A240926 -id:A240926 - cp's OEIS frontend

A246640 Sequence a(n) = 1 + A001519(n+1) appearing in a certain touching problem for three circles and a chord, together with A246638.

2, 3, 6, 14, 35, 90, 234, 611, 1598, 4182, 10947, 28658, 75026, 196419, 514230, 1346270, 3524579, 9227466, 24157818, 63245987, 165580142, 433494438, 1134903171, 2971215074, 7778742050, 20365011075, 53316291174, 139583862446, 365435296163, 956722026042, 2504730781962

Offset: 0

Wolfdieter Lang, Sep 03 2014

Essentially the same as A093467 and A032908.
This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.
a(n), together with b(n) = A246638(n), appears in a curvature c(n) =  b(n) + 4*a(n)*phi, with phi = (1+sqrt(5))/2, the golden section. This is an integer in the real quadratic field Q(sqrt(5)). c(n) is the curvature of the circle which touches i) a chord of length 2 (in some length units) of a circle of radius 5/4 which is divided by this chord in two unequal parts, and ii) the two touching circles in the smaller part which have curvatures A240926(n) and A240926(n+1). These two touching circles touch also the circle with radius 5/4 and the chord. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the smaller (upper) part are shown. c(n) is an integer in the real quadratic field  Q(sqrt(5)).
From Descartes' theorem on touching circles (see the links) one has here: c(n) = A(n) + A(n+1) + 2*sqrt(A(n)*A(n+1)),
  with A(n) = A240926(n), n >= 0. In this application the chord has curvature 0.
For the proof for the first formula for a(n) given below use the formula for the curvature A240926(n) = 2 + 2*S(n, 3) - 3* S(n-1, 3) (see the W. Lang link found in A240926, part II) in c(n) from Descartes' formula and compare it with a(n) from c(n) = A246638(n) + 4*a(n)*(1+sqrt(5))/2. This can be done by using standard S-polynomial identities like the three term recurrence and the Cassini-Simson type identity (see a comment on A246638) which implies the formula S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the above mentioned W. Lang link part III b).

			a(1) = 3 because c(1) = 0 +  5 + 9 + 2*sqrt(5*9) = 8 + 12*phi which is indeed 8 + 4*3*phi, with 8 = A246638(1).

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Descartes' Circle Theorem.
Wikipedia, Descartes' Theorem.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A246638, A240926, A049310, A001519.

I:=[2, 3, 6]; [n le 3 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{4,-4,1},{2, 3, 6}, 30] (* or *) CoefficientList[ Series[ (2-5*x+ 2*x^2)/ ((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

Vec((2-5*x+2*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

a(n) = 1 + S(n, 3) - S(n-1, 3) = 1 + A001519(n+1), n>=0, with Chebyshev's S-polynomials (see A049310), and S(-1, x) = 0.
O.g.f.: (2-5*x+2*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >= 1, with a(-2) = 3, a(-1) = 2 and a(0) = 2.
a(n) = 1+(2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/sqrt(5). - Colin Barker, Nov 02 2016

A249862 A special solution of X(n)^2 - 280Y(n)^2 = 3^(2n), n >= 0; here the X member.

Original entry on oeis.org

1, 17, 569, 19193, 647441, 21840257, 736741769, 24852657833, 838359690401, 28280555553137, 953993651593049, 32181259154185433, 1085576868377967281, 36619982192463218657, 1235309202728347728809, 41670933053031653811593, 1405693940978521100034881

