A001091 -id:A001091 - cp's OEIS frontend

A237262 Values of x in the solutions to x^2 - 8xy + y^2 + 11 = 0, where 0 < x < y.

1, 2, 6, 15, 47, 118, 370, 929, 2913, 7314, 22934, 57583, 180559, 453350, 1421538, 3569217, 11191745, 28100386, 88112422, 221233871, 693707631, 1741770582, 5461548626, 13712930785, 42998681377, 107961675698, 338527902390, 849980474799, 2665224537743

Offset: 1

Colin Barker, Feb 05 2014

The corresponding values of y are given by a(n+2).

Also values of y in the solutions to the negative Pell equation x^2 - 15*y^2 = -11. - Colin Barker, Jan 25 2017

			6 is a term because (x, y) = (6, 47) is a solution to x^2 - 8xy + y^2 + 11 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-1).

Cf. A001091, A070997, A199336, A281584.

For first and second differences see A322780, A199336.

LinearRecurrence[{0,8,0,-1},{1,2,6,15},30] (* Harvey P. Dale, Sep 06 2020 *)

Vec(-x*(x-1)*(x^2+3*x+1)/(x^4-8*x^2+1) + O(x^100))

G.f.: -x*(x-1)*(x^2 + 3*x + 1) / (x^4 - 8*x^2 + 1).
a(n) = 8*a(n-2) - a(n-4) for n > 4.
a(n) = (11*a(n-1) - 4*a(n-2))/3 if n is odd; a(n) = (11*a(n-1) - 3*a(n-2))/4 if n is even. - R. J. Mathar, Jun 18 2014

A077244 Bisection (odd part) of Chebyshev sequence with Diophantine property.

Original entry on oeis.org

3, 22, 173, 1362, 10723, 84422, 664653, 5232802, 41197763, 324349302, 2553596653, 20104423922, 158281794723, 1246149933862, 9810917676173, 77241191475522, 608118614128003, 4787707721548502, 37693543158260013, 296760637544531602, 2336391557197992803

Offset: 0

Wolfdieter Lang, Nov 08 2002

3*a(n)^2 - 5*b(n)^2 = 7, with the companion sequence b(n)= A077243(n).

The even part is A077246(n) with Diophantine companion A077245(n).

			22 = a(1) = sqrt((5*A077243(1)^2 + 7)/3) = sqrt((5*17^2 + 7)/3) = sqrt(484) = 22.

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

I:=[3,22]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 12 2015

 LinearRecurrence[{8, -1}, {3, 22}, 25] (* Vincenzo Librandi, Oct 12 2015 *)

Vec((3-2*x)/(1-8*x+x^2) + O(x^40)) \\ Colin Barker, Oct 12 2015

a(n)= (2*T(n+1, 4)+T(n, 4))/3, with T(n, x) Chebyshev's polynomials of the first kind, A053120. T(n, 4)= A001091(n).
G.f.: (3-2*x)/(1-8*x+x^2).
From Colin Barker, Oct 12 2015: (Start)
a(n) = (((4-sqrt(15))^n * (-10+3*sqrt(15)) + (4+sqrt(15))^n * (10+3*sqrt(15)))) / (2*sqrt(15)).
a(n) = 8*a(n-1) - a(n-2).
(End)

A077246 Bisection (even part) of Chebyshev sequence with Diophantine property.

Original entry on oeis.org

2, 13, 102, 803, 6322, 49773, 391862, 3085123, 24289122, 191227853, 1505533702, 11853041763, 93318800402, 734697361453, 5784260091222, 45539383368323, 358530806855362, 2822707071474573, 22223125764941222

Offset: 0

Wolfdieter Lang, Nov 08 2002

3*a(n)^2 - 5*b(n)^2 = 7, with the companion sequence b(n)= A077245(n).

The odd part is A077244(n) with Diophantine companion A077243(n).

			13 = a(1) = sqrt((5*A077245(1)^2 + 7)/3) = sqrt((5*10^2 + 7)/3) = sqrt(169) = 13.

Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

LinearRecurrence[{8,-1},{2,13},30] (* Harvey P. Dale, Apr 30 2012 *)

a(n)= 8*a(n-1) - a(n-2), a(-1) := 3, a(0)=2.
a(n)= (T(n+1, 4)+2*T(n, 4))/3, with T(n, x) Chebyshev's polynomials of the first kind, A053120. T(n, 4)= A001091(n).
G.f.: (2-3*x)/(1-8*x+x^2).

