A097315 -id:A097315 - cp's OEIS frontend

A195616 Denominators of Pythagorean approximations to 3.

12, 444, 16872, 640680, 24328980, 923860548, 35082371856, 1332206269968, 50588755886940, 1921040517433740, 72948950906595192, 2770139093933183544, 105192336618554379492, 3994538652411133237140, 151687276455004508631840

Offset: 1

Clark Kimberling, Sep 22 2011

See A195500 for a discussion and references.

Colin Barker, Table of n, a(n) for n = 1..633
Index entries for linear recurrences with constant coefficients, signature (37,37,-1).

Cf. A085447, A097314, A097315, A195500, A195617.

I:=[12, 444, 16872]; [n le 3 select I[n] else 37*Self(n-1) +37*Self(n-2) -Self(n-3): n in [1..40]]; // G. C. Greubel, Feb 13 2023

r = 3; z = 20;
p[{f_, n_}] := (#1[[2]]/#1[[
      1]] &)[({2 #1[[1]] #1[[2]], #1[[1]]^2 - #1[[
         2]]^2} &)[({Numerator[#1], Denominator[#1]} &)[
     Array[FromContinuedFraction[
        ContinuedFraction[(#1 + Sqrt[1 + #1^2] &)[f], #1]] &, {n}]]]];
{a, b} = ({Denominator[#1], Numerator[#1]} &)[
  p[{r, z}]]  (* A195616, A195617 *)
Sqrt[a^2 + b^2] (* A097315 *)
(* Peter J. C. Moses, Sep 02 2011 *)
Table[(1/20)*(LucasL[2*n+1,6] -6*(-1)^n), {n,40}] (* G. C. Greubel, Feb 13 2023 *)

Vec(12*x/((1+x)*(1-38*x+x^2)) + O(x^20)) \\ Colin Barker, Jun 04 2015

A085447=BinaryRecurrenceSequence(6,1,2,6)
[(A085447(2*n+1) - 6*(-1)^n)/20 for n in range(1,41)] # G. C. Greubel, Feb 13 2023

From Colin Barker, Jun 04 2015: (Start)
a(n) = 37*a(n-1) + 37*a(n-2) - a(n-3).
G.f.: 12*x / ((1+x)*(1-38*x+x^2)). (End)
From G. C. Greubel, Feb 13 2023: (Start)
a(n) = (3/10)*(A097314(n) + (-1)^n).
a(n) = (1/20)*(A085447(2*n+1) - 6*(-1)^n). (End)

A249457 The numerator of curvatures of touching circles inscribed in a special way in the larger segment of a unit circle divided by a chord of length sqrt(84)/5.

Original entry on oeis.org

10, 100, 2890, 96100, 3237610, 109202500, 3683712490, 124263300100, 4191798484810, 141402777864100, 4769968258260490, 160906295771812900, 5427884341892493610, 183099910962324064900, 6176546013641762558890, 208354665265158340802500, 7028469704892605715408010

Offset: 0

Kival Ngaokrajang, Oct 29 2014

The denominators are conjectured to be A005032.
Refer to comments and links of A240926. Consider a unit circle with a chord of length sqrt(84)/5. This has been chosen such that the larger sagitta has length 7/5. The input, besides the unit circle C, is the circle C_0 with radius R_0 = 7/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n).
If one considers the curvature of touching circles inscribed in the smaller segment (sagitta length 3/5), the rational sequence would be A249458/A169634. See an illustration given in the link.
For the proof and the formula for the rational curvatures of the circles in the larger segment see a comment under A249862. C_n = (5/7)*(S(n, 34/3) - (17/3)*S(n-1, 34/3) + 1), n >= 0, with Chebyshev's S polynomials (A049310). - Wolfdieter Lang, Nov 07 2014

Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Sagitta.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (37,-111,27).

Cf. A005032, A049310, A078986, A097315, A169364, A240926, A247335, A247512, A248834, A249458, A249862.