Offset: 0

Wolfdieter Lang, Nov 07 2014

The member Y(n) = A248163(n-1) with A248163(-1) = 0.
This pair of sequences (X(n), Y(n)) appears in the solution of the touching circles and chord problem proposed by Kival Ngaokrajang in A249457. The curvatures (inverse radii) b(n) (for bend) of the circles in the lower section (on the left hand side) are here considered.
The derivation of the solution follows the lines given in the Wolfdieter Lang link in A240926, part I. Now the original radius of the large circle is R = 1 l.u. (length units) and the larger sagitta is 7/5 l.u. The circle radii are R(n), n >= 0, starting with R(0) = 7/10. Then a rescaling is done by a factor of 10/7 in the lengths: r = (10/7)*R = 10/7 l.u., such that the larger sagitta has 2 l.u. and r(n) = (10/7)*R(n).
The (nonlinear) recurrence for the curvature b(n) = 1/r(n) is written for bhat(n) := 3^n*b(n) and found to be: bhat(n) = 17*bhat(n-1) - 7*3^(n-1) + 140*sqrt((bhat(n-1) - 3^(n-1))*bhat(n-1)/(7*10)), n >= 1 with input bhat(0) = 1. This looks like A249457(n)/10. We search therefore for a positive integer solution which will then be the unique solution.
Define Y(n) := sqrt((bhat(n)-3^(n))*bhat(n)/(7*10)). This means (for positive bhat(n)) bhat(n) = (3^n + sqrt(9^n + 280*Y(n)^2))/2. Isolating the root and squaring yields X(n)^2 - 289*Y(n)^2 = 9^n, with X(n) := 2*bhat(n) - 3^n for n >= 0. For fixed n there are infinitely many solutions of this Diophantine equation. Here we only need a special solution for each n, which has to be an odd positive integer X(n) and X(0) = 1. Such a solution is X(n) = 3^(n-1)*(3*S(n, 34/3) - 17*S(n-1, 34/3)) and Y(n) = 3^(n-1)*S(n-1, 34/3) given in A249863(n-1), with Chebyshev's S-polynomial S(n, x). The proof is easy (once the o.g.f. for one of the sequences X or Y has been guessed by superseeker). Inserting the given formulas one has to prove S(n, 34/3)^2 + S(n-1, 34/3)^2 = 1 + (34/3)*S(n, 34/3)*S(n-1,34/3) which reduces after use of the recurrence relation for S to the well known Cassini-Simson identity S(n-1, x)^2 = 1 + S(n, x)*S(n-2, x), with S(-2) = -1, n >= 0, here with x = 34/3.
The solution for bhat(n) is then (X(n) + 3^n)/2. This satisfies indeed the original recurrence with input due to the recurrence relation of S(n, 34/3). Therefore, A249457(n)/10 = bhat(n).

G. C. Greubel, Table of n, a(n) for n = 0..650
Index entries for linear recurrences with constant coefficients, signature (34,-9).

Cf. A249457, A240926, A248163.

I:=[1,17]; [n le 2 select I[n] else 34*Self(n-1) - 9*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 08 2014

CoefficientList[Series[(1 - 17 x) / (1 - 34 x + 9 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 08 2014 *)
LinearRecurrence[{34,-9},{1,17},30] (* Harvey P. Dale, Dec 13 2016 *)

x='x+O('x^30); Vec((1 - 17*x)/(1 - 34*x + 9*x^2)) \\ G. C. Greubel, Dec 20 2017

a(n) = 3^(n-1)*(3*S(n, 34/3) - 17*S(n-1, 34/3)), n >= 0, with the scaled Chebyshev S sequence 3^n*S(n, 34/3) given in A248163.
O.g.f.: (1 - 17*x)/(1 - 34*x + 9*x^2).
a(n) = 34*a(n-1) - 9*a(n-2), a(0) = 0, a(1) = 17.
E.g.f.: exp(17*x)*cosh(2*sqrt(70)*x). - Stefano Spezia, Mar 24 2023

A249864 A special solution of X(n)^2 - 120Y(n)^2 = 7^(2n), n >= 0. The present sequence gives the X values.

Original entry on oeis.org

1, 13, 289, 6877, 164641, 3943693, 94468609, 2262942877, 54207552961, 1298512176013, 31105146481249, 745106711887837, 17848622331502561, 427553951736562573, 10241820250907001409, 245337182888490470557

