A033890 -id:A033890 - cp's OEIS frontend

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 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 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, 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.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
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 (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

Original entry on oeis.org

1, 19, 341, 6119, 109801, 1970299, 35355581, 634430159, 11384387281, 204284540899, 3665737348901, 65778987739319, 1180356041958841, 21180629767519819, 380070979773397901, 6820097006153642399, 122381675130992165281, 2196050055351705332659, 39406519321199703822581

Offset: 0

Clark Kimberling

x(n) := 2*a(n) and y(n) := A007805(n), n >= 0, give all the positive solutions of the Pell equation x^2 - 5*y^2 = -1.
The Gregory V. Richardson formula follows from this. - Wolfdieter Lang, Jun 20 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
2*a(n) = 2 o 2 o ... o 2 (2*n+1 terms). For example, 2 o 2 = 4*sqrt(5) and 2 o 2 o 2 = 2 o 4*sqrt(5) = 38 = 2*a(1). Cf. A084068.
a(n) = U(2*n+1) where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = sqrt(20)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/4)*( (sqrt(5) + 2)^n - (sqrt(5) - 2)^n ). (End)

			Pell, n=1: (2*19)^2 - 5*17^2 = -1.

G. C. Greubel, Table of n, a(n) for n = 0..795
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.
Tanya Khovanova, Recursive Sequences.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Eric Weisstein's World of Mathematics, Lehmer Number.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Bisection of A001077 divided by 2.
Cf. A000045, A007805, A049310, A049660, A100047, A188378.
Cf. A005013, A033890, A049685, A084608.
Cf. similar sequences of the type (1/k)*sinh((2*n+1)*arcsinh(k)) listed in A097775.

[(Fibonacci(6*n+5) - Fibonacci(6*n+1))/4: n in [0..30]]; // G. C. Greubel, Dec 15 2017

with(numtheory): with(combinat):
seq((fibonacci(6*n+5)-fibonacci(6*n+1))/4,n=0..20); # Muniru A Asiru, Mar 25 2018

a[n_] := Simplify[(2 + Sqrt@5)^(2 n - 1) + (2 - Sqrt@5)^(2 n - 1)]/4; Array[a, 16] (* Robert G. Wilson v, Oct 28 2010 *)

my(x='x+O('x^30)); Vec((1+x)/(1 - 18*x + x^2)) \\ G. C. Greubel, Dec 15 2017

a(n) ~ (1/4)*(sqrt(5) + 2)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all members x of the sequence, 20*x^2 + 5 is a square. Lim_{n -> oo} a(n)/a(n-1) = 9 + 2*sqrt(20) = 9 + 4*sqrt(5). The 20 can be seen to derive from the statement "20*x^2 + 5 is a square". - Gregory V. Richardson, Oct 12 2002
a(n) = (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)). - Gregory V. Richardson, Oct 12 2002
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1+x)/(1 - 18x + x^2).
a(n) = A049660(n) + A049660(n+1). (End)
a(n) = 18*a(n-1) - a(n-2) for n>1; a(0)=1, a(1)=19. - Philippe Deléham, Nov 17 2008
a(n) = S(n,18) + S(n-1,18) with the Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jun 20 2013
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, k != 1.
The aerated sequence (b(n))n>=1 = [1, 0, 19, 0, 341, 0, 6119, 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 = -16, 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) = (A188378(n)^3 + (A188378(n)-2)^3) / 8. - Altug Alkan, Jan 24 2016
a(n) = sqrt(5 * Fibonacci(3 + 6*n)^2 - 4)/4. - Gerry Martens, Jul 25 2016
a(n) = Lucas(6*n + 3)/4. - Ehren Metcalfe, Feb 18 2017
From Peter Bala, Mar 23 2018: (Start)
a(n) = 1/4*( (sqrt(5) + 2)^(2*n+1) - (sqrt(5) - 2)^(2*n+1) ).
a(n) = 9*a(n-1) + 2*sqrt(5 + 20*a(n-1)^2).
a(n) = (1/2)*sinh((2*n + 1)*arcsinh(2)). (End)
From Peter Bala, May 09 2025: (Start)
a(n)^2 - 18*a(n)*a(n+1) + a(n+1)^2 = 20.
More generally, for real x, a(n+x)^2 - 18*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 20 with a(n) := (((9 + 4*sqrt(5))^(n+1) - (9 - 4*sqrt(5))^(n+1)) + ((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n)) / (8*sqrt(5)) as given above.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/20 (telescoping series).
Product_{n >= 1} ((a(n) + 1)/(a(n) - 1)) = sqrt(5)/2 (telescoping product). (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)