I:=[10,100,2890]; [n le 3 select I[n] else 37*Self(n-1) - 111*Self(n-2) + 27*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{37, -111, 27},{10, 100, 2890},16] (* Ray Chandler, Aug 11 2015 *)
CoefficientList[Series[10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.7;dn=7;print1(round(dn/r),", ");r1=r;
for (n=1,40,
     if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
     ac=sqrt(ab^2-r^2);
     if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
     b=acos(r/ab)-z;
     r=r*(1-cos(b))/(1+cos(b)); dn=dn*3;
     print1(round(dn/r),", ");
)
}

x='x+O('x^30); Vec(10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x))) \\ G. C. Greubel, Dec 20 2017

Empirical g.f.: -10*(30*x^2-27*x+1) /((3*x - 1)*(9*x^2-34*x+1)). - Colin Barker, Oct 29 2014
From Wolfdieter Lang, Nov 07 2014: (Start)
a(n) = 5*(A249862(n) + 3^n) = 5*3^n*(S(n, 34/3) - (17/3)*S(n-1, 34/3) + 1), n >= 0, with Chebyshev's S polynomials (A049310). See the comments on A249862 for the proof.
O.g.f.: 5*((1 - 17*x)/(1 - 34*x + 9*x^2) + 1/(1-3*x)) = 10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x)) proving the conjecture of Colin Barker above. (End)
E.g.f.: 5*exp(3*x)*(1 + exp(14*x)*cosh(2*sqrt(70)*x)). - Stefano Spezia, Mar 24 2023

Edited. Name and comment small changes, keyword easy added. - Wolfdieter Lang, Nov 07 2014

a(16) from Stefano Spezia, Mar 24 2023

A195617 Numerators b(n) of Pythagorean approximations b(n)/a(n) to 3.

Original entry on oeis.org

35, 1333, 50615, 1922041, 72986939, 2771581645, 105247115567, 3996618809905, 151766267660819, 5763121552301221, 218846852719785575, 8310417281799550633, 315577009855663138475, 11983615957233399711421, 455061829365013525895519

Offset: 1

Clark Kimberling, Sep 22 2011

See A195500 for discussion and list of related sequences; see A195616 for Mathematica program.

Colin Barker, Table of n, a(n) for n = 1..632
Index entries for linear recurrences with constant coefficients, signature (37,37,-1).

Cf. A085447, A097315, A195500, A195616.

I:=[35, 1333, 50615]; [n le 3 select I[n] else 37*Self(n-1) +37*Self(n-2) -Self(n-3): n in [1..40]]; // G. C. Greubel, Feb 13 2023

Table[(3*LucasL[2*n+1,6] +2*(-1)^n)/20, {n, 40}] (* G. C. Greubel, Feb 13 2023 *)

Vec(-x*(x^2-38*x-35)/((x+1)*(x^2-38*x+1)) + O(x^50)) \\ Colin Barker, Jun 04 2015

A085447=BinaryRecurrenceSequence(6,1,2,6)
[(3*A085447(2*n+1) + 2*(-1)^n)/20 for n in range(1,41)] # G. C. Greubel, Feb 13 2023

From Colin Barker, Jun 04 2015: (Start)
a(n) = 37*a(n-1) + 37*a(n-2) - a(n-3).
G.f.: x*(35+38*x-x^2) / ((1+x)*(1-38*x+x^2)). (End)
a(n) = (1/20)*(3*A085447(2*n+1) + 2*(-1)^n). - G. C. Greubel, Feb 13 2023

A248834 The numerator of curvature of touching circles inscribed in a special way in the smaller segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Original entry on oeis.org

15, 25, 245, 3025, 39605, 525625, 6997445, 93219025, 1242045605, 16549536025, 220514700245, 2938258798225, 39150987330005, 521669482807225, 6951013841444645, 92619168339300625, 1234109231890228805, 16443956730548563225, 219108411138085022645, 2919522145350504838225

Offset: 0

Kival Ngaokrajang, Oct 15 2014

Refer to comment of A240926. Consider a circle C of radius 1/6 (in some length units) with a chord of length sqrt(8/75). This has been chosen such that the smaller sagitta has length 2/15. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/15, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n). The denominator is A000244 for n > 0. If one considers the curvature of touching circles inscribed in the larger segment (sagitta length 1/5), the sequence would be A248833. See an illustration given in the link.

