A060645 -id:A060645 - cp's OEIS frontend

A298212 Smallest n such that A060645(a(n)) = 0 (mod n), i.e., x=A023039(a(n)) and y=A060645(a(n)) is the fundamental solution of the Pell equation x^2 - 5(ny)^2 = 1.

1, 1, 2, 1, 5, 2, 4, 2, 2, 5, 5, 2, 7, 4, 10, 4, 3, 2, 3, 5, 4, 5, 4, 2, 25, 7, 6, 4, 7, 10, 5, 8, 10, 3, 20, 2, 19, 3, 14, 10, 10, 4, 22, 5, 10, 4, 8, 4, 28, 25, 6, 7, 9, 6, 5, 4, 6, 7, 29, 10, 5, 5, 4, 16, 35, 10, 34, 3, 4, 20, 35, 2, 37, 19, 50

Offset: 1

A.H.M. Smeets, Jan 15 2018

nonn

The fundamental solution of the Pell equation x^2 - 5*(n*y)^2 = 1, is the smallest solution of x^2 - 5*y^2 = 1 satisfying y = 0 (mod n).

Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.

A.H.M. Smeets, Table of n, a(n) for n = 1..20000
H. W. Lenstra Jr., Solving the Pell Equation, Notices of the AMS, Vol.49, No.2, Feb. 2002, pp. 182-192.

Cf. A298210, A298211.

b[n_] := b[n] = Switch[n, 0, 0, 1, 4, _, 18 b[n - 1] - b[n - 2]];
a[n_] := For[k = 1, True, k++, If[Mod[b[k], n] == 0, Return[k]]];
a /@ Range[100] (* Jean-François Alcover, Nov 16 2019 *)

xf, yf = 9, 4
x, n = 2*xf, 0
while n < 20000:
    n = n+1
    y1, y0, i = 0, yf, 1
    while y0%n != 0:
        y1, y0, i = y0, x*y0-y1, i+1
    print(n, i)

a(n) <= n.
a(A000351(n)) = A000351(n).
A023039(a(n)) = A002350(5*n^2).
A060645(a(n)) = A002349(5*n^2).
if n | m then a(n) | a(m).
a(5^m) = 5^m for m>=0.
In general: if p is prime and p = 1 (mod 4) then: a(n) = n iff n = p^m, for m>=0.

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

A075796 Numbers k such that 5*k^2 + 5 is a square.

Original entry on oeis.org

2, 38, 682, 12238, 219602, 3940598, 70711162, 1268860318, 22768774562, 408569081798, 7331474697802, 131557975478638, 2360712083917682, 42361259535039638, 760141959546795802, 13640194012307284798, 244763350261984330562, 4392100110703410665318, 78813038642399407645162

Offset: 1

Gregory V. Richardson, Oct 13 2002

Bisection of A001077; a(n) = A001077(2*n-1). - Greg Dresden, Jun 08 2021
From Peter Bala, Aug 25 2022: (Start)
The aerated sequence (b(n))n>=1 = [2, 0, 38, 0, 682, 0, 1238, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). The sequence (1/2)*(b(n))n>=1 is the case P1 = 0, P2 = -16, Q = -1  of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. (End)

A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A000290, A306380, A001077.

I:=[2,38]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 30 2011

[Lucas(6*n-3)/2: n in [1..20]]; // G. C. Greubel, Feb 13 2019

with(combinat); A075796:=n->fibonacci(6*n+3)+fibonacci(6*n)/2; seq(A075796(n), n=1..50); # Wesley Ivan Hurt, Nov 29 2013

LinearRecurrence[{18, -1}, {2, 38}, 50] (* Sture Sjöstedt, Nov 29 2011; typo fixed by Vincenzo Librandi, Nov 30 2011 *)
LucasL[6*Range[20]-3]/2 (* G. C. Greubel, Feb 13 2019 *)
CoefficientList[Series[2*(1+x)/( 1-18*x+x^2 ), {x,0,20}],x] (* Stefano Spezia, Mar 02 2019 *)

vector(20, n, (fibonacci(6*n-2) + fibonacci(6*n-4))/2) \\ G. C. Greubel, Feb 13 2019

