cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A246639 Sequence a(n) = 3 + 5*A001519(n+1) appearing in a certain three circle touching problem, together with A246638.

Original entry on oeis.org

8, 13, 28, 68, 173, 448, 1168, 3053, 7988, 20908, 54733, 143288, 375128, 982093, 2571148, 6731348, 17622893, 46137328, 120789088, 316229933, 827900708, 2167472188, 5674515853, 14856075368, 38893710248, 101825055373, 266581455868, 697919312228, 1827176480813, 4783610130208, 12523653909808
Offset: 0

Views

Author

Wolfdieter Lang, Sep 01 2014

Keywords

Comments

See the comments under A246638. The curvature c(n) defined there is c(n) = A246638(n) + (4*a(n)/5)*phi with phi = (1+sqrt(5))/2, the golden section. It lives in the quadratic number field Q(sqrt(5)). Descartes' theorem on touching circles gives c(n) = -4/5 + A(n) + A(n+1) + 2*sqrt((-4/5 )*(A(n) + A(n+1)) + A(n)*A(n+1)), with A(n) = A240926(n), n >= 0. For the proof of the first formula given below one compares this a(n) with the a(n) in c(n) given above. This uses standard Chebyshev S-polynomial identities with x = 3, like the three term recurrence and the Cassini-Simson type identity S(n, x)*S(n-2, x) = -1 + S(n-1, x)^2 (here for x=3). This implies S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the W. Lang link in A240926, part III a).

Links

Crossrefs

Cf. A246638, A049310, A001519, A115032.

Programs

  • Magma
    I:=[8, 13, 28]; [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
  • Mathematica
    CoefficientList[Series[(8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* or *) LinearRecurrence[{4,-4,1}, {8, 13, 28}, 30] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    Vec((8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

Formula

a(n) = 3 + 5*(S(n, 3) - S(n-1, 3)) = 3 + 5*A001519(n+1), n >= 0, with Chebyshev S-polynomials (see A049310) with S(-1, x) = 0.
O.g.f.: (8-19*x+8*x^2)/((1-x)*(1-3*x+x^2)).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >=1, a(-2) = 13, a(-1) = 8, a(0) = 8.
a(n) = 2^(-1-n)*(3*2^(1+n)-(3-sqrt(5))^n*(-5+sqrt(5))+(3+sqrt(5))^n*(5+sqrt(5))). - Colin Barker, Nov 02 2016

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

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Sep 03 2014

Keywords

Comments

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).

Examples

			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).

Links

Crossrefs

Cf. A246638, A240926, A049310, A001519.

Programs

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

Formula

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

A240926 a(n) = 2 + L(2*n) = 2 + A005248(n), n >= 0, with the Lucas numbers (A000032).

Original entry on oeis.org

4, 5, 9, 20, 49, 125, 324, 845, 2209, 5780, 15129, 39605, 103684, 271445, 710649, 1860500, 4870849, 12752045, 33385284, 87403805, 228826129, 599074580, 1568397609, 4106118245, 10749957124, 28143753125, 73681302249, 192900153620, 505019158609
Offset: 0

Views

Author

Kival Ngaokrajang, Aug 03 2014

Keywords

Comments

This sequence also gives the curvature of touching circles inscribed in a special way in the smaller segment of a circle of radius 5/4 cut by a chord of length 2.
Consider a circle C of radius 5/4 (in some length units) with a chord of length 2. This has been chosen so that the larger sagitta also has length 2. The smaller sagitta has length 1/2. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/4, 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). See an illustration given in the link. As found by Wolfdieter Lang (see part II of the proof given by W. Lang in the link), this circle problem is related to the nonnegative solutions of the Pell equation X^2 - 5*Y^2 = 4: a(n) = 2 + X(n) = 2 + A005248(n). For the larger segment below the chord (with sagitta length 2) the sequence would be A115032, see W. Lang's proof given in part I of the link.
If the circle radius and the sagitta length were both equal to 1, the curvature sequence would be A099938.
Essentially a duplicate of A092387. - R. J. Mathar, Jul 07 2023

Links

Crossrefs

Cf. A000032, A005248, A005592, A099938, A115032, A246638, A246639, A246640, A246641, A246642, A338303.

Programs

  • Magma
    [2+Lucas(2*n): n in [0..40]]; // Vincenzo Librandi, Oct 08 2015
  • Mathematica
    Table[2 + LucasL[2 n], {n, 0, 50}] (* Vincenzo Librandi, Oct 08 2015 *)
  • PARI
    vector(100, n, n--; 2 + fibonacci(2*n-1) + fibonacci(2*n+1)) \\ Altug Alkan, Oct 08 2015

