A240926 -id:A240926 - cp's OEIS frontend

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799

Offset: 0

N. J. A. Sloane

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) =  x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying  x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Marco Abrate, Stefano Barbero, Umberto Cerruti and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Kasper K. S. Andersen, Lisa Carbone and Diego Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011), Article 11.4.3, page 9.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 25.
Nathan D. Cahill, John R. D'Errico and John P. Spence, Complex Factorizations of the Fibonacci and Lucas Numbers, Fibonacci Quarterly, 1(41):13-19, 2003.
L. Carlitz, Problem B-110, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; An Infinite Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 469-470.
Paul Curtz, A269500, SeqFan list, March 2, 2016.
Murray Elder and Arkadius Kalka, Logspace computations for rigid Garside groups, arXiv preprint arXiv:1310.0933 [math.GR], 2013.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, Vol. 5 (2014), pp. 2226-2234.
Alex Fink, Richard K. Guy and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol. 3, No. 2 (2008), pp. 76-114. See Section 13.
Dale Gerdemann, Collision of Digits "Also interesting are the two bisections of the Lucas numbers A005248 (digit minimizer) and A002878 (digit maximizer). I particularly like the multiples of A005248 because I have this image of the two digits piling up on top of each other and then spreading out like waves".
André Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding Part 1 Part 2, Fib. Quart., Vol. 9, No. 3 (1971), pp. 277-295, 298.
Tian-Xiao He and Louis W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic., Vol. 532 (2017) pp. 25-41, example p 34.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Seong Ju Kim, Ryan Stees and Laura Taalman, Sequences of Spiral Knot Determinants, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.4.
Tanya Khovanova, Recursive Sequences.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum, Vol. 19 (2019), pp. 11-16.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics, Vol. 340, No. 2 (2017), pp. 160-174.
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
Serge Perrine, Some properties of the equation x^2=5y^2-4, The Fibonacci Quarterly, Vol. 54, No. 2 (2016) pp. 172-177.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Ryan Stees, Sequences of Spiral Knot Determinants, Senior Honors Projects, Paper 84, James Madison Univ., May 2016.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 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 (3,-1).

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019

a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012

[Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011

A002878 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,4]);
    else
        3*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)

a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011

for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015

a002878 = [1, 4]
for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
print(a002878) # Gennady Eremin, Feb 05 2022

[lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A115032 Expansion of (5-14x+x^2)/((1-x)(x^2-18*x+1)).

Original entry on oeis.org

5, 81, 1445, 25921, 465125, 8346321, 149768645, 2687489281, 48225038405, 865363202001, 15528312597605, 278644263554881, 5000068431390245, 89722587501469521, 1610006506595061125, 28890394531209630721, 518417095055178291845, 9302617316461999622481

Offset: 0

Creighton Dement, Feb 26 2006

Relates squares of numerators and denominators of continued fraction convergents to sqrt(5).
Sequence is generated by the floretion A*B*C with A = + 'i - 'k + i' - k' - 'jj' - 'ij' - 'ji' - 'jk' - 'kj' ; B = - 'i + 'j - i' + j' - 'kk' - 'ik' - 'jk' - 'ki' - 'kj' ; C = - 'j + 'k - j' + k' - 'ii' - 'ij' - 'ik' - 'ji' - 'ki' (apart from a factor (-1)^n)
Floretion Algebra Multiplication Program, FAMP Code: tesseq[A*B*C].
The sequence a(n-1), n >= 0, with a(-1) = 1, is also the curvature of circles inscribed in a special way in the larger segment of a circle of radius 5/4 (in some units) cut by a chord of length 2. For the smaller segment, the analogous curvature sequence is given in A240926. For more details see comments on A240926. See also the illustration in the present sequence, and the proof of the coincidence of the curvatures with a(n-1) in part I of the W. Lang link. - Kival Ngaokrajang, Aug 23 2014

			G.f. = 5 + 81*x + 1445*x^2 + 25921*x^3 + 465125*x^4 + 8346321*x^5 + ...