Offset: 0

Wolfdieter Lang, Nov 09 2014

The member Y(n) = A249863(n-1) with A249863(-1) = 0.
This pair of sequences (X(n), Y(n)) appears in the solution of the touching circles and chord problem proposed by Kival Ngaokrajang in A249458. The curvatures (inverse radii) b(n) (for bend) of the circles in the smaller section (on the left hand side) are here considered.
The derivation of the solution follows the lines given in the Wolfdieter Lang link in A240926, part II. Now the original radius of the large circle is R = 1 l.u. (length units) and the larger sagitta h is 7/5 l.u. The circle radii are R(n), n >= 0, starting with R(0) = 3/10. Then a rescaling is done by r = (10/7)*R = 10/7 l.u., such that the larger sagitta has 2 l.u. and r(n) = (10/7)*R(n).
The (nonlinear) recurrence for the curvature b(n) = 1/r(n) is written for bhat(n) := 3*7^(n-1)*b(n) and found to be: bhat(n) = (1/7)*(91*bhat(n-1) - 3*7^n + 2*sqrt(210)*sqrt((7*bhat(n-1) - 7^n)*bhat(n-1))), n >= 1, with input bhat(0) = 1. This looks like A249458(n)/10. We search therefore for a positive integer solution which will then be the unique solution.
Define Y(n) := sqrt((bhat(n)-7^(n))*bhat(n)/(30)). This means (for positive bhat(n)) bhat(n) = (7^n + sqrt(7^(2*n) + 120*Y(n)^2))/2. Isolating the root and squaring yields X(n)^2 - 120*Y(n)^2 = 7^(2*n), with X(n) := 2*bhat(n) - 7^n for n >= 0. For fixed n there are infinitely many solutions of this Diophantine equation. Here we only need a special solution for each n, which has to be an odd positive integer X(n) and X(0) = 1. This solution is X(n) = 7^(n-1)*(7*S(n, 26/7) - 13*S(n-1, 26/7)) and Y(n) = 7^(n-1)*S(n-1, 26/7) given in A249863(n-1), with Chebyshev's S-polynomial S(n, x). The proof is easy (once the o.g.f. for one of the sequences X or Y has been guessed, e.g., by superseeker). Inserting the given formulas one has to prove S(n, 26/7)^2 + S(n-1, 26/7)^2 = 1 + (26/7)*S(n, 26/7)*S(n-1, 26/7) which reduces after use of the recurrence relation for S to the well known Cassini-Simson identity S(n-1, x)^2 = 1 + S(n, x)*S(n-2, x), with S(-2) = -1, n >= 0, here with x = 26/7.
The solution for bhat(n) is then (X(n) + 7^n)/2. This satisfies indeed the original recurrence with input due to the recurrence relation of S(n, 26/7). Therefore, A249458(n)/10 = bhat(n).

G. C. Greubel, Table of n, a(n) for n = 0..720
Index entries for sequences related to Chebyshev polynomials
Index entries for linear recurrences with constant coefficients, signature (26,-49)

Cf. A249457, A249458, A249863.

I:=[1,13]; [n le 2 select I[n] else 26*Self(n-1)-49*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 09 2014

f:= gfun:-rectoproc({a(n)=26*a(n-1)-49*a(n-2),a(0)=1,a(1)=13},a(n),remember):
seq(f(n),n=0..50); # Robert Israel, Nov 18 2014

LinearRecurrence[{26, -49}, {1, 13} , 30] (* or *) CoefficientList[ Series[(1 - 13*x)/(1 - 26*x + (7*x)^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

x='x+O('x^30); Vec((1 - 13*x)/(1 - 26*x + (7*x)^2)) \\ G. C. Greubel, Dec 20 2017

a(n) = (7^n)*(S(n, 26/7) - (13/7)*S(n-1, 26/7)), n >= 0, with the scaled Chebyshev S sequence 7^n*S(n, 26/7) given in A249863.
O.g.f.: (1 - 13*x)/(1 - 26*x + (7*x)^2).
a(n) = 26*a(n-1) - 49*a(n-2), a(0) = 1, a(1) = 13.
a(n) = (r^n + s^n)/2 where r,s are the roots of x^2 - 26*x + 49. - Robert Israel, Nov 18 2014

A246641 Sequence a(n) = (1 + A007805(n))/2, appearing in a certain touching problem for three circles and a chord, together with A007805.

