393, 98209, 317811, 514229, 416020, 1346269, 2178309, 1762289, 5702887, 9227465, 7465176, 24157817, 39088169, 31622993

Offset: 0

Reinhard Zumkeller, Nov 12 2009

Define a sequence c(n) by c(0)=0, c(1)=1; thereafter c(n) = (c(n-2)*c(n-1)-1)/(c(n-2)+c(n-1)+2). Then it appears that (apart from signs), a(n) is the denominator of c(n). - Jonas Holmvall, Jun 21 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,1).

Cf. A000045, A130196 (denominator).

The a(2*n) appear in A179135. - Johannes W. Meijer, Jul 01 2010

a:=[0,1,1,1,3,5];; for n in [7..40] do a[n]:=4*a[n-3]+a[n-6]; od; a; # Muniru A Asiru, Oct 16 2018

nmax:=39; x(0):=0: x(1):=1/2:for n from 2 to nmax do x(n):=x(n-1)+x(n-2) od: for n from 0 to nmax do a(n):= numer(x(n)) od: seq(a(n),n=0..nmax); # Johannes W. Meijer, Jul 01 2010
with(combinat):f:=n->fibonacci(n):L:=n->f(n)+2*f(n-1):seq(numer(f(n)/L(n)), n=0..39); # Gary Detlefs, Dec 11 2010

f[n_]:=Numerator[Fibonacci[n]/Fibonacci[n+3]];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011*)
Numerator[LinearRecurrence[{1,1},{0,1/2},40]] (* Harvey P. Dale, Aug 08 2014 *)
CoefficientList[Series[-x (1 + x + x^2 - x^3 + x^4)/((x^2 + x - 1) (x^4 - x^3 + 2 x^2 + x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 08 2014 *)
LinearRecurrence[{0, 0, 4, 0, 0, 1},{0, 1, 1, 1, 3, 5},40] (* Ray Chandler, Aug 03 2015 *)
a[n_]:=If[Mod[n,3]==0, Fibonacci[n]/2, Fibonacci[n]]; Array[a, 40, 0] (* Stefano Spezia, Oct 16 2018 *)

a(n) = (a(n-1)*A131534(n) + a(n-2)*A131534(n+2))/A131534(n+1) for n > 1.
a(3*n) = A001076(n) = (a(3*n-1) + a(3*n-2))/2;
a(3*n+1) = A033887(n) = 2*a(3*n-1) + a(3*n-2);
a(3*n+2) = A015448(n+1) = a(3*n-1) + 2*a(3*n-2).
From Johannes W. Meijer, Jul 01 2010: (Start)
a(2*n) = A001906(n)/A131534(n+1) for n >= 0 and a(2*n+1) = A179131(n)/5 for n >= 1.
a(6*n+2) - 2*a(6*n) = A134493(n);
2*a(6*n+1) - a(6*n+3) = A023039(n);
2*a(6*n+4) - a(6*n+2) = A134497(n);
a(6*n+5) - 2*a(6*n+3) = A103134(n);
2*a(6*n+4) - a(6*n+6) = A075796(n).
(End)
From Gary Detlefs, Dec 11 2010: (Start)
a(n) = numerator(A000045(n)/A000032(n)).
If n mod 3 = 0 then a(n) = Fibonacci(n)/2, else a(n)= Fibonacci(n). (End)
G.f.: -x*(1 + x + x^2 - x^3 + x^4) / ( (x^2 + x - 1)*(x^4 - x^3 + 2*x^2 + x + 1) ). - R. J. Mathar, Mar 08 2011
a(n) = 4*a(n-3) + a(n-6). - Muniru A Asiru, Oct 16 2018

Typo in title corrected by Johannes W. Meijer, Jun 26 2010

A221762 Numbers m such that 11*m^2 + 5 is a square.

Original entry on oeis.org

1, 2, 22, 41, 439, 818, 8758, 16319, 174721, 325562, 3485662, 6494921, 69538519, 129572858, 1387284718, 2584962239, 27676155841, 51569671922, 552135832102, 1028808476201, 11015040486199, 20524599852098, 219748673891878, 409463188565759

Offset: 1

Bruno Berselli, Jan 24 2013

Corresponding squares are: 16, 49, 5329, 18496, 2119936, 7360369, 843728209, 2929407376, ... (subsequence of A016778).
The Diophantine equation 11*x^2+k = y^2, for |k|<11, has integer solutions with the following k values:
k = -10, the nonnegative x values are in A198947;
k =  -8,            "                    2*A075839;
k =  -7,            "                    A221763;
k =  -2,            "                    A075839;
k =   1,            "                    A001084;
k =   4,            "                    A075844;
k =   5,            "                    this sequence;
k =   9,            "                    3*A001084.
Also, the Diophantine equation h*x^2+5 = y^2 has infinitely many integer solutions for h = 5, 11, 19, 20, 29, 31, 41, 44, 55, 59, ...
a(n+1)/a(n) tends alternately to (1+sqrt(11))^2/10 and (4+sqrt(11))^2/5.
a(n+2)/a(n) tends to A176395^2/2.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

Cf. A001084, A075839, A075844, A198947, A198949, A221763.