G. C. Greubel, Table of n, a(n) for n = 0..795
Creighton Dement, Floretions associated with A115032.
Wolfdieter Lang, A proof for the touching circle problem (part I).
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Kival Ngaokrajang, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001076, A001077, A007805, A023039, A097924.

Cf. also A000045, A049660, A049310, A023039. - Wolfdieter Lang, Aug 22 2014

seq((9*combinat:-fibonacci(6*(n+1)) - combinat:-fibonacci(6*n) + 8)/16, n = 0 .. 20); # Robert Israel, Aug 25 2014

LinearRecurrence[{19,-19,1},{5,81,1445},30] (* Harvey P. Dale, Nov 14 2014 *)
CoefficientList[Series[(5 - 14*x + x^2)/((1 - x)*(x^2 - 18*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

Vec((5-14*x+x^2)/((1-x)*(x^2-18*x+1)) + O(x^20)) \\ Michel Marcus, Aug 23 2014

sqrt(a(2*n)) = sqrt(5)*A007805(n) = sqrt(5)*Fibonacci(6*n+3)/2 = sqrt(5)*A001076(2*n+1); sqrt(a(2*n+1)) = A023039(2*n+1) = A001077(2*n).
From Wolfdieter Lang, Aug 22 2014: (Start)
O.g.f.: (5-14*x+x^2)/((1-x)*(x^2-18*x+1)) (see the name).
a(n) = (9*F(6*(n+1)) - F(6*n) + 8)/16, n >= 0 with F(n) = A000045(n) (Fibonacci). From the partial fraction decomposition of the o.g.f.: (1/2)*((9 - x)/(1 - 18*x + x^2) + 1/(1 - x)). For F(6*n)/8 see A049660(n). a(-1) = 1 with F(-6) = -F(6) = -8.
a(n) = (9*S(n, 18) - S(n-1, 18) + 1)/2, with the Chebyshev S-polynomials (see A049310). From A049660.
a(n) = (A023039(n+1) + 1)/2.
(End)
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3). - Colin Barker, Aug 23 2014
From Wolfdieter Lang, Aug 24 2014: (Start)
a(n) = 18*a(n-1) - a(n-2) - 8, n >= 1, a(-1) = 1, a(0) = 5. See the Chebyshev S-polynomial formula above.
The o.g.f. for the sequence a(n-1) with a(-1) = 1, n >= 0, [1, 5,  81, 1445, ..] is (1-14*x+5*x^2)/((1-x)*(1-18*x+x^2)).
(See the Colin Barker formula from Aug 04 2014 in the history of A240926.) (End)

More terms from Michel Marcus, Aug 23 2014
Edited (comment by Kival Ngaokrajang rewritten, Chebyshev index link added) by Wolfdieter Lang, Aug 26 2014
Partially edited by Jon E. Schoenfield and N. J. A. Sloane, Mar 15 2024

A247512 The curvature (rounded down) of touching circles inscribed in a special way in the smaller segment of circle of radius 10/9 divided by a chord of length 4/3.

Original entry on oeis.org

9, 10, 13, 20, 35, 64, 119, 224, 428, 821, 1576, 3030, 5828, 11215, 21584, 41545, 79968, 153931, 296306, 570371, 1097933, 2113463, 4068308, 7831289, 15074840, 29018319, 55858826, 107525476, 206981225, 398428629, 766955420, 1476351286, 2841903278, 5470523390

Offset: 0

Kival Ngaokrajang, Sep 18 2014

Refer to comment of A240926. This is the companion of A247335. After the first two terms, the curvatures seem to be non-integer.

The actual rational curvatures can be computed. See part II of the W. Lang link for the proofs of the statements given in the formula section.

			The first curvatures r(n) are 9, 10, 121/9, 1690/81, 25921/729, 420250/6561, 7027801/59049, 119508490/531441,... - _Wolfdieter Lang_, Sep 30 2014

G. C. Greubel, Table of n, a(n) for n = 0..1000
Kival Ngaokrajang, Illustration of initial terms
Wolfdieter Lang, Curvature computation for A247335 and A247512.
Index entries for sequences related to Chebyshev polynomials.

Cf. A240926, A247335.