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

A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

Original entry on oeis.org

1, 9, 71, 559, 4401, 34649, 272791, 2147679, 16908641, 133121449, 1048062951, 8251382159, 64962994321, 511452572409, 4026657584951, 31701808107199, 249587807272641, 1965000650073929, 15470417393318791, 121798338496476399

Offset: 0

Wolfdieter Lang, Aug 04 2000

a(n) = L(n,-8)*(-1)^n, where L is defined as in A108299; see also A070997 for L(n,+8). - Reinhard Zumkeller, Jun 01 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
The primes in this sequence are 71, 34649, 16908641, 8251382159, 31701808107199,... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 9, 0, 71, 0, 559, 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 = -6, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*a(n)^2 - 5*b(n)^2 = -2. The corresponding b(n) are A070997(n). Note that (a(n)*a(n+2) - a(n+1)^2)/2 = -5 and (b(n)*b(n+2) - b(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (a(n+1) - b(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also a(n)*b(n+1) - a(n+1)*b(n) = -2.
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)) as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0 where (t) is a sequence satisfying t(i+3) = 9*t(i+2) - 9*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i), regardless of the initial values and including this sequence itself. (End)

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.
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.
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
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=10.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A033890, A100047.

a:=[1,9];; for n in [3..30] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,9]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A057080 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,9]);
    else
        8*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1+x)/(1-8x+x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Vec((1+x)/(1-8*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,8,1)-lucas_number2(n-1,8,1))/6 for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all elements x of the sequence, 15*x^2 + 10 is a square. Lim. n-> Inf. a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
a(n) = 8*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 8) + S(n-1, 8) = S(2*n, sqrt(10)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 8) = A001090(n).
G.f.: (1+x)/(1-8*x+x^2).
a(n) = ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)). - Gregory V. Richardson, Oct 13 2002
a(n) = sqrt((5*A070997(n)^2 - 2)/3) (cf. Richardson comment).
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = (-1)^n*q(n,-10). - Benoit Cloitre, Nov 10 2002
a(n) = Jacobi_P(n,1/2,-1/2,4)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry, Feb 03 2006
a(n+1) = 4*a(n) + sqrt(5*(3*a(n)^2 + 2)). - Richard Choulet, Aug 30 2007
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbritrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 4), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 8*a(n)*a(n+1) + a(n+1)^2 = 10.
More generally, for arbitrary x, a(n+x)^2 - 8*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 10 with a(n) := ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)) as given above.
a(n+1/2) = sqrt(10) * A001090(n+1).
a(n+3/4) + a(n+1/4) = sqrt(10)*sqrt(sqrt(10) + 2) * A001090(n+1).
a(n+3/4) - a(n+1/4) = sqrt((sqrt(40) - 4)/3) * A001091(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/10 (telescoping series: for n >= 1, 10/(a(n) - 1/a(n)) = 1/A001090(n) + 1/A001090(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5/3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 5/3 * (1 - 2/(1 + A001091(k+1)))). (End)

A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

Original entry on oeis.org

1, 10, 89, 791, 7030, 62479, 555281, 4935050, 43860169, 389806471, 3464398070, 30789776159, 273643587361, 2432002510090, 21614379003449, 192097408520951, 1707262297685110, 15173263270645039, 134852107138120241, 1198495700972437130, 10651609201613813929

Offset: 0

Wolfdieter Lang, Aug 04 2000

This is the m=11 member of the m-family of sequences S(n,m-2)+S(n-1,m-2) = S(2*n,sqrt(m)) (for S(n,x) see Formula). The m=4..10 instances are A005408, A002878, A001834, A030221, A002315, A033890 and A057080, resp. The m=1..3 (signed) sequences are: A057078, A057077 and A057079, resp.
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim_{n->oo} 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
The primes in this sequence are 89, 389806471, 192097408520951, 7477414486269626733119, ... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 10, 0, 89, 0, 791, 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 = -7, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

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.
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.
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
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=11.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A005408, A002878, A001834, A030221, A002315, A033890, A057080.
Cf. A057078, A057077, A057079.
Cf. A018913, A049310.
Cf. A100047.