[(fibonacci(6*n-2) + fibonacci(6*n-4))/2 for n in (1..20)] # G. C. Greubel, Feb 13 2019

a(n) = (((9 + 4*sqrt(5))^n - (9 - 4*sqrt(5))^n) + ((9 + 4*sqrt(5))^(n-1) - (9 - 4*sqrt(5))^(n-1)))/(4*sqrt(5)).
a(n) = 18*a(n-1) - a(n-2).
a(n) = 2*A049629(n-1).
Limit_{n->oo} a(n)/a(n-1) = 8*phi + 1 = 9 + 4*sqrt(5).
a(n+1) = 9*a(n) + 4*sqrt(5)*sqrt((a(n)^2+1)). - Richard Choulet, Aug 30 2007
G.f.: 2*x*(1 + x)/(1 - 18*x + x^2). - Richard Choulet, Oct 09 2007
From Johannes W. Meijer, Jul 01 2010: (Start)
a(n) = A000045(6*n+3) + A000045(6*n)/2.
a(n) = 2*A167808(6*n+4) - A167808(6*n+6).
Limit_{k->oo} a(n+k)/a(k) = A023039(n)*A060645(n)*sqrt(5).
(End)
5*A007805(n)^2 - 1 = a(n+1)^2. - Sture Sjöstedt, Nov 29 2011
From Peter Bala, Nov 29 2013: (Start)
a(n) = Lucas(6*n - 3)/2.
Sum_{n >= 1} 1/(a(n) + 5/a(n)) = 1/4. Compare with A002878, A005248, A023039. (End)
Limit_{n->oo} a(n)/A007805(n-1) = sqrt(5). - A.H.M. Smeets, May 29 2017
E.g.f.: (exp((9 - 4*sqrt(5))*x)*(- 5 + 2*sqrt(5) + (5 + 2*sqrt(5))*exp(8*sqrt(5)*x)))/(2*sqrt(5)). - Stefano Spezia, Feb 13 2019
Sum_{n > 0} 1/a(n) = (1/log(9 - 4*sqrt(5)))*(- 17 - 38/sqrt(5))*sqrt(5*(9 - 4*sqrt(5)))*(- 9 + 4*sqrt(5))*(psi_{9 - 4*sqrt(5)}(1/2) - psi_{9 - 4*sqrt(5)}(1/2 - (I*Pi)/log(9 - 4*sqrt(5)))) approximately equal to 0.527868600269500798938265500122302016..., where psi_q(x) is the q-digamma function. - Stefano Spezia, Feb 25 2019
a(n) = sinh((6*n - 3)*arccsch(2)). - Peter Luschny, May 25 2022

A134493 a(n) = Fibonacci(6*n+1).

Original entry on oeis.org

1, 13, 233, 4181, 75025, 1346269, 24157817, 433494437, 7778742049, 139583862445, 2504730781961, 44945570212853, 806515533049393, 14472334024676221, 259695496911122585, 4660046610375530309, 83621143489848422977, 1500520536206896083277, 26925748508234281076009

Offset: 0

Artur Jasinski, Oct 28 2007

For positive n, a(n) equals (-1)^n times the permanent of the (6n)X(6n) tridiagonal matrix with ((-1)^(1/6))'s along the three central diagonals. - John M. Campbell, Jul 12 2011

a(n) = x + y where those two values 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
Index entries for linear recurrences with constant coefficients, signature (18,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000032, A000045, A049660, A134492, A134494, A134495, A103134, A134497.