Kival Ngaokrajang, Illustration of initial terms

Cf. A240926, A078986, A097315, A247512, A247335, A247512, A248833.

{
r=0.4;print1(round(6/r),", ");r1=r;dn=1;
for (n=1,40,
if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
ac=sqrt(ab^2-r^2);
if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
b=acos(r/ab)-z;
r=r*(1-cos(b))/(1+cos(b));
print1(round((6/r)*dn),", ");
dn=dn*3
)
}

Conjecture: a(n) = 17*a(n-1) - 51*a(n-2) + 27*a(n-3) for n > 3. - Colin Barker, Oct 15 2014

Empirical g.f.: 5*(54*x^3-117*x^2+46*x-3) / ((3*x-1)*(9*x^2-14*x+1)). - Colin Barker, Oct 15 2014

A249458 The numerators of curvatures of touching circles inscribed in a special way in the smaller segment of unit circle divided by a chord of length sqrt(84)/5.

Original entry on oeis.org

10, 100, 1690, 36100, 835210, 19802500, 472931290, 11318832100, 271066588810, 6492762648100, 155527144782490, 3725543446072900, 89243180863948810, 2137770243127864900, 51209104645650371290, 1226685938180259902500

Offset: 0

Kival Ngaokrajang, Oct 29 2014

The denominators are conjectured to be A169634.
Refer to comments and links of A240926. Consider a unit circle with a chord of length sqrt(84)/5. This has been chosen such that the smaller sagitta has length 3/5. The input, besides the circle C, is the circle C_0 with radius R_0 = 3/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the larger segment (sagitta length 7/5), the sequence would be A249457/A005032. See an illustration given in the link.
For the proof and the formula for the rational curvatures of the circles in the smaller segment see a comment under A249864. C_n = (5/(3*7))*(7*S(n, 26/7) - 13*S(n-1, 26/7) + 7), n >= 0, with Chebyshev's S polynomials (A049310). - Wolfdieter Lang, Nov 08 2014

Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Sagitta.
Index entries for linear recurrences with constant coefficients, signature (33,-231,343).
Index entries for sequences related to Chebyshev polynomials.

Cf. A240926, A078986, A097315, A247512, A247335, A247512, A248834, A169634, A249457, A049310, A249863, A249864.

I:=[10, 100, 1690]; [n le 3 select I[n] else 33*Self(n-1) - 231*Self(n-2) + 343*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{33, -231, 343},{10, 100, 1690},16] (* Ray Chandler, Aug 11 2015 *)
CoefficientList[Series[10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.3;dn=3;print1(round(dn/r),", ");r1=r;
for (n=1,40,
     if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
     ac=sqrt(ab^2-r^2);
     if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
     b=acos(r/ab)-z;
     r=r*(1-cos(b))/(1+cos(b)); dn=dn*7;
     print1(round(dn/r),", ");
)
}

x='x+O('x^30); Vec(10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x))) \\ G. C. Greubel, Dec 20 2017

Empirical g.f.: -10*(70*x^2-23*x+1) / ((7*x-1)*(49*x^2-26*x+1)). - Colin Barker, Oct 29 2014
From Wolfdieter Lang, Nov 09 2014 (Start)
a(n) = 5*(A249864(n) + 7^n) = (5*7^n)*(S(n, 26/7) - (13/7)*S(n-1, 26/7) + 1), n >= 0, with Chebyshev's S polynomials (A049310). See the comments on A249864 for the proof.
O.g.f.: 5*((1 - 13*x)/(1 - 26*x + (7*x)^2) + 1/(1-7*x)) = 10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x)) proving the conjecture of Colin Barker above. (End)

Edited. In name and comment small changes, keyword easy and crossrefs added. - Wolfdieter Lang, Nov 08 2014

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

A157881 Expansion of 152x^2 / (-x^3+1443x^2-1443*x+1).