Original entry on oeis.org

1, 9, 153, 2737, 49105, 881145, 15811497, 283725793, 5091252769, 91358824041, 1639367579961, 29417257615249, 527871269494513, 9472265593285977, 169972909409653065, 3050040103780469185, 54730748958638792257, 982103441151717791433, 17623131191772281453529, 316234258010749348372081, 5674593513001715989243921

Offset: 0

Wolfdieter Lang, Sep 05 2014

This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.
a(n), together with b(n) = A007805(n), appears in a sequence of curvatures c(n) = 4*(b(n) + a(n)*phi), with phi = (1+sqrt(5))/2, the golden section, and n >= 0. These are integers in the real quadratic number field Q(sqrt(5)).
The circle with curvature c(n) touches i) the chord of length 2 (in some length units) bisecting a circular disk of radius 5/4, and ii) two touching circles in the larger section with curvatures given by c1(n) and c1(n+1), where c1(n) = A115032(n-1), with c1(0) = 1. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the larger (lower) section are shown.
From Descartes' theorem on touching circles (see the links), one has here: c(n) = c1(n) + c1(n+1) + 2*sqrt(c1(n)*c1(n+1)), with c1(n) = (1 + S(n, 18) - 9*S(n-1, 18))/2, n >= 0, where Chebyshev's S-polynomials (see A049310) appear. See also the W. Lang link in A240926, part I. In this application curvature 0 for the chord is used.
For the proof for the first a(n) formula given below use the above given curvature c1(n) in Descartes' formula and compare it with a(n) from c(n) = 4*(A007805(n) + a(n)* (1+sqrt(5))/2). This can be done by using standard S-polynomial identities like the three term recurrence for S(n+1, 18) and the Cassini-Simson type identity (see a comment on A246638) which implies the formula S(n, 18)*S(n-1, 18) = (-1 + S(n, 18)^2 + S(n-1, 18)^2)/18. See also the W. Lang link in A240926 part IV a).
Also the indices of centered pentagonal numbers which are also centered square numbers. - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 4*x^2 - 5*y^2 - 4*x + 5*y = 0. - Colin Barker, Jan 01 2015

			a(1) = 9 because c(1) = 5 + 81 + 2*sqrt(5*81) = 68 + 36*phi, which is indeed 4*(17 + 9*phi), with 17 = A007805(1).

Colin Barker, Table of n, a(n) for n = 0..797
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Eric Weisstein's World of Mathematics, Descartes' Circle Theorem.
Wikipedia, Descartes' Theorem.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A007805, A049310, A246638, A240926, A115032.

I:=[1, 9, 153]; [n le 3 select I[n] else 19*Self(n-1) - 19*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{19,-19,1},{1, 9, 153}, 30] (* or *) CoefficientList[ Series[(1 - 10*x + x^2)/((1-x)*(1 - 18*x + x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

Vec((1-10*x+x^2)/((1-x)*(1-18*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

a(n) = (1 + S(n, 18) - S(n-1, 18))/2 = (1 + A007805(n))/2, n >= 0.
O.g.f.: (1 - 10*x + x^2)/((1-x)*(1 - 18*x + x^2)).
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3), n >= 1, with a(-2) = 9, a(-1) = 1 and a(0) = 1.
a(n) = (1/2+1/20*(9+4*sqrt(5))^(-n)*(5-2*sqrt(5)+(5+2*sqrt(5))*(9+4*sqrt(5))^(2*n))). - Colin Barker, Mar 04 2016

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

A338303 Decimal expansion of Sum_{k>=0} 1/(L(2*k) + 2), where L(k) is the k-th Lucas number (A000032).

