A032908 -id:A032908 - cp's OEIS frontend

A104457 Decimal expansion of 1 + phi = phi^2 = (3 + sqrt(5))/2.

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

Offset: 1

Eric W. Weisstein, Mar 08 2005

Only first term differs from the decimal expansion of phi.
Zelo extends work of D. Roy by showing that the square of the golden ratio is the optimal exponent of approximation by algebraic numbers of degree 4 with bounded denominator and trace. - Jonathan Vos Post, Mar 02 2009 (Cf. last sentence in the Zelo reference. - Joerg Arndt, Jan 04 2014)
Hawkes asks: "What two numbers are those whose product, difference of their squares, and the ratio or quotient of their cubes, are all equal to each other?". - Charles R Greathouse IV, Dec 11 2012
This is the case n=10 in (Gamma(1/n)/Gamma(3/n))*(Gamma((n-1)/n)/Gamma((n-3)/n)) = 1+2*cos(2*Pi/n). - Bruno Berselli, Dec 14 2012
An algebraic integer of degree 2, with minimal polynomial x^2 - 3x + 1. - Charles R Greathouse IV, Nov 12 2014 [The other root is 2 - phi = A132338 - Wolfdieter Lang, Aug 29 2022]
To eight digits: 5*(((Pi+1)/e)-1) = 2.61803395481182... - Dan Graham, Nov 21 2017
The ratio diagonal/side of the second smallest diagonal in a regular 10-gon. - Mohammed Yaseen, Nov 04 2020
phi^2/10 is the moment of inertia of a solid regular icosahedron with a unit mass and a unit edge length (see A341906). - Amiram Eldar, Jun 08 2021

			2.6180339887498948482045868343656381177203091798...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.17.1, p. 153.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 138-139.
Damien Roy. Diophantine Approximation in Small Degree. Centre de Recherches Mathématiques. CRM Proceedings and Lecture Notes. Volume 36 (2004), 269-285.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 45.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Murray Berg, Phi, the golden ratio (to 4599 decimal places) and Fibonacci numbers, Fibonacci Quarterly, Vol. 4, No. 2 (1966), pp. 157-162.
John Hawkes et al., Question 1029, The Mathematical Questions Proposed in the Ladies' Diary (1817), p. 339. Originally published 1798 and answered in 1799.
Casey Mongoven, Phi^2 number 1; electronic music created using Phi^2.
Hideyuki Ohtsuka, Problem B-1237, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 56, No. 4 (2018), p. 366; A Telescoping Product, Solution to Problem B-1237 by Steve Edwards, ibid., Vol. 57, No. 4 (2019), pp. 369-370.
Damien Roy, Diophantine Approximation in Small Degree, arXiv:math/0303150 [math.NT], 2003.
Eric Weisstein's World of Mathematics, Chromatic Polynomial.
Eric Weisstein's World of Mathematics, Fibonacci Hyperbolic Functions.
Wikipedia, Perron number.
Dmitrij Zelo, Simultaneous Approximation to Real and p-adic Numbers, arXiv:0903.0086 [math.NT], 2009.
Index entries for algebraic numbers, degree 2.
Index entries for sequences related to moment of inertia.

Cf. A001622, A094214, A094885, A132338, A229780, A239798, A341906.
Cf. A001519, A001906, A032908, A049310.
2 + 2*cos(2*Pi/n): A116425 (n = 7), A332438 (n = 9), A296184 (n = 10), A019973 (n = 12).

SetDefaultRealField(RealField(100)); (1+Sqrt(5))^2/4; // G. C. Greubel, Nov 23 2018

RealDigits[N[GoldenRatio+1,200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2011 *)