Cf. A049629 (numbers m such that 20*m^2 + 5 is a square), A075796 (numbers m such that 5*m^2 + 5 is a square).

m:=24; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+2*x^2+x^3)/(1-20*x^2+x^4)));

I:=[1,2,22,41]; [n le 4 select I[n] else 20*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 18 2013

A221762:=proc(q)
local n;
for n from 1 to q do if type(sqrt(11*n^2+5), integer) then print(n);
fi; od; end:
A221762(1000); # Paolo P. Lava, Feb 19 2013

LinearRecurrence[{0, 20, 0, -1}, {1, 2, 22, 41}, 24]
CoefficientList[Series[(1 + 2 x + 2 x^2 + x^3)/(1 - 20 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(((-11*(-1)^n+4*sqrt(11))*(10+3*sqrt(11))^floor(n/2)-(11*(-1)^n+4*sqrt(11))*(10-3*sqrt(11))^floor(n/2))/22), n, 1, 24);

G.f.: x*(1+2*x+2*x^2+x^3)/(1-20*x^2+x^4).
a(n) = -a(1-n) = ((-11*(-1)^n+4*t)*(10+3*t)^floor(n/2)-(11*(-1)^n+4*t)*(10-3*t)^floor(n/2))/22, where t=sqrt(11).
a(n) = 20*a(n-2) - a(n-4) for n>4, a(1)=1, a(2)=2, a(3)=22, a(4)=41.
a(n)*a(n-3)-a(n-1)*a(n-2) = -(3/2)*(9-7*(-1)^n).
a(n+1) + a(n-1) =  A198949(n), with a(0)=-1.
2*a(n-1) - a(n) =  A001084(n/2-1) for even n.

A382209 Numbers k such that 10+k and 10*k are perfect squares.

Original entry on oeis.org

90, 136890, 197402490, 284654260890, 410471246808090, 591899253243012090, 853518312705176632890, 1230772815021611461622490, 1774773545742851022483004890, 2559222222188376152809031436090, 3690396669622092669499600847844090, 5321549438372835441042271613559748890

Offset: 1

Emilio Martín, Mar 18 2025

The limit of a(n+1)/a(n) is 1441.99930651839... = 721+228*sqrt(10) = (19+6*sqrt(10))^2.

If 10*A158490(n) is a perfect square, then A158490(n) is a term.

			90 is a term because 10+90=100 is a square and 10*90=900 is a square.
(3,1) is a solution to x^2 - 10*y^2 = -1, from which a(n) = 100*y^2-10 = 10*x^2 = 90.

Emilio Martín, Table of n, a(n) for n = 1..100
Wikipedia, Negative Pell equation (in German)
Wikipedia, Pell's equation
Index entries for linear recurrences with constant coefficients, signature (1443,-1443,1).

Subsequence of A158490.
Cf. A097315, A005667, A173127, A081071.
Cf. A383734 = 2*A008843 (2+k and 2*k are squares).
Cf. 5*A075796^2 (5+k and 5*k are squares).
Cf. 5*A081071 (20+k and 20*k are squares).
Cf. A245226 (m such that k+m and k*m are squares).

CoefficientList[Series[ 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)),{x,0,11}],x] (* or *) LinearRecurrence[{1443,-1443,1},{90,136890,197402490},12] (* James C. McMahon, May 08 2025 *)

from itertools import islice
def A382209_gen(): # generator of terms
    x, y = 30, 10
    while True:
        yield x**2//10
        x, y = x*19+y*60, x*6+y*19
A382209_list = list(islice(A382209_gen(),30)) # Chai Wah Wu, Apr 24 2025

a(n) = 10 * ((1/2) * (3+sqrt(10))^(2*n-1) + (1/2) * (3-sqrt(10))^(2*n-1))^2.
a(n) = 10 * (sinh((2n-1) * arcsinh(3)))^2.
a(n) = 10 * A173127(n)^2 = 100 * A097315(n)^2 - 10 (negative Pell's equation solutions).
a(n+2) = 1442 * a(n+1) - a(n) + 7200.
G.f.: 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)). - Stefano Spezia, Apr 24 2025

A107075 Centered square numbers that are also centered pentagonal numbers.

