A164581 -id:A164581 - cp's OEIS frontend

A001077 Numerators of continued fraction convergents to sqrt(5).

1, 2, 9, 38, 161, 682, 2889, 12238, 51841, 219602, 930249, 3940598, 16692641, 70711162, 299537289, 1268860318, 5374978561, 22768774562, 96450076809, 408569081798, 1730726404001, 7331474697802, 31056625195209

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A001076(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = -1.
a(2*n) with b(2*n) := A001076(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,9) = A023039(n), n >= 0 and a(2*n+1) = 2*S(2*n, 2*sqrt(5)) = A075796(n+1), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
From Greg Dresden, May 21 2023: (Start)
For n >= 2, 8*a(n) is the number of ways to tile this T-shaped figure of length n-1 with four colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 8*a(6) = 23112 different tilings.
    _
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(End)

			1  2  9  38  161  (A001077)
-, -, -, --, ---, ...
0  1  4  17   72  (A001076)
1 + 2*x + 9*x^2 + 38*x^3 + 161*x^4 + 682*x^5 + 2889*x^6 + 12238*x^7 + ... - _Michael Somos_, Aug 11 2009

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).
V. Thébault, Les Récréations Mathématiques, Gauthier-Villars, Paris, 1952, p. 282.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Ex. 1, pp. 237-238.
Tanya Khovanova, Recursive Sequences
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
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A001076, A023039, A049629, A052924, A078343, A164581, A179237, A180148, A329723, A374439.

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001077:=(-1+2*z)/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
with(combinat): a:=n->fibonacci(n+1, 4)-2*fibonacci(n, 4): seq(a(n), n=0..30); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{4, 1}, {1, 2}, 30]
Join[{1},Numerator[Convergents[Sqrt[5],30]]] (* Harvey P. Dale, Mar 23 2016 *)
CoefficientList[Series[(1-2*x)/(1-4*x-x^2), {x, 0, 30}], x] (* G. C. Greubel, Dec 19 2017 *)
LucasL[3*Range[0,30]]/2 (* Rigoberto Florez, Apr 03 2019 *)
a[ n_] := LucasL[n, 4]/2; (* Michael Somos, Nov 02 2021 *)

{a(n) = fibonacci(3*n) / 2 + fibonacci(3*n - 1)}; /* Michael Somos, Aug 11 2009 */

a(n)=if(n<2,n+1,my(t=4);for(i=1,n-2,t=4+1/t);numerator(2+1/t)) \\ Charles R Greathouse IV, Dec 05 2011

x='x+O('x^30); Vec((1-2*x)/(1-4*x-x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, May 14 2009

G.f.: (1-2*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) + a(n-2), a(0)=1, a(1)=2.
a(n) = ((2 + sqrt(5))^n + (2 - sqrt(5))^n)/2.
a(n) = A014448(n)/2.
Limit_{n->infinity} a(n)/a(n-1) = phi^3 = 2 + sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = ((-i)^n)*T(n, 2*i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
Binomial transform of A084057. - Paul Barry, May 10 2003
E.g.f.: exp(2x)cosh(sqrt(5)x). - Paul Barry, May 10 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)*5^k*2^(n-2k). - Paul Barry, Nov 15 2003
a(n) = 4*a(n-1) + a(n-2) when n > 2; a(1) = 1, a(2) = 2. - Alex Vinokur (alexvn(AT)barak-online.net), Oct 25 2004
a(n) = A001076(n+1) - 2*A001076(n) = A097924(n) - A015448(n+1); a(n+1) = A097924(n) + 2*A001076(n) = A097924(n) + 2(A048876(n) - A048875(n)). - Creighton Dement, Mar 19 2005
a(n) = F(3*n)/2 + F(3*n-1) where F() = Fibonacci numbers A000045. - Gerald McGarvey, Apr 28 2007
a(n) = A000032(3*n)/2.
For n >= 1: a(n) = (1/2)*Fibonacci(6*n)/Fibonacci(3*n) and a(n) = integer part of (2 + sqrt(5))^n. - Artur Jasinski, Nov 28 2011
a(n) = Sum_{k=0..n} A201730(n,k)*4^k. - Philippe Deléham, Dec 06 2011
a(n) = A001076(n) + A015448(n). - R. J. Mathar, Jul 06 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) is the (1,1)-entry of the matrix W^n with W=[2, sqrt(5); sqrt(5), 2]. - Carmine Suriano, Mar 21 2014
From Rigoberto Florez, Apr 03 2019: (Start)
a(n) = A099919(n) + A049651(n) if n > 0.
a(n) = 1 + Sum_{k=0..n-1} L(3*k + 1) if n >= 0, L(n) = n-th Lucas number (A000032). (End)
From Christopher Hohl, Aug 22 2021: (Start)
For n >= 2, a(2n-1) = A079962(6n-9) + A079962(6n-3).
For n >= 1, a(2n) = sqrt(20*A079962(6n-3)^2 + 1). (End)
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*4^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*4^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 4^n*Sum_{k=0..n} A374439(n, k)*(-1/4)^k. - Peter Luschny, Jul 26 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) + 5*(-1)^(n+1)/a(n)) = 3/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = b(n) - b(n+1), where b(n) = (1/4) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5*(-1)^(n+1)/a(n)) = 1/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = c(n) + c(n+1), where c(n) = (1/4) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A052924 Expansion of g.f.: (1-x)/(1 - 3*x - x^2).