Original entry on oeis.org

0, 152, 219336, 316282512, 456079163120, 657665836936680, 948353680783529592, 1367525350024012735136, 1971970606380945580536672, 2843580246875973503121146040, 4100440744024547410555112053160, 5912832709303150490046968459510832

Offset: 1

Paul Weisenhorn, Mar 08 2009, Jun 25 2009

This sequence is part of a solution of a more general problem involving two equations, three sequences a(n), b(n), c(n) and a constant A:
    A    * c(n)+1 = a(n)^2,
   (A+1) * c(n)+1 = b(n)^2, for details see comment in A157014.
A157881 is the c(n) sequence for A=9.

Colin Barker, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (1443,-1443,1).

8*A157881(n)+1 = A097315(n-1)^2.

9*A157881(n)+1 = A097314(n-1)^2.

LinearRecurrence[{1443,-1443,1},{0,152,219336},20] (* Harvey P. Dale, Jul 18 2019 *)

concat(0, Vec(152*x^2/(-x^3+1443*x^2-1443*x+1) + O(x^20))) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = round(-((721+228*sqrt(10))^(-n)*(-1+(721+228*sqrt(10))^n)*(19+6*sqrt(10)+(-19+6*sqrt(10))*(721+228*sqrt(10))^n))/360) \\ Colin Barker, Jul 25 2016

G.f.: 152*x^2/(-x^3+1443*x^2-1443*x+1).
c(1) = 0, c(2) = 152, c(3) = 1443*c(2), c(n) = 1443 * (c(n-1)-c(n-2)) + c(n-3) for n>3.
a(n) = -((721+228*sqrt(10))^(-n)*(-1+(721+228*sqrt(10))^n)*(19+6*sqrt(10)+(-19+6*sqrt(10))*(721+228*sqrt(10))^n))/360. - Colin Barker, Jul 25 2016

Edited by Alois P. Heinz, Sep 09 2011

A248833 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Original entry on oeis.org

10, 25, 160, 1225, 9610, 75625, 595360, 4687225, 36902410, 290532025, 2287353760, 18008298025, 141779030410, 1116223945225, 8788012531360, 69187876305625, 544714997913610, 4288532107003225, 33763541858112160, 265819802757894025, 2092794880205040010, 16476539238882426025

Offset: 0

Kival Ngaokrajang, Oct 15 2014

Refer to comment of A240926. Consider a circle C of radius 1/6 (in some length units) with a chord of length sqrt(8/75). This has been chosen such that the larger sagitta has length 1/5. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle, C_n = 1/R_n, n >= 0, is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the smaller segment (sagitta length 2/15), the sequence would be A248834. See an illustration given in the link.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Sagitta.
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A240926, A078986, A097315, A247512, A247335, A247512, A248834.

I:=[10,25,160]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Oct 29 2014

CoefficientList[Series[- 5 (5 x^2 - 13 x + 2)/((x - 1) (x^2 - 8 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 29 2014 *)
LinearRecurrence[{9,-9,1}, {10,25,160}, 30] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.6;print1(round(6/r),", ");r1=r;
for (n=1,40,
     if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
     ac=sqrt(ab^2-r^2);
     if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
     b=acos(r/ab)-z;
     r=r*(1-cos(b))/(1+cos(b));
     print1(round(6/r),", ");
)
}

Vec(-5*(5*x^2-13*x+2)/((x-1)*(x^2-8*x+1)) + O(x^100)) \\ Colin Barker, Oct 15 2014

From Colin Barker, Oct 15 2014: (Start)
a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3).
G.f.: -5*(5*x^2-13*x+2) / ((x-1)*(x^2-8*x+1)). (End)
a(n) = 5*(2+(4-sqrt(15))^n+(4+sqrt(15))^n)/2. - Colin Barker, Mar 03 2016
E.g.f.: 5*exp(x)*(1 + exp(3*x)*cosh(sqrt(15)*x)). - Stefano Spezia, Aug 27 2025

A226694 Pell equation solutions (32a(n))^2 - 41(5*b(n))^2 = -1 with b(n) := A226695(n), n>=0.