A057081 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,10]);
    else
        9*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1 + x)/(1 - 9*x + x^2), {x,0,50}], x] (* or *) LinearRecurrence[{9,-1}, {1,10}, 50] (* G. C. Greubel, Apr 12 2017 *)

Vec((1+x)/(1-9*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,9,1)-lucas_number2(n-1,9,1))/7 for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

a(n) = 9*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 9) + S(n-1, 9) = S(2*n, sqrt(11)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 9) = A018913(n).
G.f.: (1+x)/(1-9*x+x^2).
Let q(n, x) = Sum{i=0..n} x^(n-i)*binomial(2*n-i, i), a(n) = (-1)^n*q(n, -11). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-9)*(-1)^n, where L is defined as in A108299; see also A070998 for L(n,+9). - Reinhard Zumkeller, Jun 01 2005
From Peter Bala, Jun 08 2025: (Start)
a(n) = (1/sqrt(7)) * ( ((sqrt(11) + sqrt(7))/2)^(2*n+1) - ((sqrt(11) - sqrt(7))/2)^(2*n+1) ).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/11 (telescoping series: 11/(a(n) - 1/a(n)) = 1/A018913(n+1) + 1/A018913(n)).
Conjecture: 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).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(11/7) [telescoping product: ((a(n) + 1)/(a(n) - 1))^2 = (1 - 4/b(n+1))/(1 - 4/b(n)), where b(n) = 2 + A056918(n)]. (End)

A269254 To find a(n), define a sequence by s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n + 1; then a(n) is the smallest index k such that s(k) is prime, or -1 if no such k exists.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, -1, 2, 2, 1, 2, 1, 2, -1, 2, 1, 3, 1, 2, 2, 2, 1, -1, 2, 6, 2, 3, 1, 3, 1, 2, 9, 9, -1, 2, 1, 6, 2, 2, 1, 2, 1, 5, 2, 2, 1, -1, 2, 5, 2, 9, 1, 2, 2, 2, 2, 6, 1, 2, 1, 14, -1, 5, 2, 2, 1, 5, 2, 3, 1, 6, 1, 8, 3, 6, 2, 3, 1, -1, 3, 18, 1, 2, 3, 2, 2, 3, 1, 2, 9, 3, 5, 2, 2, 96, 1, 3, -1, 5, 1, 2, 1, 2, 15, 14, 1, 44, 1, 3, -1

Offset: 1

Arkadiusz Wesolowski, Jul 09 2016

sign

The s(k) sequences can be viewed in A294099, where they appear as rows. - Peter Munn, Aug 31 2020
For n >= 3, a(n) is that positive integer k yielding the smallest prime of the form (x^y - 1/x^y)/(x - 1/x), where x = (sqrt(n+2) +- sqrt(n-2))/2 and y = 2*k + 1, or -1 if no such k exists.
Every positive term belongs to A005097.
When n=7, the sequence {s(k)} is A033890, which is Fibonacci(4i+2), and since x|y <=> F_x|F_y, and 2i+1|4i+2, A033890 is never prime, and so a(7)=-1. For the other -1 terms below 100, see the theorem below and the Klee link - N. J. A. Sloane, Oct 20 2017 and Oct 22 2017
Theorem (Brad Klee): For all n > 2, a(n^2 - 2) = -1. See Klee link for a proof. - L. Edson Jeffery, Oct 22 2017
Theorem (Based on work of Hans Havermann, L. Edson Jeffery, Brad Klee, Don Reble, Bob Selcoe, and N. J. A. Sloane) a(110) = -1. [For proof see link. - N. J. A. Sloane, Oct 23 2017]
From Bob Selcoe, Oct 24 2017, edited by N. J. A. Sloane, Oct 27 2017: (Start)
Suppose n = m^2 - 2, where m >= 3, and let j = m-2, with j >= 1.
For this value of n, the sequence s(k) satisfies s(k) = (c(k) + d(k))*(c(k) - d(k)), where c(0) = 1, d(0) = 0; and for k >= 1: c(k) = (j+2)*c(k-1) - d(k-1), and d(k) = c(k-1). So (as Brad Klee already proved) a(n) = -1 .
We have s(0) = 1 and s(1) = n+1 = j^2 + 4j + 3. In general, the coefficients of s(k) when expanded in powers of j are given by the (4k+2)-th row of A011973 (the triangle of coefficients of Fibonacci polynomials) in reverse order. For example, s(2) = j^4 + 8j^3 + 21j^2 + 20j + 5, s(3) = j^6 + 12j^5 + 55j^4 + 120j^3 + 126j^2 + 56j + 7, etc.
Perhaps the above comments could be generalized to apply to a(110) or to other n for which a(n) = -1?
(End)
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

