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 1-6 of 6 results.

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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.
_
|| _
|||_|||
|_|
(End)

Examples

			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

References

  • 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.

Links

Crossrefs

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

Programs

  • Magma
    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
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    {a(n) = fibonacci(3*n) / 2 + fibonacci(3*n - 1)}; /* Michael Somos, Aug 11 2009 */
  • PARI
    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
  • PARI
    x='x+O('x^30); Vec((1-2*x)/(1-4*x-x^2)) \\ G. C. Greubel, Dec 19 2017
  • Sage
    [lucas_number2(n,4,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, May 14 2009

Formula

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)

Extensions

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

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Author

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

Keywords

Comments

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

References

  • 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.

Links

Crossrefs

A108300 (first differences), A006190 (partial sums), A355981 (primes).
Cf. A000032, A001077, A078343, A164581, A179237, A180148, A329723.

Programs

  • GAP
    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
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x)/(1-3*x-x^2) )); // G. C. Greubel, Jun 09 2019
  • Maple
    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
  • Mathematica
    CoefficientList[Series[(1-x)/(1-3*x-x^2), {x,0,30}], x] (* G. C. Greubel, Jun 09 2019 *)
  • PARI
    Vec((1-x)/(1-3*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011
  • Sage
    ((1-x)/(1-3*x-x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jun 09 2019

Formula

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)

Extensions

More terms from James Sellers, Jun 06 2000

A228469 a(n) = 6*a(n-2) + a(n-4), where a(0) = 2, a(1) = 8, a(2) = 13, a(3) = 49.

Original entry on oeis.org

2, 8, 13, 49, 80, 302, 493, 1861, 3038, 11468, 18721, 70669, 115364, 435482, 710905, 2683561, 4380794, 16536848, 26995669, 101904649, 166354808, 627964742, 1025124517, 3869693101, 6317101910, 23846123348, 38927735977, 146946433189, 239883517772, 905524722482
Offset: 0

Views

Author

Clark Kimberling, Aug 22 2013

Keywords

Comments

The classical Euclidean algorithm iterates the mapping u(x,y) = (y, (x mod y)) until reaching g = GCD(x,y) in a pair ( . , g). In much the same way, the modified algorithm (A228247) iterates the mapping v(x,y) = (y, y - (x mod y)). The accelerated Euclidean algorithm uses w(x,y) = min(u(x,y),v(x,y)). Let s(x,y) be the number of applications of u, starting with (x,y) -> u(x,y) needed to reach ( . , g), and let u'(x,y) be the number of applications of w to reach ( . , g). Then u'(x,y) <= u(x,y) for all (x,y).
Starting with a pair (x,y), at each application of w, record 0 if w( . , . ) = u( . , . ) and 1 otherwise. The trace of (x,y), denoted by trace(x,y), is the resulting 01 word, whose length is the total number of applications of w.
Conjecture: a(n) is the least positive integer c for which there is a positive integer b for which trace(b,c) consists of the first n letters of 01010101010101...

Examples

			a(3) = 13 because trace(18/13) = 010, and 13 is the least c for which there is a number b such that trace(b/c) = 010.  Successive applications of w are indicated by (18,13)->(13,5)->(5,2)->(2,1).  Whereas w finds GCD in 3 steps, u takes 4 steps, as indicated by (18,3)->(13,5)->(5,3)->(3,2)->(2,1).

Links

Crossrefs

Cf. A179237 (bisection), A228470, A228471, A228487, A228488.

Programs

  • Mathematica
    c1 = CoefficientList[Series[(2 + 8 x + x^2 + x^3)/(1 - 6 x^2 - x^4), {x, 0, 40}], x]; c2 = CoefficientList[Series[(3 + 11 x + 2 x^3)/(1 - 6 x^2 - x^4), {x, 0, 40}], x]; pairs = Transpose[CoefficientList[Series[{-((3 + 11 x + 2 x^3)/(-1 + 6 x^2 + x^4)), -((2 + 8 x + x^2 + x^3)/(-1 + 6 x^2 + x^4))}, {x, 0, 20}], x]]; t[{x_, y_, }] := t[{x, y}]; t[{x, y_}] := Prepend[If[# > y - #, {y - #, 1}, {#, 0}], y] &[Mod[x, y]]; userIn2[{x_, y_}] := Most[NestWhileList[t, {x, y}, (#[[2]] > 0) &]]; Map[Map[#[[3]] &, Rest[userIn2[#]]] &, pairs] (* Peter J. C. Moses, Aug 20 2013 *)
LinearRecurrence[{0, 6, 0, 1}, {2, 8, 13, 49}, 30] (* T. D. Noe, Aug 23 2013 *)
  • PARI
    Vec((x^3+x^2+8*x+2)/(1-6*x^2-x^4)+O(x^99)) \\ Charles R Greathouse IV, Jun 12 2015