A084764 a(n) = 2*a(n-1)^2 - 1, a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 31, 1921, 7380481, 108942999582721, 23737154316161495960243527681, 1126904990058528673830897031906808442930637286502826475521

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jun 04 2003

Product_{k=1..n} (1 + 1/a(k)) converges to sqrt(5/3).

The next term has 115 digits. - Harvey P. Dale, Oct 31 2013

G. C. Greubel, Table of n, a(n) for n = 0..11
H. S. Wilf, Limit of a sequence, Elementary Problem E 1093, Amer. Math. Monthly 61 (1954), 424-425.
Jeffrey Shallit, Rational numbers with non-terminating, non-periodic modified Engel-type expansions, Fib. Quart., 31 (1993), 37-40.

Cf. A001091, A005828 (essentially the same).

[n le 2 select 4^(n-1) else 2*Self(n-1)^2 - 1: n in [1..10]]; // G. C. Greubel, May 16 2023

a[n_]:= a[n]= If[n<2, 4^n, 2 a[n-1]^2 -1]; Table[a[n], {n,0,10}]
Join[{1},NestList[2#^2-1&,4,10]] (* Harvey P. Dale, Oct 31 2013 *)

def A084764(n): return 1 if n==0 else chebyshev_T(2^(n-1), 4)
[A084764(n) for n in range(11)] # G. C. Greubel, May 16 2023

With x=4+sqrt(15), y=4-sqrt(15): a(n+1) = (x^(2^n) + y^(2^n))/2.
a(n) = A005828(n-1), n>0. - R. J. Mathar, Sep 17 2008
a(n) = A001091(2^(n-1)) with a(0) = 1; i.e. a(n) = ChebyshevT(2^(n-1), 4) with a(0) = 1. - G. C. Greubel, May 16 2023

A086903 a(n) = 8*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 8.

Original entry on oeis.org

2, 8, 62, 488, 3842, 30248, 238142, 1874888, 14760962, 116212808, 914941502, 7203319208, 56711612162, 446489578088, 3515205012542, 27675150522248, 217885999165442, 1715412842801288, 13505416743244862, 106327921103157608

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 21 2003

a(n+1)/a(n) converges to (4+sqrt(15)) = 7.872983... a(0)/a(1)=2/8; a(1)/a(2)=8/62; a(2)/a(3)=62/488; a(3)/a(4)=488/3842; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.127016... = 1/(4+sqrt(15)) = (4-sqrt(15)).
Twice A001091. - John W. Layman, Sep 25 2003
Except for the first term, positive values of x (or y) satisfying x^2 - 8xy + y^2 + 60 = 0. - Colin Barker, Feb 13 2014

			a(4) = 3842 = 8*a(3) - a(2) = 8*488 - 62 = (4+sqrt(15))^4 + (4-sqrt(15))^4 = 3841.9997397 + 0.0002603 = 3842.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A086594, A058316, A006245, A009271, A005248, A174502.

I:=[2,8]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 15 2014

a[0] = 2; a[1] = 8; a[n_] := 8a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
CoefficientList[Series[(2 - 8 x)/(1 - 8 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 15 2014 *)
LinearRecurrence[{8,-1},{2,8},30] (* Harvey P. Dale, Jan 18 2015 *)

[lucas_number2(n,8,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.
G.f.: (2-8*x)/(1-8*x+x^2). [Philippe Deléham, Nov 02 2008]
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = product {n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 4 - sqrt(15). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.12474 84992 41370 33639 ... = 2 + 1/(8 + 1/(62 + 1/(488 + ...))).  Cf. A174502 and A005248.
Also F(-alpha) = 0.87474 74663 84045 35032 ... has the continued fraction representation 1 - 1/(8 - 1/(62 - 1/(488 - ...))) and the simple continued fraction expansion 1/(1 + 1/((8-2) + 1/(1 + 1/((62-2) + 1/(1 + 1/((488-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((8^2-4) + 1/(1 + 1/((62^2-4) + 1/(1 + 1/((488^2-4) + 1/(1 + ...))))))).
(End)

A095004 a(n) = 9a(n-1) - 9a(n-2) + a(n-3); given a(1) = 1, a(2) = 10, a(3) = 81.