[Fibonacci(6*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Mathematica
```
Table[Fibonacci[6n+1], {n, 0, 30}]
```

a(n)=fibonacci(6*n+1) \\ Charles R Greathouse IV, Jul 15 2011

Vec((1-5*x)/(1-18*x+x^2) + O(x^100)) \\ Altug Alkan, Jan 24 2016

From R. J. Mathar, Apr 17 2011: (Start)
G.f.: ( 1-5*x ) / ( 1-18*x+x^2 ).
a(n) = A049660(n+1) - 5*A049660(n). (End)
a(n) = Fibonacci(3*n+1)^2 + Fibonacci(3*n)^2. - Gary Detlefs, Oct 12 2011
a(n) = 18*a(n-1) - a(n-2). - Richard R. Forberg, Sep 05 2013
a(n) = A060645(n) + A023039(n), as derives from comment above. - Richard R. Forberg, Sep 05 2013
a(n) = ((5-sqrt(5)+(5+sqrt(5))*(9+4*sqrt(5))^(2*n)))/(10*(9+4*sqrt(5))^n). - Colin Barker, Jan 24 2016
2*a(n) = Fibonacci(6*n) + Lucas(6*n). - Bruno Berselli, Oct 13 2017
a(n) = S(n, 18) - 5*S(n-1, 18), n >= 0, with the Chebyshev S-polynomials S(n-1, 18) = A049660(n). (See the g.f.) - Wolfdieter Lang, Jul 10 2018

Offset changed to 0 by Vincenzo Librandi, Apr 16 2011

A033314 Least D in the Pellian x^2 - D*y^2 = 1 for which x has least solution n.

Original entry on oeis.org

3, 2, 15, 6, 35, 12, 7, 5, 11, 30, 143, 42, 195, 14, 255, 18, 323, 10, 399, 110, 483, 33, 23, 39, 27, 182, 87, 210, 899, 60, 1023, 17, 1155, 34, 1295, 38, 1443, 95, 1599, 105, 1763, 462, 215, 506, 235, 138, 47, 96, 51, 26, 2703, 78, 2915, 21, 3135, 203, 3363

Offset: 2

Eric W. Weisstein

nonn

The i-th solution pair V(i) = [x(i), y(i)] to the Pellian x^2 - D*y^2 = 1 for a given least solution x(1) = n may be generated through the recurrence V(i+2) = 2*n*V(i+1) - V(i) taking V(0) = [1, 0] and V(1) = [n, sqrt((n^2-1)/a(n))]. V(i) stands for the numerator and denominator of the 2i-th convergent of the continued fraction expansion of sqrt(D).

Thus setting n = 3, for instance, we have D = a(3) = 2 and V(1) = [3, 2] so that along with V(0) = [1, 0] recurrence V(i+2) = 6*V(i+1) - V(i) generates [A001333(2k), A000129(2k)]. Similarly, setting n = 9 generates [A023039, A060645], respectively the numerator and denominator of the 2i-th convergent of sqrt(a(9)), i.e., sqrt(5). - Lekraj Beedassy, Feb 26 2002

Ray Chandler, Table of n, a(n) for n = 2..1001
Eric Weisstein's World of Mathematics, Pell Equation.

Cf. A000037, A033313, A033318.

squarefreepart[n_] :=
  Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n]);
pellminx[d_] := Module[{q, p, z}, {q, p} = ContinuedFraction[Sqrt[d]];
  If[OddQ[p // Length], p = Join[p, p]];
  z = FromContinuedFraction[Join[{q}, Drop[p, -1]]]; Numerator[z]]
NMAX = 60; a = {};
For[n = 2, n <= NMAX, n++, s = squarefreepart[n^2 - 1];
sd = s Divisors[Sqrt[(n^2 - 1)/s]]^2;
t = Sort[Transpose[{sd, pellminx[#] & /@ sd}]];
AppendTo[a, Select[t, #[[2]] == n &, 1][[1, 1]]]
]; a (* Herbert Kociemba, Jun 05 2022 *)