			Let b(k) be the recursive sequence defined by the initial conditions b(0) = 1, b(1) = 16, and the recursive equation b(k) = 15*b(k-1) - b(k-2). a(15) = 2 because b(2) = 239 is the smallest prime in b(k).
Let c(k) be the recursive sequence defined by the initial conditions c(0) = 1, c(1) = 18, and the recursive equation c(k) = 17*c(k-1) - c(k-2). a(17) = 3 because c(3) = 5167 is the smallest prime in c(k).

Hans Havermann, Table of n, a(n) for n = 1..946
Hans Havermann, Table of n, a(n) for n = 1..10000
C. K. Caldwell, Top Twenty page, Lehmer number
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.
Brad Klee, Proof for A269254, Sequence Fans Mailing List, October 2017.
N. J. A. Sloane et al., Proof that a(110) = -1
Wikipedia, Lehmer number.

Cf. A005097, A011973, A269251, A269252, A269253.
Cf. A294099 (array used to compute this sequence).
Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A298675, A298677, A298878, A299045, A299071.

lst:=[]; for n in [1..85] do if n gt 2 and IsSquare(n+2) then Append(~lst, -1); else a:=n+1; c:=1; t:=1; if IsPrime(a) then Append(~lst, t); else repeat b:=n*a-c; c:=a; a:=b; t+:=1; until IsPrime(a); Append(~lst, t); end if; end if; end for; lst;

kmax = 100;
a[1] = a[2] = 1;
a[n_ /; IntegerQ[Sqrt[n+2]]] = -1;
a[n_] := Module[{s}, s[0] = 1; s[1] = n+1; s[k_] := s[k] = n s[k-1] - s[k-2]; For[k=1, k <= kmax, k++, If[PrimeQ[s[k]], Return[k]]]; Print["For n = ", n, ", k = ", k, " exceeds the limit kmax = ", kmax]; -1];
Array[a, 110] (* Jean-François Alcover, Aug 05 2018 *)

allocatemem(2^30);
default(primelimit,(2^31)+(2^30));
s(n,k) = if(0==k,1,if(1==k,(1+n),((n*s(n,k-1)) - s(n,k-2))));
A269254(n) = { my(k=1); if((n>2)&&issquare(2+n),-1,while(!isprime(s(n,k)),k++);(k)); }; \\ Antti Karttunen, Oct 20 2017

If n is prime then a(n-1) = 1.

a(86)-a(94) from Antti Karttunen, Oct 20 2017

a(95)-a(109) appended by L. Edson Jeffery, Oct 22 2017

A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))n^(k-j), n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 5, -1, 1, 5, 11, 7, -2, 1, 6, 19, 29, 9, -1, 1, 7, 29, 71, 76, 11, 1, 1, 8, 41, 139, 265, 199, 13, 2, 1, 9, 55, 239, 666, 989, 521, 15, 1, 1, 10, 71, 377, 1393, 3191, 3691, 1364, 17, -1, 1, 11, 89, 559, 2584, 8119, 15289, 13775, 3571, 19, -2

Offset: 1

L. Edson Jeffery, Bob Selcoe and Andrew Hone, Oct 22 2017

This array is used to compute A269254: A269254(n) = least k such that A(n,k) is a prime, or -1 if no such k exists.
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

			Array begins:
  1   2    1    -1     -2      -1        1         2          1          -1
  1   3    5     7      9      11       13        15         17          19
  1   4   11    29     76     199      521      1364       3571        9349
  1   5   19    71    265     989     3691     13775      51409      191861
  1   6   29   139    666    3191    15289     73254     350981     1681651
  1   7   41   239   1393    8119    47321    275807    1607521     9369319
  1   8   55   377   2584   17711   121393    832040    5702887    39088169
  1   9   71   559   4401   34649   272791   2147679   16908641   133121449
  1  10   89   791   7030   62479   555281   4935050   43860169   389806471
  1  11  109  1079  10681  105731  1046629  10360559  102558961  1015229051

Robert Price, Table of n, a(n) for n = 1..5050
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.
Rows: A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, ...
Columns: A000012, A000027, A028387, ...