Original entry on oeis.org

6, 4, 4, 5, 2, 1, 7, 8, 3, 0, 6, 7, 2, 7, 4, 4, 4, 2, 0, 9, 9, 2, 7, 3, 1, 1, 9, 0, 3, 8, 0, 1, 6, 9, 0, 2, 9, 2, 8, 9, 0, 8, 1, 2, 3, 8, 7, 7, 9, 9, 1, 8, 5, 7, 6, 5, 1, 4, 2, 5, 5, 2, 7, 5, 7, 7, 6, 8, 6, 8, 6, 1, 6, 8, 3, 6, 7, 8, 7, 4, 3, 3, 4, 1, 4, 0, 8

Offset: 0

Amiram Eldar, Oct 21 2020

Backstrom (1981) found that the sum is approximately equal to 1/8 + 1/(4*log(phi)), where phi is the golden ratio (A001622). The difference is less than 1/10^7. He called this difference "a tantalizing problem".
Almkvist (1986) added a term to Backstrom's formula to get an even better approximation which differs from the exact value by less than 1/10^33: 1/8 + 1/(4*log(phi)) + Pi^2/(log(phi)^2 * (exp(Pi^2/log(phi)) - 2)). He found an exact formula from a quotient of two Jacobi theta functions (see the FORMULA section), and showed that both approximations are just the first terms in a rapidly converging series.
Since exp(-Pi^2/log(phi)) = 1.23...*10^(-9) is small, the convergence is rapid: the number of terms needed in each of the two series in the formula to get 10^2, 10^3 and 10^4 decimal digits are merely 3, 10 and 32, respectively.
Andre-Jeannin (1991) noted that if the summand 2 that is added to the Lucas numbers is replaced with sqrt(5), then the sum is Sum_{k>=0} 1/(L(2*k) + sqrt(5)) = 1/phi (A094214).

			0.64452178306727444209927311903801690292890812387799...

Jonathan M. Borwein and Peter B. Borwein, Pi and the AGM, Wiley, 1987, p. 99.

Gert Almkvist, A solution to a tantalizing problem, The Fibonacci Quarterly, Vol. 24, No. 4 (1986), pp. 316-322.
Richard Andre-Jeannin, Summation of certain reciprocal series related to Fibonacci and Lucas numbers, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 200-204.
Robert P. Backstrom, On reciprocal series related to Fibonacci numbers with subscripts in arithmetic progression, The Fibonacci Quarterly, Vol. 19, No. 1 (1981), pp. 14-21. See section 6, pp. 19-20.
Daniel Duverney and Iekata Shiokawa, On series involving Fibonacci and Lucas numbers I, AIP Conference Proceedings, Vol. 976, No. 1 (2008), pp. 62-76, alternative link.

Cf. A000032, A001622, A002390, A094214, A160509, A240926.

With[{lg = Log[GoldenRatio], kmax = 3}, sum[m_, k_] := (-1)^k*k^m*Exp[-Pi^2*k^2/lg]; RealDigits[1/8 + ( 1/(4*lg))*(1 - (4*Pi^2/lg)*(Sum[sum[2, k], {k, 1, kmax}]/(1 + 2*Sum[sum[0, k], {k, 1, kmax}]))), 10, 100][[1]]]

Equals Sum_{k>=0} 1/A240926(k).
Equals 1/4 + Sum_{k>=1} q^(2*k)/(1 + q^(2*k))^2, where q = 1/phi.
Equals 1/8 + (1/(4*log(phi))) * (1 - (4*Pi^2/log(phi)) * (S(2)/(1 + 2*S(0)))), where S(m) = Sum_{k>=1} (-1)^k * k^m * exp(-Pi^2*k^2/log(phi)) (Almkvist, 1986).

cp's OEIS Frontend

A246640 Sequence a(n) = 1 + A001519(n+1) appearing in a certain touching problem for three circles and a chord, together with A246638.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A249862 A special solution of X(n)^2 - 280*Y(n)^2 = 3^(2*n), n >= 0; here the X member.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A249864 A special solution of X(n)^2 - 120*Y(n)^2 = 7^(2*n), n >= 0. The present sequence gives the X values.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A246641 Sequence a(n) = (1 + A007805(n))/2, appearing in a certain touching problem for three circles and a chord, together with A007805.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

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).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

A338303 Decimal expansion of Sum_{k>=0} 1/(L(2*k) + 2), where L(k) is the k-th Lucas number (A000032).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Formula

A249862 A special solution of X(n)^2 - 280Y(n)^2 = 3^(2n), n >= 0; here the X member.

A249864 A special solution of X(n)^2 - 120Y(n)^2 = 7^(2n), n >= 0. The present sequence gives the X values.