Cf. A246643, A049310, A246645. - Wolfdieter Lang, Sep 30 2014

r[0] := 9; r[n_] := r[n] = (11*r[n - 1] - 9 + 20*Sqrt[(r[n - 1] - 9)*r[n - 1]/10])/9; Table[Floor[r[n]], {n, 0, 30}] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.1;print1(floor(9/(10*r)),", ");r1=r;
for (n=1,50,
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(floor(9/(10*r)),", ")
)
}

From Wolfdieter Lang, Sep 30 2014 (Start)
a(n) = floor(r(n)) with the rational curvatures r(n) satisfying the one step nonlinear recurrence relation r(n) = (11*r(n-1) - 9 + 20*sqrt((r(n-1) - 9)*r(n-1)/10))/9 with input r(0) = 9. (In the link r(n) is called b'(n).)
r(n) =  A246643(n)/9^(n-1)  = (9/2)*(1 + S(n, 22/9) - (11/9)*S(n-1, 22/9)), n >= 0, with Chebyshev/s S-polynomials (see A049310). 9^n*S(n, 22/9) = A246645(n). See A246643 for more details. (End)

Edited: Keyword easy and Chebyshev index link added. Wolfdieter Lang, Sep 30 2014

A247335 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 10/9 divided by a chord of length 4/3.

Original entry on oeis.org

1, 10, 361, 13690, 519841, 19740250, 749609641, 28465426090, 1080936581761, 41047124680810, 1558709801289001, 59189925324301210, 2247658452522156961, 85351831270517663290, 3241121929827149048041, 123077281502161146162250

Offset: 0

Kival Ngaokrajang, Sep 18 2014

Refer to comment of A240926. Consider a circle C of radius 10/9 (in some length units) with a chord of length 4/3. This has been chosen such that the larger sagitta has length 2. The smaller sagitta has length 2/9. The input, besides the circle C is the circle C_0 with radius R_0 = 1, 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, the sequence would be A247512. See an illustration given in the link.
a(n) also seems to be A078986(i)^2 and 10*A097315(j)^2 interleaved; where i = n/2 for n even, j = n/2 - 1/2 for n odd; as following:
   1          = 1^2
   10         = 10*1^2
   361        = 19^2
   13690      = 10*37^2
   519841     = 721^2
   19740250   = 10*1405^2
   749609641  = 27379^2
   ...
A078986; Chebyshev... polynomial: 1, 19, 721, 27379, ...
A097315; Pell equation...       : 1, 37, 1405, 53353, ...

Colin Barker, Table of n, a(n) for n = 0..600
Kival Ngaokrajang, Illustration of initial terms
Wolfdieter Lang, Curvature computation for A247335 and A247512.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (39,-39,1).

Cf. A240926, A078986, A097315, A247512.

I:=[39,-39,1]; [n le 3 select I[n] else Self(n-1) - 10*Self(n-2) + 361*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{39,-39,1}, {1, 10, 361}, 50] (* or *) Table[Round[((19 + 6*Sqrt[10])^(-n)*(1 + (19 + 6*Sqrt[10])^n)^2)]/4, {n, 0, 30}] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.9;print1(1,", ");r1=r;
for (n=1,50,
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));
an=floor(9/(10*r));
print1(if(an>9,an,10),", ")
)
}

Vec(-(10*x^2-29*x+1)/((x-1)*(x^2-38*x+1)) + O(x^20)) \\ Colin Barker, Mar 03 2016