A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

Original entry on oeis.org

0, 6, 228, 8658, 328776, 12484830, 474094764, 18003116202, 683644320912, 25960481078454, 985814636660340, 37434995712014466, 1421544022419889368, 53981237856243781518, 2049865494514843808316, 77840907553707820934490, 2955904621546382351702304

Offset: 0

Benoit Cloitre, May 10 2003

nonn

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 10*y^2 = 1. The corresponding x values are in A078986. - Vincenzo Librandi, Aug 08 2010 [edited by Jon E. Schoenfield, May 04 2014]

			G.f. = 6*x + 228*x^2 + 8658*x^3 + 328776*x^4 + ... - _Michael Somos_, Feb 24 2023

Indranil Ghosh, Table of n, a(n) for n = 0..632
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A001078, A001109, A001353, A001653, A005668, A060645, A084068, A084069, A221874.

Cf. A078986. - Vincenzo Librandi, Apr 14 2010

a:=[0,6];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2020

I:=[0,6]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Jan 12 2020

seq( simplify(6*ChebyshevU(n-1, 19)), n=0..20); # G. C. Greubel, Jan 12 2020

LinearRecurrence[{38,-1},{0,6},30] (* Harvey P. Dale, Nov 01 2011 *)
6*ChebyshevU[Range[20]-2, 19] (* G. C. Greubel, Jan 12 2020 *)

u=0; v=6; for(n=2,20, w=38*v-u; u=v; v=w; print1(w,","))

vector(21, n, 6*polchebyshev(n-2, 2, 19) ) \\ G. C. Greubel, Jan 12 2020

[6*chebyshev_U(n-1, 19) for n in (0..20)] # G. C. Greubel, Jan 12 2020

Numbers k such that 10*k^2 = floor(k*sqrt(10)*ceiling(k*sqrt(10))).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 37*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 39*(a(n-1) - a(n-2)) + a(n-3). (End)
From R. J. Mathar, Feb 19 2008: (Start)
O.g.f.: 6*x/(1 - 38*x + x^2).
a(n) = 6*A078987(n-1). (End)
a(n) = 6*ChebyshevU(n-1, 19). - G. C. Greubel, Jan 12 2020
a(n) = A005668(2*n). - Michael Somos, Feb 24 2023

A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

Original entry on oeis.org

0, 2, 36, 646, 11592, 208010, 3732588, 66978574, 1201881744, 21566892818, 387002188980, 6944472508822, 124613502969816, 2236098580947866, 40125160954091772, 720016798592704030

Offset: 0

Gary Detlefs, Feb 20 2012

Denote as {a,b,c,d} the second-order linear recurrence a(n) = c*a(n-1) + d*a(n-2) with initial terms a, b. The following sequences and recurrence formulas are related to integer solutions of k*x^2 + 1 = y^2.
.
   k             x                        y
   -  -----------------------  -----------------------
A001542 {0,2,6,-1}       A001541 {1,3,6,-1}
A001353 {0,1,4,-1}       A001075 {1,2,4,-1}
A060645 {0,4,18,-1}      A023039 {1,9,18,-1}
A001078 {0,2,10,-1}      A001079 {1,5,10,-1}
A001080 {0,3,16,-1}      A001081 {1,8,16,-1}
A001109 {0,1,6,-1}       A001541 {1,3,6,-1}
A084070 {0,1,38,-1}      A078986 {1,19,38,-1}
A001084 {0,3,20,-1}      A001085 {1,10,20,-1}
A011944 {0,2,14,-1}      A011943 {1,7,14,-1}
A075871 {0,180,1298,-1}  A114047 {1,649,1298,-1}
A068204 {0,4,30,-1}      A069203 {1,15,30,-1}
A001090 {0,1,8,-1}       A001091 {1,4,8,-1}
A121740 {0,8,66,-1}      A099370 {1,33,66,-1}
A202299 {0,4,34,-1}      A056771 {1,17,34,-1}
A174765 {0,39,340,-1}    A114048 {1,179,340,-1}
a(n)    {0,2,18,-1}      A023039 {1,9,18,-1}
A174745 {0,12,110,-1}    A114049 {1,55,110,-1}
A174766 {0,42,394,-1}    A114050 {1,197,394,-1}
A174767 {0,5,48,-1}      A114051 {1,24,48,-1}
A004189 {0,1,10,-1}      A001079 {1,5,10,-1}
A174768 {0,10,102,-1}    A099397 {1,51,102,-1}
The sequence of the c parameter is listed in A180495.