Formula

Conjectures (proved in the next entry) from Colin Barker, Aug 25 2014 (and Aug 27 2014): (Start)
a(n) = (2 + ((1/2)*(3-sqrt(5)))^n + ((1/2)*(3+sqrt(5)))^n).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: -(5*x^2-11*x+4) / ((x-1)*(x^2-3*x+1)). (End)
From Wolfdieter Lang, Aug 26 2014: (Start)
a(n) = 2 + S(n, 3) - S(n-2, 3) = 2 + 2*S(n, 3) - 3*S(n-1, 3).
a(n) = 3*a(n-1) - a(n-2) - 2, n >= 1, with a(-1)= 5 and a(0) = 4 (from the S(n, 3) recurrence or from A005248).
The first of the Colin Barker conjectures above is true because of the Binet-de Moivre formula for L(2*n) (see the Jul 24 2003 Dennis P. Walsh comment on A005248). With phi = (1+sqrt(5))/2, use 1/phi = phi-1, phi^2 = phi+1, (phi-1)^2 = 2 - phi.
His third conjecture (the g.f.) follows from the g.f. of A005248 by adding 2/(1-x).
His second conjecture (recurrence) with input a(-3) = 20, a(-2) = 9 and a(-1) = 5 (from the above given recurrence) leads to his g.f. with the expanded denominator. Thus all three conjectures are true. (End)
a(n) = A005592(n) + 3, with n > 0. - Zino Magri, Feb 16 2015
a(n) = (phi^n + phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Jun 10 2020
Sum_{k>=0} 1/a(k) = A338303. - Amiram Eldar, Oct 22 2020

Extensions

Edited: name changed (after proof has been given in part II of the W. Lang link), comments rewritten, cross refs. and link to Chebyshev index added. - Wolfdieter Lang, Aug 26 2014

A246642 Sequence appearing in the curvature of a certain four-circle touching problem: (-3 + 5*A007805(n))/2.

Original entry on oeis.org

1, 41, 761, 13681, 245521, 4405721, 79057481, 1418628961, 25456263841, 456794120201, 8196837899801, 147086288076241, 2639356347472561, 47361327966429881, 849864547048265321, 15250200518902345921, 273653744793193961281, 4910517205758588957161, 88115655958861407267641
Offset: 0

Views

Author

Wolfdieter Lang, Sep 05 2014

Keywords

Comments

This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.
a(n) appears in a curvature c(n) = (4/5)*(2*(a(n) + 2) + a(n)*phi), with phi = (1+sqrt(5))/2, the golden section. c(n) is the curvature of the circle which touches (i) the larger part of a circle of radius 5/4 (in some length units), obtained from the bisection of the circle with a chord of length 2 and (ii) two touching circles in the larger part of this bisected disk of radius 5/4 having curvatures c1(n) and c1(n+1) with c1(n) = A115032(n-1) and c1(0) = 1, n >= 0. (See the illustration of Kival Ngaokrajang's link given in A115032, where the first circles in the larger (lower) part are shown.)
From Descartes's theorem on touching circles (see the links) one has here: c(n) = -4/5 + c1(n) + c1(n+1) + 2*sqrt((-4/5)*( c1(n) + c1(n+1)) + 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.
For the proof for the first a(n) formula given below use the curvature c1(n) = (1 + S(n, 18) - 9*S(n-1, 18))/2 (see the W. Lang link in A240926, part I) in c(n) from Descartes's formula and compare it with a(n) in c(n) = (4/5)*(2*(a(n) + 2) + 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 mentioned W. Lang link part IV b).
Also the first of four consecutive positive integers the sum of the squares of which is equal to the sum of the squares of five consecutive positive integers. For example 41^2 + ... + 44^2 = 7230 = 36^2 + ... + 40^2. - Colin Barker, Sep 08 2015

Examples

			a(1) = 41 because the two curvatures of the circles in the larger part are c1(1) = 5 and c1(2) = 81 (from A115032), and c(1) = -4/5 + 5 + 81 + 2*sqrt((-4/5)*(5 + 81) + 5*81) = (4/5)*(213 + 41*sqrt(5))/2 = (4/5)*(86 + 41*phi) (by Descartes). This is indeed (4/5)*(2*(a(1) + 2) + a(1)*phi).

Links

Crossrefs

Cf. A007805, A049310, A115032, A240926.

Programs

  • Magma
    I:=[1, 41, 761]; [n le 5 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
  • Mathematica
     CoefficientList[Series[(1+22*x+x^2)/((1-x)*(1-18*x+x^2)), {x, 0, 50}], x] (* or *) LinearRecurrence[{19,-19,1}, {1,41,761}, 30] (* G. C. Greubel, Dec 20 2017 *)
  • PARI
    Vec((1+22*x+x^2)/((1-x)*(1-18*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 08 2015

Formula

a(n) = (-3 + 5*(S(n, 18) - S(n-1, 18)))/2 = (-3 + 5* A007805(n))/2, n >= 0, with Chebyshev's S-polynomials (see A049310).
O.g.f.: (1+22*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)=41, a(-1)=1 and a(0)=1.
a(n) = (-6+(5-2*sqrt(5))*(9+4*sqrt(5))^(-n) + (5+2*sqrt(5))*(9+4*sqrt(5))^n)/4. - Colin Barker, Mar 03 2016

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

Views

Author

Wolfdieter Lang, Sep 05 2014

Keywords

Comments

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

Examples

			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).

Links

Crossrefs

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

Programs

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

Formula

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
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