(* Array: *)
Grid[Table[LinearRecurrence[{n, -1}, {1, 1 + n}, 10], {n, 10}]]
(* Array antidiagonals flattened (gives this sequence): *)
A294099[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] n^(k - j), {j, 0, k}]; Flatten[Table[A294099[n - k, k], {n, 11}, {k, 0, n - 1}]]

{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*n^(k-j))}; /* Michael Somos, Jun 19 2023 */

A(n,0) = 1, A(n,1) = n + 1, A(n,k) = n*A(n,k-1) - A(n,k-2), n >= 1, k >= 2.
G.f. for row n: (1 + x)/(1 - n*x + x^2), n >= 1.
A(n, k) = B(-n, k) where B = A299045. - Michael Somos, Jun 19 2023

A269253 Smallest prime in the sequence s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n + 1 (or -1 if no such prime exists).

Original entry on oeis.org

2, 3, 11, 5, 29, 7, -1, 71, 89, 11, 131, 13, 181, -1, 239, 17, 5167, 19, 379, 419, 461, 23, -1, 599, 251894449, 701, 20357, 29, 25171, 31, 991, 36002209323169, 47468744103199, -1, 1259, 37, 2625505273, 1481, 1559, 41, 1721, 43, 150103799, 1979, 2069, 47, -1, 2351, 287762399

Offset: 1

Arkadiusz Wesolowski, Jul 09 2016

sign

For n >= 3, smallest prime of the form (x^y - 1/x^y)/(x - 1/x), where x = (sqrt(n+2) +/- sqrt(n-2))/2 and y is an odd positive integer, or -1 if no such prime exists.
When n=7, the sequence {s(k)} is A033890, which is Fibonacci(4i+2), and since x|y <=> F_x|F_y, and 2i+1|4i+2, A033890 is never prime, and so a(7) = -1. What is the proof for the other entries that are -1? Answer: See the Comments in A269254. - N. J. A. Sloane, Oct 22 2017
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018

Hans Havermann, Table of n, a(n) for n = 1..300
C. K. Caldwell, Top Twenty page, Lehmer number
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.
Wikipedia, Lehmer number

Cf. A117522, A269251, A269252, A269254.

Cf. A294099, A298675, A298677, A298878, A299045, A299071, A285992, A299107, A299109, A088165, A299100, A299101, A113501.

lst:=[]; for n in [1..49] do if n gt 2 and IsSquare(n+2) then Append(~lst, -1); else a:=n+1; c:=1; if IsPrime(a) then Append(~lst, a); else repeat b:=n*a-c; c:=a; a:=b; until IsPrime(a); Append(~lst, a); end if; end if; end for; lst;

terms = 172;
kmax = 120;
a[n_] := Module[{s, k}, s[k_] := s[k] = n s[k-1] - s[k-2]; s[0] = 1; s[1] = n+1; For[k = 1, k <= kmax, k++, If[PrimeQ[s[k]], Return[s[k]]]]];
Array[a, terms] /. Null -> -1 (* Jean-François Alcover, Aug 30 2018 *)

If n is prime then a(n-1) = n.

A064170 a(1) = 1; a(n+1) = product of numerator and denominator in Sum_{k=1..n} 1/a(k).