Original entry on oeis.org

1, 181, 58141, 18721081, 6028129801, 1941039074701, 625008553923781, 201250813324382641, 64802136881897286481, 20866086825157601864101, 6718815155563865902953901, 2163437614004739663149291881

Offset: 1

Richard Choulet, Aug 30 2007, Sep 20 2007

The centered square numbers are n^2 + (n+1)^2 while the centered pentagonal numbers are (5*r^2 + 5*r + 2)/2. A number has both properties iff 5*(2*r+1)^2 = (4*n+2)^2 + 1. We solve the equation 5*Y^2 - 1 = X^2 whose solutions in positive integers are given by A075796 and A007805 respectively. The r values are 0,8,..., i.e., A053606. The n values define A119032.

Harvey P. Dale, Table of n, a(n) for n = 1..399
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A007805, A053606, A075796, A119032.

a:= n-> (Matrix([181,1,1]). Matrix([[323,1,0], [ -323,0,1], [1,0,0]])^n)[1,3]: seq(a(n), n=1..20); # Alois P. Heinz, Aug 14 2008

LinearRecurrence[{323,-323,1},{1,181,58141},20] (* Harvey P. Dale, Nov 15 2018 *)

G.f.: (z*(1-142*z+z^2))/((1-z)*(1-322*z+z^2)).
a(n+2) = 322*a(n+1)-a(n)-140 with a(1)=1 and a(2)=181.
a(n+1) = 161*a(n)-70+18*(80*a(n)^2-70*a(n)+15)^0.5.
a(n) = (14+(9-4*sqrt(5))^(2*n-1)+(9+4*sqrt(5))^(2*n-1))/32. - Gerry Martens, Jun 06 2015

More terms from Alois P. Heinz, Aug 14 2008

A306380 Squares of the form 5*k^2 + 5.

Original entry on oeis.org

25, 7225, 2325625, 748843225, 241125192025, 77641562988025, 25000342156951225, 8050032532975305625, 2592085475275891459225, 834643473006304074564025, 268752606222554636118156025, 86537504560189586525971675225

Offset: 1

Stefano Spezia, Feb 13 2019

All terms of this sequence are odd, hence they end with 5.

Stefano Spezia, Table of n, a(n) for n = 1..390
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Cf. A000290, A075796 (associated k).

a:=[25,7225,2325625];; for n in [4..20] do a[n]:=323*a[n-1]-323*a[n-2]+a[n-3]; od; a;

I:=[25, 7225, 2325625]; [n le 3 select I[n] else 323*Self(n-1)-323*Self(n-2)+Self(n-3): n in [1..20]];

a := n ->(5/4)*(2+(9-4*sqrt(5))^(2*n-2)*(9+sqrt(5))+(9+4*sqrt(5))^(2*n-2)*(9-sqrt(5))): op(map(simplify, [seq(a(n), n = 1 .. 20)]))

LinearRecurrence[{323, -323, 1}, {25, 7225, 2325625}, 30]

a[1]:25$ a[2]:7225$ a[3]:2325625$ a[n]:=323*a[n-1]-323*a[n-2]+a[n-3]$ create_list(a[n], n, 1, 20);

Vec(25*x*(1-34*x+x^2)/((1-x)*(1-322*x+x^2)) + O(x^20))

O.g.f.: 25*x*(1 - 34*x + x^2)/((1 - x)*(1 - 322*x + x^2)).
E.g.f.: (5/4)*x*(2*exp(x) + (9 - 4*sqrt(5))*exp((9 - 4*sqrt(5))^2*x) + (9 + 4*sqrt(5))*exp((9 + 4*sqrt(5))^2*x)).
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 3.
a(n) = (5/4)*(2 + (9 - 4*sqrt(5))^(2*n)*(9 + 4*sqrt(5)) + (9 - 4*sqrt(5))*(9 + 4*sqrt(5))^(2*n)).
a(n) = 5*A000290(A075796(n)) + 5.

cp's OEIS Frontend

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

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A001077 Numerators of continued fraction convergents to sqrt(5).

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A007805 a(n) = Fibonacci(6*n + 3)/2.

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A023039 a(n) = 18*a(n-1) - a(n-2).

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A103134 a(n) = Fibonacci(6n+4).

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