Original entry on oeis.org

1, 10, 81, 640, 5041, 39690, 312481, 2460160, 19368801, 152490250, 1200553201, 9451935360, 74414929681, 585867502090, 4612525087041, 36314333194240, 285902140466881, 2250902790540810, 17721320183859601, 139519658680336000, 1098435949258828401, 8647967935390291210

Offset: 1

Gary W. Adamson, May 27 2004

A sequence derived from A076765, with a(n)/a(n-1) tending to 4 + sqrt(15).
a(n)/a(n-1) tends to C = 4 + sqrt(15) = 7.87298334... (C having the property that C + 1/C = 8). Eigenvalues of M (1, C, 1/C) are roots to x^3 - 9x^2 + 9x - 1.
This is the r=10 member of the r-family of sequences S_r(n), n>=1, defined in A092184, where more information can be found.

			a(4) = 640 = 568 + 72 = A076765(3) + A076765(2).
a(4) = 640 = 9*81 - 9*10 + 1.
a(4) = 640, rightmost term in M^4 * [1 0 0]: [145 352 640] = [A095002(4) A095003(4) A095004(4)].

Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A095002, A095003, A076765.

a:= n-> (<<1|1|1>, <1|2|3>, <1|3|6>>^n)[1, 3]:
seq(a(n), n=1..23);  # Alois P. Heinz, Jun 06 2021

a[n_] := (MatrixPower[{{1, 1, 1}, {1, 2, 3}, {1, 3, 6}}, n].{{1}, {0}, {0}})[[3, 1]]; Table[ a[n], {n, 20}]; (* Robert G. Wilson v, May 29 2004 *)

a(n) = A076765(n-1) + A076765(n-2).
Let M be the 3 X 3 matrix [1 1 1 / 1 2 3 / 1 3 6]; then M^n * [1 0 0] = [A095002(n) A095003(n) a(n)].
a(n)= (T(n, 4)-1)/3 with Chebyshev's polynomials of the first kind evaluated at x=4: T(n, 4)=A001091(n). a(0):=0. - Wolfdieter Lang, Oct 18 2004
G.f.: x*(1+x)/((1-x)*(1-8*x+x^2)) = x*(1+x)/(1-9*x+9*x^2-x^3).

Edited and extended by Robert G. Wilson v, May 29 2004

Definition aligned with A095002, A095003 by Georg Fischer, Jun 06 2021

A098306 Unsigned member r=-6 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 6, 49, 384, 3025, 23814, 187489, 1476096, 11621281, 91494150, 720331921, 5671161216, 44648957809, 351520501254, 2767515052225, 21788599916544, 171541284280129, 1350541674324486, 10632792110315761, 83711795208201600

Offset: 0

Wolfdieter Lang, Oct 18 2004

((-1)^(n+1))*a(n) = S_{-6}(n), n>=0, defined in A092184.

This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. Case P1 = 6, P2 = -16, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Bala, Linear divisibility sequences and Chebyshev polynomials
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 (7,7,-1).

Cf. A001090, A001091, A070997, A092184, A100047.

a[n_] := 1/10*((4 + Sqrt[15])^n + (4 - Sqrt[15])^n - 2*(-1)^n) // Simplify; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Apr 28 2017 *)
LinearRecurrence[{7, 7, -1}, {0, 1, 6, 49, 384, 3025}, 50] (* G. C. Greubel, Aug 08 2017 *)