(3+sqrt(5))/2 \\ Charles R Greathouse IV, Aug 21 2012

numerical_approx(golden_ratio^2, digits=100) # G. C. Greubel, Nov 23 2018

Equals 2 + A094214 = 1 + A001622. - R. J. Mathar, May 19 2008
Satisfies these three equations: x-sqrt(x)-1 = 0; x-1/sqrt(x)-2 = 0; x^2-3*x+1 = 0. - Richard R. Forberg, Oct 11 2014
Equals the nested radical sqrt(phi^2+sqrt(phi^4+sqrt(phi^8+...))). For a proof, see A094885. - Stanislav Sykora, May 24 2016
From Christian Katzmann, Mar 19 2018: (Start)
Equals Sum_{n>=0} (5*(2*n)!+8*n!^2)/(2*n!^2*3^(2*n+1)).
Equals 3/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
Equals 1/A132338 = 2*A239798 = 5*A229780. - Mohammed Yaseen, Nov 04 2020
Equals Product_{k>=1} 1 + 1/(phi + phi^k), where phi is the golden ratio (A001622) (Ohtsuka, 2018). - Amiram Eldar, Dec 02 2021
c^n = phi * A001906(n) + A001519(n), where c = phi^2. - Gary W. Adamson, Sep 08 2023
Equals lim_{n->oo} S(n, 3)/S(n-1, 3) with the S-Chebyshev polynomials (see A049310), S(3, n) = A000045(2*(n+1)) = A001906(n+1). - Wolfdieter Lang, Nov 15 2023
From Peter Bala, May 08 2024: (Start)
Constant c = 2 + 2*cos(2*Pi/5).
The linear fractional transformation z -> c - c/z has order 5, that is, z = c - c/(c - c/(c - c/(c - c/(c - c/z)))). (End)
Equals Product_{k>=1} (1 + 1/A032908(k)). - Amiram Eldar, Nov 28 2024

A101265 a(1) = 1, a(2) = 2, a(3) = 6; a(n) = 5a(n-1) - 5a(n-2) + a(n-3) for n > 3.

Original entry on oeis.org

1, 2, 6, 21, 77, 286, 1066, 3977, 14841, 55386, 206702, 771421, 2878981, 10744502, 40099026, 149651601, 558507377, 2084377906, 7779004246, 29031639077, 108347552061, 404358569166, 1509086724602, 5631988329241, 21018866592361, 78443478040202, 292755045568446

Offset: 1

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 25 2005

Let M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1 ]; then [1,0,0]*M^n = [a(n), A001353(n), A061278(n-1)] for n > 1. Further, A001353 consists of the first differences of {a(n)}, and since a(n) = A061278(n) + 1, A001353 is also the first differences of A061278. Let v(n) = [1,0,0]*M^n; then, for n >= 0, sum(v_i(n)) = A001075(n) and v_1(n) + v_3(n) = A001835(n). The characteristic polynomial of M is x^3 - 5x^2 + 5x - 1. a(n)/a(n-1) tends to 2 + sqrt(3) = 3.732.... (see A019973) (a root of the polynomial and an eigenvalue of the matrix).
Numbers k such that the RootMeanSquare([1..6*k-5]) is an integer. - Ctibor O. Zizka, Dec 17 2008
Place a(n) blue and b(n) red balls in an urn. Draw 3 balls without replacement. Then Probability(3 red balls) = Probability(1 red and 2 blue balls); binomial(b(n),3) = binomial(b(n),1)*binomial(a(n),2); b(n) = A179167(n). - Paul Weisenhorn, Jul 01 2010
Conjecture: consecutive terms of this sequence and consecutive terms of A032908 provide all the positive integer solutions of (a+b)*(a+b+1) == 0 (mod (a*b)). - Robert Israel, Aug 26 2015
Conjecture is true: see StackExchange link. - Robert Israel, Sep 06 2015
Values of a unitary Y-frieze pattern associated to the linearly oriented quiver K3 (i.e., the quiver whose underlying graph is the complete graph on the vertices {1,2,3}, oriented such that i -> j whenever i < j). - Antoine de Saint Germain, Dec 30 2024

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
R. Israel, W. Jagy et al., Diophantine equation (x+y)(x+y+1)-kxy=0, Math StackExchange, Sep 1 2015.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001075, A001353, A001835, A005246, A019973, A032908, A061278, A211955.

a:=[1,2,6];; for n in [4..20] do a[n]:=5a[n-1]-5*a[n-2]+a[n-3]; od; a; #
G. C. Greubel, Dec 23 2019

a101265 n = a101265_list !! (n-1)
a101265_list = 1 : 2 : 6 : zipWith (+) a101265_list
    (map (* 5) $ tail $ zipWith (-) (tail a101265_list) a101265_list)
-- Reinhard Zumkeller, May 18 2014

I:=[1,2,6]; [n le 3 select I[n] else 5*Self(n-1) - 5*Self(n-2) + Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 07 2015

r:=sqrt(3): for n from 1 to 100 do a[n]:=(6+(3+r)*(2+r)^(n-1)+(3-r)*(2-r)^(n-1))/12: end do: # Paul Weisenhorn, Jul 01 2010
r:=sqrt(3): a[n]:=round((6+(3+r)*(2+r)^(n-1))/12): # Paul Weisenhorn, Jul 01 2010
f:= proc(n)
  option remember; local x;
  x:= procname(n-1);
  2*x + (sqrt(12*x^2 - 12*x + 1) - 1)/2
end proc:
f(1):= 1:
map(f, [$1..30]); # Robert Israel, Aug 26 2015
seq( simplify((ChebyshevU(n,2) - Chebyshev(n-1,2) + 1)/2), n=0..20); # G. C. Greubel, Dec 23 2019