Original entry on oeis.org

1, 1, 2, 10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026, 71778070001175617, 491974210728665290

Offset: 1

Leroy Quet, Sep 19 2001

The numerator and denominator in the definition have no common divisors >1.
Also denominators in a system of Egyptian fraction for ratios of consecutive Fibonacci numbers: 1/2 = 1/2, 3/5 = 1/2 + 1/10, 8/13 = 1/2 + 1/10 + 1/65, 21/34 = 1/2 + 1/10 + 1/65 + 1/442 etc. (Rossi and Tout). - Barry Cipra, Jun 06 2002
a(n)-1 is a square. - Sture Sjöstedt, Nov 04 2011
From Wolfdieter Lang, May 26 2020: (Start)
Partial sums of the reciprocals: Sum_{k=1..n} 1/a(k) equal 1 for n=1, and F(2*n - 1)/F(2*n - 3) for n >= 2, with F = A000045. Proof by induction. Hence a(n) = 1 for n=1, and F(2*n - 3)*F(2*n - 5) for n >= 2, with F(-1) = 1 (gcd(F(n), F(n+1)) = 1). See the comment by Barry Cipra.
Thus a(n) = 1, for n = 1, and a(n) = 1 + F(2*(n-2))^2, for n >= 2 (by Cassini-Simson for even index, e.g., Vajda, p. 178 eq.(28)). See the Sture Sjöstedt comment.
The known G.f. of {F(2*n)^2} from A049684 leds then to the conjectured formula by R. J. Mathar below, and this proves also the recurrence given there..
From the partial sums the series Sum_{k>=1} 1/a(k) converges to 1 + phi, with phi = A001622. See the formulas by Gary W. Adamson and Diego Rattaggi below. (End)

			1/a(1) + 1/a(2) + 1/a(3) + 1/a(4) = 1 + 1 + 1/2 + 1/10 = 13/5. So a(5) = 13 * 5 = 65.

S. Vajda, Fibonacci & Lucas Numbers, and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Michael De Vlieger, Table of n, a(n) for n = 1..1199
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
C. Rossi and C. A. Tout, Were the Fibonacci Series and the Golden Section Known in Ancient Egypt?, Historia Mathematica, vol. 29 (2002), 101-113.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A059929, A058038.
Cf. A033890 (first differences). - R. J. Mathar, Jul 03 2009
Cf. A001906, A001622.

A064170[1] := 1; A064170[n_] := A064170[n] = Module[{temp = Sum[1/A064170[i], {i, n - 1}]}, Numerator[temp] Denominator[temp]]; Table[A064170[n], {n, 20}](* Alonso del Arte, Sep 05 2013 *)
Join[{1}, LinearRecurrence[{8, -8, 1}, {1, 2, 10}, 23]] (* Jean-François Alcover, Sep 22 2017 *)

a(n) = Fibonacci(2*n-5)*Fibonacci(2*n-3), for n >= 3. - Barry Cipra, Jun 06 2002
Sum_{n>=3} 1/a(n) = 2/(1+sqrt(5)) = phi - 1, with phi = A001622. - Gary W. Adamson, Jun 07 2003
Conjecture: a(n) = 8*a(n-1)-8*a(n-2)+a(n-3), n>4. G.f.: -x*(2*x^2+x^3-7*x+1)/((x-1)*(x^2-7*x+1)). - R. J. Mathar, Jul 03 2009 [For a proof see the W. Lang comment above.]
a(n+1) = (A005248(n)^2 - A001906(n)^2)/4, for n => 0. - Richard R. Forberg, Sep 05 2013
From Diego Rattaggi, Apr 21 2020: (Start)
a(n) = 1 + A049684(n-2) for n>1.
Sum_{n>=2} 1/a(n) = phi = (1+sqrt(5))/2 = A001622.
Sum_{n>=1} 1/a(n) = phi^2 = 1 + phi. (End) [See a comment above for the proof]
a(n) = F(2*n - 3)*F(2*n - 5) = 1 + F(2*(n - 2))^2, for n >= 2, with F(-1) = 1. See the W. Lang comments above. - Wolfdieter Lang, May 26 2020

cp's OEIS Frontend

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

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A049629 a(n) = (F(6*n+5) - F(6*n+1))/4 = (F(6*n+4) + F(6*n+2))/4, where F = A000045.

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

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

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A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

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A057081 Even-indexed Chebyshev U-polynomials evaluated at sqrt(11)/2.

A049629 a(n) = (F(6n+5) - F(6n+1))/4 = (F(6n+4) + F(6n+2))/4, where F = A000045.

A294099 Rectangular array read by (upward) antidiagonals: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))n^(k-j), n >= 1, k >= 0.