Original entry on oeis.org

1, 2, 7, 23, 76, 251, 829, 2738, 9043, 29867, 98644, 325799, 1076041, 3553922, 11737807, 38767343, 128039836, 422886851, 1396700389, 4612988018, 15235664443, 50319981347, 166195608484, 548906806799, 1812916028881

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Euler encountered this sequence when finding the largest root of z^2 - 3z - 1 = 0. - V. Frederick Rickey (fred-rickey(AT)usma.edu), Aug 20 2003
Let M = a triangle with the Pell series A000129 (1, 2, 5, 12, ...) in each column, with the leftmost column shifted upwards one row. A052924 starting (1, 2, 7, 23, ...) = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 31 2010
a(n) is the number of compositions of n when there are 2 types of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals partial sums of A108300 prefaced with a 1: (1, 1, 5, 16, 53, 175, 578, ...). - Gary W. Adamson, Feb 15 2012

L. Euler, Introductio in analysin infinitorum, 1748, section 338. English translation by John D. Blanton, Introduction to Analysis of the Infinite, 1988, Springer, p. 286.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Sergio Falcón, The k-Fibonacci difference sequences, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.
Sergio Falcón and Ángel Plaza, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 909
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,1).

A108300 (first differences), A006190 (partial sums), A355981 (primes).

Cf. A000032, A001077, A078343, A164581, A179237, A180148, A329723.

a:=[1,2];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Jun 09 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x)/(1-3*x-x^2) )); // G. C. Greubel, Jun 09 2019

spec:= [S,{S=Sequence(Prod(Sequence(Z),Union(Z,Z,Prod(Z,Z))))}, unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
seq(coeff(series((1-x)/(1-3*x-x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 16 2019

CoefficientList[Series[(1-x)/(1-3*x-x^2), {x,0,30}], x] (* G. C. Greubel, Jun 09 2019 *)

Vec((1-x)/(1-3*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

((1-x)/(1-3*x-x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jun 09 2019

a(n) = 3*a(n-1) + a(n-2).
a(n) = Sum_{alpha=RootOf(-1+3*x+x^2)} (1/13)*(1+5*alpha)*alpha^(-1-n).
With offset 1: a(1)=1; for n > 1, a(n) = Sum_{i=1..3*n-4} a(ceiling(i/3)). - Benoit Cloitre, Jan 04 2004
Binomial transform of A006130. a(n) = (1/2 - sqrt(13)/26)*(3/2 - sqrt(13)/2)^n + (1/2 + sqrt(13)/26)*(3/2 + sqrt(13)/2)^n. - Paul Barry, Jul 20 2004
From Creighton Dement, Nov 04 2004: (Start)
a(n) = A006190(n+1) - A006190(n);
4*a(n) = 9*A006190(n+1) - A006497(n+1) - 2*A003688(n+1). (End)
Numerators in continued fraction [1, 2, 3, 3, 3, ...], where the latter = 0.69722436226...; the length of an inradius of a right triangle with legs 2 and 3. - Gary W. Adamson, Dec 19 2007
If p[1]=2, p[i]=3, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i<=j), A[i,j] = -1, (i=j+1), and A[i,j]=0 otherwise. Then, for n >= 1, a(n-1) = det A. - Milan Janjic, Apr 29 2010
a(n) = A006190(n) + A003688(n). - R. J. Mathar, Jul 06 2012
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*3^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*3^k, n>0. - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) + 3*(-1)^(n+1)/a(n)) = 1/2, since 1/(a(n) + 3*(-1)^(n+1)/a(n)) = b(n) - b(n+1), where b(n) = (1/3) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 3*(-1)^(n+1)/a(n)) = 1/6, since 1/(a(n) + 3*(-1)^(n+1)/a(n)) = c(n) + c(n+1), where c(n) = (1/3) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

More terms from James Sellers, Jun 06 2000

A213894 Fixed points of a sequence h(n) defined by the minimum number of 5's in the relation n*[n,5,5,...,5,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 3, 11, 19, 31, 43, 47, 79, 127, 131, 163, 211, 251, 271, 307, 311, 331, 367, 379, 443, 503, 563, 599, 607, 659, 743, 751, 823, 839, 859, 887, 907, 911, 947, 967, 1063, 1087, 1091, 1123, 1163, 1171, 1187, 1259, 1279, 1291, 1303, 1307, 1319, 1423, 1447, 1471, 1487

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,5,5,...,5,n] and increase the number of 5's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 5, 5, 2] = [4, 2, 1, 1, 2, 4],
3 * [3, 5, 5, 5, 3] = [9, 1, 1, 2, 1, 2, 1, 1, 9],
4 * [4, 5, 5, 5, 5, 5, 4] = [16, 1, 3, 2, 1, 4, 1, 2, 3, 1, 16] ,
5 * [5, 5, 5] = [25, 1, 25].
The number of 5's needed defines the sequence h(n) = 2, 3, 5, 1, 11, 5, 5, 3, 5, 11, 11, ... (n >= 2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the generalized Fibonacci sequences satisfying f(n) = 5*f(n-1) + f(n-2), A052918, A015449, A164581, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 5*f(n-1) + f(n-2).
The above sequence h() is recorded as A262215. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891 - A213893, A213895 - A213899, A261311; A213358.

Cf. A213648, A262212 - A262220, A213900, A262211.

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[5, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)

{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,5), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

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A001077 Numerators of continued fraction convergents to sqrt(5).

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A052924 Expansion of g.f.: (1-x)/(1 - 3*x - x^2).

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A213894 Fixed points of a sequence h(n) defined by the minimum number of 5's in the relation n*[n,5,5,...,5,n] = [x,...,x] between simple continued fractions.

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