LinearRecurrence[{5,-5,1},{1,2,6},25] (* Ray Chandler, Jan 27 2014 *)
CoefficientList[Series[(1-3x+x^2)/((1-x)(1-4x+x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 07 2015 *)
Table[(ChebyshevU[n, 2] - ChebyshevU[n-1, 2] + 1)/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)

M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1,30,print1(([1,0,0]*M^i)[1],","))

{a(n)=polcoeff(x*(1-3*x+x^2)/((1-x)*(1-4*x+x^2)+x*O(x^n)),n)}

{a(n)=if(n==0,1,if(n==1,1,a(n-1)*(a(n-1)+1)/a(n-2)))} /* Paul D. Hanna, Apr 08 2012 */

vector(21, n, (polchebyshev(n, 2, 2) - polchebyshev(n-1, 2, 2) + 1)/2 ) \\ G. C. Greubel, Dec 23 2019

[(chebyshev_U(n,2) - chebyshev_U(n-1,2) + 1)/2 for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = A005246(n)*A005246(n+1). a(n+1) = a(n)*(a(n)+1)/a(n-1). - Franklin T. Adams-Watters, Apr 24 2006
a(n) = (A001835(n) + 1) / 2. - Ralf Stephan, May 16 2007
O.g.f.: x*(1-3*x+x^2)/((1-x)*(1-4*x+x^2)). - R. J. Mathar, Aug 22 2008
a(n) = 1 + A061278(n). - Ctibor O. Zizka, Dec 17 2008
a(n) = 4*a(n-1) - a(n-2) - 1. - N. Sato, Jan 21 2010
a(n) = (6+(3+r)*(2+r)^(n-1) + (3-r)*(2-r)^(n-1))/12; r=sqrt(3). - Paul Weisenhorn, Jul 01 2010
a(n+1) = a(n) * (a(n) + 1) / a(n-1) for n>1 with a(0)=1, a(1)=1. - Paul D. Hanna, Apr 08 2012
From Peter Bala, May 01 2012: (Start)
a(n+1) = 1 + Sum {k = 1..n} 2^(k-1)*binomial(n+k,2*k).
Row sums of A211955.
a(n) = T(n,u)*T(n+1,u)/u with u = sqrt(3) and T(n,x) denotes the Chebyshev polynomial of the first kind.
Sum_{n >= 0} 1/a(n) = sqrt(3). In fact, 3 - (Sum_{n = 0..2*N} 1/a(n))^2 = 2/(A001835(N+1))^2 and 3 - (Sum_{n = 0..2*N+1} 1/a(n))^2 = 3/(A001075(N+1))^2. (End)
From Robert Israel, Aug 26 2015: (Start)
(a(n) + a(n+1))*(a(n) + a(n+1) + 1) = 6 * a(n) * a(n+1).
a(n+1) = 2*a(n) + (sqrt(12*a(n)^2 - 12*a(n) + 1) - 1)/2. (End)
a(n) = (ChebyshevU(n, 2) - ChebyshevU(n-1, 2) + 1)/2 = (ChebyshevT(n, 2) + ChebyshevU(n, 2) + 2)/4. - G. C. Greubel, Dec 23 2019
a(n) = (1+a(n-1))*(1+a(n-2))/a(n-3) for n > 3. - Antoine de Saint Germain, Dec 30 2024

a(26)-a(27) from Vincenzo Librandi, Sep 07 2015

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).

Original entry on oeis.org

1, 1, 2, 6, 21, 77, 286, 1066, 3977, 14841, 55386, 206702, 771421, 2878981, 10744502, 40099026, 149651601, 558507377, 2084377906, 7779004246, 29031639077, 108347552061, 404358569166, 1509086724602, 5631988329241, 21018866592361, 78443478040202, 292755045568446

Offset: 0

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 28 2005

Consider the matrix M=[1,1,0; 1,3,1; 0,1,1]; characteristic polynomial of M is x^3 - 5*x^2 + 5*x - 1. Use (M^n)[1,1] to define the recursion a(0) = 1, a(1) = 1, a(2) = 2, for n>2 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
a(n+1)/a(n) converges to 2 + sqrt(3) as n goes to infinity, the largest root of the characteristic polynomial. a(n) = A061278(n) + 1; (M^n)[1,2] = A001353(n); (M^n)[1,3] = A061278(n-1) for n>0; all with the same recursive properties.
Consecutive terms of this sequence and consecutive terms of A032908 provide all positive integer pairs for which K=(a+1)/b+(b+1)/a is an integer. For this sequence K=4. - Andrey Vyshnevyy, Sep 18 2015
The two-page Reid Barton article was sent to me around 2002, but for some reason it was not included in the OEIS at that time. I recently rediscovered it in my files. - N. J. A. Sloane, Sep 08 2018

Seiichi Manyama, Table of n, a(n) for n = 0..1750
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A061278, A001353, A001835, A005246, A276122, A276271.