x='x+O('x^50); Vec(x*(1-x)/((1+x)*(1-8*x+x^2))) \\ G. C. Greubel, Aug 08 2017

a(n) = (T(n, 4)-(-1)^n)/5, with Chebyshev's polynomials of the first kind evaluated at x=4: T(n, 4)=A001091(n)=((4+sqrt(15))^n + (4-sqrt(15))^n)/2.
a(n) = 8*a(n-1) - a(n-2) + 2*(-1)^(n+1), n>=2, a(0)=0, a(1)=1.
a(n) = 7*a(n-1) + 7*a(n-2) - a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=6.
G.f.: x*(1-x)/((1+x)*(1-8*x+x^2)) = x*(1-x)/(1-7*x-7*x^2+x^3) (from the Stephan link, see A092184).
From Peter Bala, Mar 25 2014: (Start)
a(2*n) = 6*A001090(n)^2; a(2*n+1) = A070997(n)^2.
a(n) = |u(n)|^2, where {u(n)} is the Lucas sequence in the quadratic integer ring Z[sqrt(-6)] defined by the recurrence u(0) = 0, u(1) = 1, u(n) = sqrt(-6)*u(n-1) - u(n-2) for n >= 2.
Equivalently, a(n) = U(n-1,sqrt(-6)/2)*U(n-1,-sqrt(-6)/2), where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = 1/10*( (4 + sqrt(15))^n + (4 - sqrt(15))^n - 2*(-1)^n ).
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 4; 1, 3] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)

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

A145607 Numbers k such that (3(2k + 1)^2 + 2)/5 is a square.

Original entry on oeis.org

0, 4, 35, 279, 2200, 17324, 136395, 1073839, 8454320, 66560724, 524031475, 4125691079, 32481497160, 255726286204, 2013328792475, 15850904053599, 124793903636320, 982500325036964, 7735208696659395, 60899169248238199

Offset: 1

Richard Choulet, Oct 14 2008

Square roots of (3*(2*k+1)^2+2)/5 are listed in A070997, therefore (3*(2*a(n) + 1)^2 + 2)/5 = A070997(n-1)^2.

Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A070997, A131751.

Cf. A001091 (first differences).

a(n+2) = 8*a(n+1) - a(n) + 3.
From R. J. Mathar, Oct 24 2008: (Start)
G.f.: x^2*(4 - x)/((1 - x)*(1 - 8*x + x^2)).
a(n) = (A057080(n-1)-1)/2. (End)

a(4) corrected, extended, definition corrected by R. J. Mathar, Oct 24 2008

Offset changed by Bruno Berselli, Apr 06 2018

A118894 Numbers m such that the Pell equation x^2-m*y^2=1 has fundamental solution with x even.

Original entry on oeis.org

3, 7, 11, 15, 19, 23, 27, 31, 35, 43, 47, 51, 59, 63, 67, 71, 75, 79, 83, 87, 91, 99, 103, 107, 115, 119, 123, 127, 131, 135, 139, 143, 151, 159, 163, 167, 171, 175, 179, 187, 191, 195, 199, 211, 215, 219, 223, 227, 231, 235, 239, 243, 247, 251, 255, 263, 267

Offset: 1

T. D. Noe, May 04 2006

nonn

Numbers m such that A002350(m) is even. These m can be used to generate consecutive odd powerful numbers, as in A076445. As shown by Lang, the solution of Pell's equation is greatly simplified by Chebyshev polynomials of the first kind T(n,x), which is illustrated in A001075 for the case m=3. In that case, the solutions are x=T(n,2), for integer n>0. For any m in this sequence, let E(k)=T(m+2mk,A002350(m)). Then E(k)-1 and E(k)+1 are consecutive odd powerful numbers for k=0,1,2,...

Wolfdieter Lang, Chebyshev Polynomials and Certain Quadratic Diophantine Equations
H. W. Lenstra Jr., Solving the Pell equation, Notices AMS, 49 (2002), 182-192.

Cf. A001075, A001091, A023038, A001081, A001085, A077424, A097310 (x solutions for m=3, 15, 35, 63, 99, 143, 195).

cp's OEIS Frontend

A237262 Values of x in the solutions to x^2 - 8*x*y + y^2 + 11 = 0, where 0 < x < y.

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A077244 Bisection (odd part) of Chebyshev sequence with Diophantine property.

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A077246 Bisection (even part) of Chebyshev sequence with Diophantine property.

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A084764 a(n) = 2*a(n-1)^2 - 1, a(0)=1, a(1)=4.

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A086903 a(n) = 8*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 8.

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A095004 a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3); given a(1) = 1, a(2) = 10, a(3) = 81.

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A098306 Unsigned member r=-6 of the family of Chebyshev sequences S_r(n) defined in A092184.

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A237262 Values of x in the solutions to x^2 - 8xy + y^2 + 11 = 0, where 0 < x < y.

A095004 a(n) = 9a(n-1) - 9a(n-2) + a(n-3); given a(1) = 1, a(2) = 10, a(3) = 81.

A145607 Numbers k such that (3(2k + 1)^2 + 2)/5 is a square.