Original entry on oeis.org

1, 4099, 16797701, 68836974599, 282093905109001, 1156020754299711499, 4737372769026312613901, 19413752451449074792054799, 79557552808665539471527952401, 326026831996158929305246756884499, 1336057877962706483627361738184724501

Offset: 0

Wolfdieter Lang, Jun 20 2013

			Pell n=0: 32^2 - 41*5^2 = -1.
Pell n=1: (32*4099)^2 - 41*(5*4097)^2 = -1.

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4098,-1)

Cf. A097314, A097315 (Pell -1 with D = 10), A226695.

LinearRecurrence[{4098,-1},{1,4099},20] (* Harvey P. Dale, Sep 23 2017 *)

a(n) = S(n,4098)+ S(n-1,4098), n>=0, with the Chebyshev S-polynomials (A049310). 4098 = 17*241 is the smallest positive integer x solution of x^2 - 41*y^2 = +4 with y also positive.
O.g.f.: (1 + x)/(1 - 4098*x + x^2).
a(n) = 4098*a(n-1) - a(n-2), a(-1) = -1 , a(0) = 1.

A309330 Numbers k such that 10*k^2 + 40 is a square.

Original entry on oeis.org

6, 234, 8886, 337434, 12813606, 486579594, 18477210966, 701647437114, 26644125399366, 1011775117738794, 38420810348674806, 1458979018131903834, 55402781878663670886, 2103846732371087589834, 79890773048222664742806

Offset: 1

Greg Dresden, Jul 23 2019

Sequence of all positive integers k such that the continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(10)).

As 10*n^2 + 40 = 10 * (n^2 + 4), n == 6 (mod 10) or n == 4 (mod 10) alternately. - Bernard Schott, Jul 24 2019

			a(2) = 234, and 10*234^2 + 40 is indeed a perfect square (it's 740^2) and furthermore the continued fraction [234, 234, 234, 234, ...] equals 117 + 37*sqrt(10), which is indeed in Q(sqrt(10)).

Colin Barker, Table of n, a(n) for n = 1..600
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A097315.

LinearRecurrence[{38, -1}, {6, 234}, 15]

Vec(6*x*(1 + x) / (1 - 38*x + x^2) + O(x^20)) \\ Colin Barker, Jul 24 2019

a(n) = 38*a(n-1) - a(n-2); a(1) = 6, a(2) = 234.
a(n) = 2*sqrt(10*A097315(n-1)^2-1).
a(n) = (3-sqrt(10))*(19-6*sqrt(10))^(n-1) + (3+sqrt(10))*(19+6*sqrt(10))^(n-1). - Jinyuan Wang, Jul 24 2019
G.f.: 6*x*(1 + x) / (1 - 38*x + x^2). - Colin Barker, Jul 24 2019
a(n) = 6*A097314(n-1). - R. J. Mathar, Sep 06 2020

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A249457 The numerator of curvatures of touching circles inscribed in a special way in the larger segment of a unit circle divided by a chord of length sqrt(84)/5.

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A195617 Numerators b(n) of Pythagorean approximations b(n)/a(n) to 3.

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A248834 The numerator of curvature of touching circles inscribed in a special way in the smaller segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

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A249458 The numerators of curvatures of touching circles inscribed in a special way in the smaller segment of unit circle divided by a chord of length sqrt(84)/5.

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

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A157881 Expansion of 152*x^2 / (-x^3+1443*x^2-1443*x+1).

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