Conjectures from Colin Barker, Sep 18 2014: (Start)
a(n) = 39*a(n-1) - 39*a(n-2) + a(n-3).
G.f.: -(10*x^2-29*x+1) / ((x-1)*(x^2-38*x+1)). (End)
From Wolfdieter Lang, Sep 30 2014 (Start)
See the W. Lang link for proofs of the following statements.
One step nonlinear recurrence: a(n) = -9 + 19*a(n-1) + 60*sqrt(a(n-1)*(a(n-1) - 1)/10), n>=1, with a(0) = 1.
a(n) =  (1 + A078986(n))/2  = (2 + S(n, 38) - S(n-2, 38))/4 =
  (1 + S(n, 38) -19*S(n-1, 38))/2 for n>=0, with Chebyshev's S-polynomials (see A049310).  S(n, 38) = A078987(n).
The G.f. conjectured by Colin Barker above follows from the one for Chebyshev's T(n, 19) = A078986(n):  (1/(1-x) + (1-19*x)/(1-38*x+x^2))/2 = (1-29*x+10*x^2)/((1-x)* (1-38*x+x^2)).
The four term recurrence conjectured by Colin Barker above follows from the expanded g.f. denominator: (1-x)* (1-38*x+x^2) = 1- 39*x + 39*x^2 - x^3.
(End)
a(n) = ((19+6*sqrt(10))^(-n)*(1+(19+6*sqrt(10))^n)^2)/4. - Colin Barker, Mar 03 2016

A246638 Sequence a(n) = 2 + 3*A001519(n+1) appearing in a certain four circle touching problem together with A246639.

Original entry on oeis.org

5, 8, 17, 41, 104, 269, 701, 1832, 4793, 12545, 32840, 85973, 225077, 589256, 1542689, 4038809, 10573736, 27682397, 72473453, 189737960, 496740425, 1300483313, 3404709512, 8913645221, 23336226149, 61095033224, 159948873521, 418751587337, 1096305888488, 2870166078125, 7514192345885

Offset: 0

Wolfdieter Lang, Aug 31 2014

This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.
a(n), together with b(n) = A246639(n) appears as the curvature c(n) =  a(n) + (4*b(n)/5)*phi (phi = (1+sqrt(5))/2, golden section) of the circle which touches i) a circle of radius 5/4 (in some length units) divided by a chord of length 2 into two unequal parts, and ii) the two touching circles in the smaller part which have curvatures A240926(n) and  A240926(n+1), both also touching the circle with radius 5/4. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the smaller (upper) part are shown. The present circles will lie in the region between the large circle and two of these circles in the upper part.
Descartes' theorem on touching circles (see the links) is applied here as  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 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 in A240926, part II) in c(n) and compare with a(n) from c(n) =  a(n) + (4*b(n)/5)*phi. This is done by using standard S-polynomial identities 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 mentioned link, part III a).
a(n) appears also in the curvature for the touching circles and chord problem in the smaller part of a circle with radius 5/4 dissected by a chord of length 2, together with A246640, where details are given.

			a(1) = 8 because c(n) = -4/5 + 5 + 9 + 2*sqrt((-4/5 )*(5 + 9) + 5*9) = 4*(2+(13/5)*phi). This is also 8 + (4*13/5)*phi with A246639(1) = 13.

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. A246639, A049310, A001519, A115032.

I:=[5,8,17]; [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}, {5, 8, 17}, 30] (* or *)  CoefficientList[ Series[(5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

Vec((5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

a(n) = 2 + 3*(S(n,3) - S(n-1,3)) = 2 + 3*A001519(n+1), n >= 0, with Chebyshev's S-polynomials (see A049310).
O.g.f.: (5-12*x+5*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) = 8, a(-1) = 5, a(0) = 5.
a(n) = (2^(-1-n)*(5*2^(2+n)-3*(3-sqrt(5))^n*(-5+sqrt(5))+3*(3+sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Nov 02 2016

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

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

Wolfdieter Lang, Sep 05 2014

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

			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).

Colin Barker, Table of n, a(n) for n = 0..796
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, A115032, A240926.

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

 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 *)

Vec((1+22*x+x^2)/((1-x)*(1-18*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 08 2015

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

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

cp's OEIS Frontend

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

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A115032 Expansion of (5-14*x+x^2)/((1-x)*(x^2-18*x+1)).

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A247512 The curvature (rounded down) of touching circles inscribed in a special way in the smaller segment of circle of radius 10/9 divided by a chord of length 4/3.

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A247335 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 10/9 divided by a chord of length 4/3.

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A246638 Sequence a(n) = 2 + 3*A001519(n+1) appearing in a certain four circle touching problem together with A246639.

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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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A115032 Expansion of (5-14x+x^2)/((1-x)(x^2-18*x+1)).