Bruno Berselli, Table of n, a(n) for n = 0..500
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A023039, A049660, A079962.

m:=16; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2*x/(1-18*x+x^2))); // Bruno Berselli, Jun 19 2019

readlib(issqr):for x from 1 to 720016798592704030 do if issqr(20*x^2+1) then print(x) fi od;

LinearRecurrence[{18, -1}, {0, 2}, 16] (* Bruno Berselli, Feb 21 2012 *)
Table[2 ChebyshevU[-1 + n, 9], {n, 0, 16}]  (* Herbert Kociemba, Jun 05 2022 *)

makelist(expand(((2+sqrt(5))^(2*n)-(2-sqrt(5))^(2*n))/(4*sqrt(5))), n, 0, 15); /* Bruno Berselli, Jun 19 2019 */

a(n) = 18*a(n-1) - a(n-2).
From Bruno Berselli, Feb 21 2012: (Start)
G.f.: 2*x/(1-18*x+x^2).
a(n) = -a(-n) = 2*A049660(n) = ((2 + sqrt(5))^(2*n)-(2 - sqrt(5))^(2*n))/(4*sqrt(5)). (End)
a(n) = Fibonacci(6*n)/4. - Bruno Berselli, Jun 19 2019
For n>=1, a(n) = A079962(6n-3). - Christopher Hohl, Aug 22 2021

A084069 Numbers k such that 7k^2 = floor(ksqrt(7)ceiling(ksqrt(7))).

Original entry on oeis.org

1, 3, 17, 48, 271, 765, 4319, 12192, 68833, 194307, 1097009, 3096720, 17483311, 49353213, 278635967, 786554688, 4440692161, 12535521795, 70772438609, 199781794032, 1127918325583, 3183973182717, 17975920770719, 50743789129440

Offset: 1

Benoit Cloitre, May 10 2003

nonn

This is a strong divisibility sequence, that is, GCD(a(n),a(m)) = a(GCD(n,m)) for all positive integers n and m. Consequently, this is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Sep 01 2019

Wikipedia, Lehmer sequence
Index entries for linear recurrences with constant coefficients, signature (0,16,0,-1).

Cf. A159678, A001080, A001653, A001353, A060645, A001078, A001109, A084068, A084070, A143643.

CoefficientList[Series[(1+3x+x^2)/(1-16x^2+x^4),{x,0,30}],x] (* or *) LinearRecurrence[{0,16,0,-1},{1,3,17,48},31] (* Harvey P. Dale, Oct 31 2011 *)

a(1)=1, a(2)=3, a(2n) = 6*a(2n-1)-a(2n-2); a(2n+1) = 3*a(2n)-a(2n-1).
a(n)*a(n+3) = -3 + a(n+1)*a(n+2).
G.f.: x*(1+3*x+x^2)/(1-16*x^2+x^4). [corrected by Harvey P. Dale, Oct 31 2011]
a(n) = 16*a(n-2) - a(n-4), n > 4. - Harvey P. Dale, Oct 31 2011
a(n) = U_n(sqrt(18),1) = (alpha^n - beta^n)/(alpha - beta) for n odd and a(n) = 3*U_n(sqrt(18),1) = (sqrt(2)/2)*(alpha^n - beta^n)/(alpha - beta) for n even, where U_n(sqrt(R),Q) denotes the Lehmer sequence with parameters R and Q and alpha = (sqrt(3) + sqrt(14))/2 and beta = (sqrt(3) - sqrt(14))/2. - Peter Bala, Sep 01 2019

A085348 Ratio-determined insertion sequence I(0.264) (see the link below).