I:=[1,1,2]; [n le 3 select I[n] else 5*Self(n-1)-5*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015

LinearRecurrence[{5, -5, 1}, {1, 1, 2}, 30] (* Vincenzo Librandi, Sep 18 2015 *)
CoefficientList[Series[(1 - 4 x + 2 x^2)/((1 - x) (1 - 4 x + x^2)), {x, 0, 27}], x] (* Michael De Vlieger, Aug 11 2016 *)
a[ n_] := If[ n < 1, a[1 - n], SeriesCoefficient[ (1/(1 - x) + (1 - 3 x)/(1 - 4 x + x^2)) / 2, {x, 0, n}]]; (* Michael Somos, Jul 09 2017 *)

M=[1,1,0; 1,3,1; 0,1,1]; for(i=0,40,print1((M^i)[1,1],","))

{a(n) = if( n<1, a(1-n), polcoeff( (1/(1 - x) + (1 - 3*x)/(1 - 4*x + x^2)) / 2 + x * O(x^n), n))}; /* Michael Somos, Jul 09 2017 */

a(n) = A101265(n), n>0. - R. J. Mathar, Aug 30 2008
a(n) = A079935(n+1) - A001571(n). - Gerry Martens, Jun 05 2015
a(0) = a(1) = 1, for n>1 a(n) = (a(n-1) + a(n-1)^2) / a(n-2). - Seiichi Manyama, Aug 11 2016
From Ilya Gutkovskiy, Aug 11 2016: (Start)
G.f.: (1 - 4*x + 2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = (6+(3-sqrt(3))*(2+sqrt(3))^n + (2-sqrt(3))^n*(3+sqrt(3)))/12. (End)
a(n) = 4*a(n-1) - a(n-2) - 1. - Seiichi Manyama, Aug 26 2016
From Seiichi Manyama, Sep 03 2016: (Start)
a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).
a(n) = A005246(n)*A005246(n+1). (End)
From Michael Somos, Jul 09 2017: (Start)
0 = +a(n)*(+1 +a(n) -4*a(n+1)) +a(n+1)*(+1 +a(n+1)) for all n in Z.
a(n) = a(1 - n) = (1 + A001835(n)) / 2 for all n in Z. (End)

a(26)-a(27) from Vincenzo Librandi, Sep 18 2015

A350919 a(0) = 9, a(1) = 9, and a(n) = 3*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

9, 9, 14, 29, 69, 174, 449, 1169, 3054, 7989, 20909, 54734, 143289, 375129, 982094, 2571149, 6731349, 17622894, 46137329, 120789089, 316229934, 827900709, 2167472189, 5674515854, 14856075369, 38893710249, 101825055374, 266581455869, 697919312229, 1827176480814, 4783610130209, 12523653909809, 32787351599214, 85838400887829, 224727851064269

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A032908, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350920, A350921, A350922, A350923, A350924, A350925, A350926.

nxt[{a_,b_}]:={b,3b-a-4}; NestList[nxt,{9,9},40][[;;,1]] (* or *) LinearRecurrence[{4,-4,1},{9,9,14},40] (* Harvey P. Dale, Jul 19 2024 *)

a(n) = 5*A032908(n) - 1. - Hugo Pfoertner, Jan 22 2022
G.f.: (3 - 2*x)*(3 - 7*x)/((1 - x)*(1 - 3*x + x^2)). - Stefano Spezia, Jan 22 2022
a(n) = 5*A001519(n) +4. - R. J. Mathar, Feb 07 2022

A182756 Numbers k > 1 such that are sequences B_k of type: {b(1) = 1, b(2) = k, for n >= 3; b(n) = the smallest number h > b(n-1) such that [[b(n-2) + b(n-1)] * [b(n-2) + h] * [b(n-1) + h]] / [b(n-2) * b(n-1) * h] is an integer}.

Original entry on oeis.org

2, 3, 6, 14, 21, 35, 77, 90, 234, 286, 611

Offset: 1

Jaroslav Krizek, Nov 27 2010

With 1 complement of A182757.