Original entry on oeis.org

1, 4, 19, 72, 341, 1292, 6119, 23184, 109801, 416020, 1970299, 7465176, 35355581, 133957148, 634430159, 2403763488, 11384387281, 43133785636, 204284540899, 774004377960, 3665737348901, 13888945017644, 65778987739319

Offset: 0

John W. Layman, Jun 24 2003

nonn

This is one of the "twin" ratio-determined insertion sequences (RDIS) that are "children" in the next generation below the "parent" sequences I(0.25024) (A004253) and I(0.26816) (A001353) in the recurrence tree of RDIS sequences. The RDIS twin of this sequence is A085349. See the link for an explanation of RDIS twin. See A082630 or A082981 for other recent examples of RDIS sequences.

Assuming that a(n) = 18a(n-2) - a(n-4) is true: For n >= 2, a(n) = (t(i+2n+2) - t(i))/(t(i+n+2) + t(i+n)*(-1)^(n-1)), where (t) is any recurrence of the form (4,1) without regard to initial values. With an additional initional 0 is this sequence the union of A060645 for even n and A049629 for odd n. - Klaus Purath, Sep 22 2024

John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [Broken link]
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [local copy, corrected]
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006 [Local copy]

Cf. A001353, A004253, A082630, A082981, A085349.

It appears that a(n)=18a(n-2)-a(n-4).
Apparently a(n)a(n+3) = -4 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004
From Klaus Purath, Sep 22 2024: (Start)
Assuming that a(n) = 18a(n-2) - a(n-4) is true:
a(2n) = 5a(2n-1) - a(2n-2), n >= 1.
a(2n+1) = 4a(2n) - a(2n-1), n >= 1. (End)

A295282 a(n) > n is chosen to minimize the difference between ratios a(n):n and n:(a(n) - n), so that they are matching approximations to the golden ratio.

Original entry on oeis.org

2, 3, 5, 7, 8, 10, 11, 13, 15, 16, 18, 19, 21, 23, 24, 26, 28, 29, 31, 32, 34, 36, 37, 39, 40, 42, 44, 45, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 65, 66, 68, 70, 71, 73, 74, 76, 78, 79, 81, 83, 84, 86, 87, 89, 91, 92, 94, 95, 97, 99, 100, 102, 104, 105, 107, 108

Offset: 1

Peter Munn, Nov 19 2017

nonn

The difference between the matching ratios is evaluated by dividing the larger by the smaller.
Take a rectangle A with sides n and n+m; remove a square of side n from one end to form rectangle B with sides n and m; scale B in the ratio (n+m):n to form rectangle C with a side n+m. Place A and C alongside, with edges of length n+m coinciding, to form rectangle D. For n > 0, let m_n be the m that has the coincident edges dividing D in nearest to equal proportions, then a(n) = n + m_n.
Compared with other neighboring values of n, the resulting proportions can be made most nearly equal when n is a Fibonacci number F(k) = A000045(k), k > 1, in which case a(n) is F(k+1). In contrast, if 2n is a Fibonacci number F(k), then a relatively good choice for rectangle A's longer side would be F(k+1)/2, except that F(k+1) is odd when F(k) is even, so F(k+1)/2 is halfway between integers.
a(n) is usually the same as A007067(n), but when 2n is a Fibonacci number, they sometimes differ. The first differences are a(4) = 7 = A007067(4) + 1 and a(72) = 117 = A007067(72) + 1. The author expects a(n) to differ from A007067(n) if and only if n is in A060645. The terms of A060645 are half the value of alternate even Fibonacci numbers.
More specifically, for k > 0: F(3k) is an even Fibonacci number, F(3k+1) is odd and a(F(3k)/2) = F(3k+1)/2 + 1/2; whereas A007067(F(6k+3)/2) = F(6k+4)/2 + 1/2, but A007067(F(6k)/2) = F(6k+1)/2 - 1/2.

			The matching ratios and the differences between them begin:
   2:1          1:1         2.0
   3:2          2:1         1.3333...
   5:3          3:2         1.1111...
   7:4          4:3         1.3125
   8:5          5:3         1.0416...
  10:6          6:4         1.1111...
  11:7          7:4         1.1136...
  13:8          8:5         1.015625
  15:9          9:6         1.1111...
  16:10        10:6         1.0416...
  18:11        11:7         1.0413...
  19:12        12:7         1.0827...
  21:13        13:8         1.0059...
  23:14        14:9         1.0561...
  24:15        15:9         1.0416...
  26:16        16:10        1.015625
  28:17        17:11        1.0657...
  29:18        18:11        1.0156...
  31:19        19:12        1.0304...
  32:20        20:12        1.0416...
  34:21        21:13        1.0022...
  ...
For n = 4:
if a(4) = 5, the matching ratios would be a(4):4 = 5:4 and 4:(a(4)-4) = 4:1, with the difference between them (larger divided by smaller) = (4/1) / (5/4) = 16/5 = 3.2;
if a(4) = 6, ratios would be 6:4 and 4:2, with difference = (4/2) / (6/4) = 16/12 = 1.333...;
if a(4) = 7, ratios would be 7:4 and 4:3, with difference = (7/4) / (4/3) = 21/16 = 1.3125;
if a(4) = 8, ratios would be 8:4 and 4:4, with difference = (8/4) / (4/4) = 32/16 = 2.0.
Any larger value for a(4) would give a difference between the ratios that exceeded 2.0, so a(4) = 7, as this achieves the minimum difference.
This example translates as follows into the geometry described early in the comments:
                      n            4           4
                      m            2           3
                     n+m           6           7
Rectangle A       n X (n+m)      4 X 6       4 X 7
Rectangle B         m X n        2 X 4       3 X 4
Scaling ratio      n:(n+m)        4:6         4:7
m scaled up       m*(n+m)/n      2*6/4       3*7/4
= side of C           l            3          5.25
Rectangle C       l X (n+m)      3 X 6      5.25 X 7
Rectangle D     (n+l) X (n+m)    7 X 6      9.25 X 7
proportion C/D     l/(n+l)        3/7      5.25/9.25
- as decimal                    0.4285...   0.5675...
- its difference from 0.5       0.0714...   0.0675...

A001622 gives the value of the golden ratio.

Cf. A000045, A007067, A022342, A060645.

a(n) = (m+n) > n so as to minimize (max((m+n)/n, n/m) / min((m+n)/n, n/m)).

a(n+1) = a(n) + 2 - floor((a(n)+2) * (a(n)+1-n) * (a(n)+1) * (a(n)-n) / (n+1)^4), with a(0) = 0 for the purpose of this calculation.

cp's OEIS Frontend

A298212 Smallest n such that A060645(a(n)) = 0 (mod n), i.e., x=A023039(a(n)) and y=A060645(a(n)) is the fundamental solution of the Pell equation x^2 - 5*(n*y)^2 = 1.

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

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A075796 Numbers k such that 5*k^2 + 5 is a square.

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A134493 a(n) = Fibonacci(6*n+1).

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A033314 Least D in the Pellian x^2 - D*y^2 = 1 for which x has least solution n.

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A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

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A298212 Smallest n such that A060645(a(n)) = 0 (mod n), i.e., x=A023039(a(n)) and y=A060645(a(n)) is the fundamental solution of the Pell equation x^2 - 5(ny)^2 = 1.

A084069 Numbers k such that 7k^2 = floor(ksqrt(7)ceiling(ksqrt(7))).