Is this a union/superset of A032908 and A101879? - Ralf Stephan, Nov 29 2010

			For n =1; a(n) = 2; B_2 = A038754(n): 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, …
For n =2; a(n) = 3; B_3 = A182751(n): 1, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486, …
For n =3; a(n) = 6; B_6 = A182752(n): 1, 6, 14, 84, 196, 1176, 2744, 16464, 38416, …
For n =4; a(n) = 14; B_14 = A182753(n): 1, 14, 35, 490, 1225, 17150, 42875, …
For n =5; a(n) = 21; B_21 = A182754(n): 1, 21, 77, 1617, 5929, 124509, 456533, …
For n =6; a(n) = 35; B_35 = A182755(n): 1, 35, 90, 3150, 8100, 283500, 729000, …

Cf. A182751, A182752, A182753, A182754, A182755, A182757, A038754.

A093467 a(1) = 1, a(2) = 2; for n >= 2, a(n+1) = a(n) + Sum_{i = 1..n} (a(i) - a(1)).

Original entry on oeis.org

1, 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

Offset: 1

Amarnath Murthy, Apr 07 2004

If the "man or boy" program A(k, x1, x2, x3) from the program section is run with k > 0 and arbitrary x1, x2, and x3, the result is A055588(k-1)*x1 + A001519(k-1)*x2. - Eric M. Schmidt, Jun 24 2021

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
K. Kuhapatanakul, On the Sums of Reciprocal Generalized Fibonacci Numbers, J. Int. Seq. 16 (2013) #13.7.1. See Theorem 3 p.3.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A093468.
Essentially the same as A032908.
Cf. A001519, A055588.

I:=[2,3,6]; [1] cat [n le 3 select I[n] else  4*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Feb 08 2017

a[1] = 1; a[2] = 2; a[n_] := a[n] = a[n - 1] + Sum[a[i] - a[1], {i, n - 1}]; Table[ a[n], {n, 30}]
Join[{1}, LinearRecurrence[{4, -4, 1}, {2, 3, 6}, 30]] (* Vincenzo Librandi, Feb 08 2017 *)

a(n)=if(n==1,1,if(n==2,2,a(n-1)+sum(i=1,n-1,a(i)-a(1)))) \\ Edward Jiang, Sep 06 2014
(ALGOL 60) begin integer procedure A(k, x1, x2, x3);
    value k; integer k;
    integer x1, x2, x3;
    begin integer procedure b;
        begin
            k:= k - 1;
            B:= A := A (k, B, x1, x2);
        end;
        A := if k <= 0 then x2 + x3 else B;
    end;
    integer i;
    for i:= 0 step 1 until 20 do
        print (A (i, 1, 1, 0));
end
comment The above is a simplified Man or Boy Test program (cf. A132343), omitting the negative parameters from the original. - Leonid Broukhis, Feb 07 2017

a(n) = 3*a(n-1) - a(n-2) - 1, n > 3. - Robert G. Wilson v, Apr 08 2004
G.f.: x - x^2*(2*x-1)*(x-2) / ( (x-1)*(x^2-3*x+1) ). - R. J. Mathar, Sep 06 2014
a(n) = A055588(n-2) + A001519(n-2), n > 1. - Eric M. Schmidt, Jun 24 2021

More terms from Robert G. Wilson v, Apr 08 2004

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

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

cp's OEIS Frontend

A104457 Decimal expansion of 1 + phi = phi^2 = (3 + sqrt(5))/2.

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A101265 a(1) = 1, a(2) = 2, a(3) = 6; a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) for n > 3.

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A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).

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A350919 a(0) = 9, a(1) = 9, and a(n) = 3*a(n-1) - a(n-2) - 4 for n >= 2.

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A182756 Numbers k > 1 such that are sequences B_k of type: {b(1) = 1, b(2) = k, for n >= 3; b(n) = the smallest number h > b(n-1) such that [[b(n-2) + b(n-1)] * [b(n-2) + h] * [b(n-1) + h]] / [b(n-2) * b(n-1) * h] is an integer}.

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A093467 a(1) = 1, a(2) = 2; for n >= 2, a(n+1) = a(n) + Sum_{i = 1..n} (a(i) - a(1)).

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A101265 a(1) = 1, a(2) = 2, a(3) = 6; a(n) = 5a(n-1) - 5a(n-2) + a(n-3) for n > 3.

A101879 a(0) = 1, a(1) = 1, a(2) = 2; for n > 2, a(n) = 5a(n-1) - 5a(n-2) + a(n-3).