Formula

G.f.: (x^3 + x^2 + 8*x + 2)/(1 - 6*x^2 - x^4). - Ralf Stephan, Aug 24 2013

A164581 a(n) = 5*a(n - 1) + a(n - 2), with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 57, 296, 1537, 7981, 41442, 215191, 1117397, 5802176, 30128277, 156443561, 812346082, 4218173971, 21903215937, 113734253656, 590574484217, 3066606674741, 15923607857922, 82684645964351, 429346837679677, 2229418834362736, 11576441009493357
Offset: 0

Views

Author

Vincenzo Librandi, Aug 17 2009

Keywords

Links

Crossrefs

Cf. A000032, A001077, A052924, A078343, A179237, A180148, A329723.

Programs

  • Magma
    [ n le 2 select (n) else 5*Self(n-1)+Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Sep 12 2013
  • Mathematica
    LinearRecurrence[{5, 1}, {1, 2}, 40] (* or *) Rest[CoefficientList[Series [x (1 - 3 x) / (1 - 5 x - x^2), {x, 0, 40}], x]] (* Harvey P. Dale, May 02 2011 *)
  • PARI
    Vec((1-3*x)/(1-5*x-x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

Formula

a(n) = 5*a(n-1)+a(n-2) = A052918(n)-3*A052918(n-1).
G.f.: (1-3*x)/(1-5*x-x^2).
a(n) = A052918(n) + A015449(n). - R. J. Mathar, Jul 06 2012
a(n) = (2^(-1-n)*((5-sqrt(29))^n*(1+sqrt(29))+(-1+sqrt(29))*(5+sqrt(29))^n))/sqrt(29). - Colin Barker, Oct 13 2015
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*5^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*5^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = A052918(n) -3*A052918(n-1). - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) - 7*(-1)^n/a(n)) = 3/10, since 1/(a(n) - 7*(-1)^n/a(n)) = b(n) - b(n+1), where b(n) = (1/5) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 7*(-1)^n/a(n)) = 1/10, since 1/(a(n) - 7*(-1)^n/a(n)) = c(n) + c(n+1), where c(n) = (1/5) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

A213895 Fixed points of a sequence h(n) defined by the minimum number of 6's in the relation n*[n,6,6,...,6,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

7, 11, 23, 47, 127, 139, 211, 223, 251, 331, 367, 379, 383, 463, 487, 499, 607, 619, 691, 727, 739, 743, 811, 823, 863, 887, 967, 971, 983, 1051, 1063, 1087, 1171, 1291, 1303, 1327, 1367, 1423, 1447, 1451, 1459
Offset: 1

Views

Author

Art DuPre, Jun 23 2012

Keywords

Comments

In a variant of A213891, multiply n by a number with simple continued fraction [n,6,6,...,6,n] and increase the number of 6's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 6, 2] = [4, 3, 4],
3 * [3, 6, 3] = [9, 2, 9],
4 * [4, 6, 6, 6, 4] = [16, 1, 1, 1, 5, 1, 1, 1, 16],
5 * [5, 6, 6, 6, 6, 5] = [25, 1, 4, 3, 3, 4, 1, 25],
6 * [6, 6, 6] = [36, 1, 36],
7 * [7, 6, 6, 6, 6, 6, 6, 6, 7] = [50, 7, 2, 1, 4, 4, 4, 1, 2, 7, 50].
The number of 6's needed defines the sequence h(n) = 1, 1, 3, 4, 1, 7, 7, 5, 9, ... (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 numbers is 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) = 6*f(n-1) + f(n-2), A005668, A015451, A179237, 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) = 6*f(n-1) + f(n-2).
The above sequence h() is recorded as A262216. - M. F. Hasler, Sep 15 2015

Crossrefs

Cf. A000057, A213891 - A213894, A213896 - A213899, A261311; A213358.
Cf. A213648, A262212 - A262220, A213900, A262211.

Programs

  • Mathematica
    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[6, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
  • PARI
    {a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,6), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A179238 Numerators in convergents to infinitely repeating period 3 palindromic continued fraction [1,2,1,...].

Original entry on oeis.org

1, 2, 3, 5, 13, 18, 31, 80, 111, 191, 493, 684, 1177, 3038, 4215, 7253, 18721, 25974, 44695, 115364, 160059, 275423, 710905, 986328, 1697233, 4380794, 6078027, 10458821, 26995669, 37454490, 64450159, 166354808, 230804967, 397159775
Offset: 1

Views

Author

Gary W. Adamson, Jul 04 2010

Keywords

Comments

If n == 2 (mod 3), a(n) is in A179237, prefaced with a 1: (1, 2, 13, 80, 493, ...).
Conjecture relating palindromic continued fractions to the tangents of angles A and 2A:
Tan 2A is rational when tan(A) is irrational, iff tan(A) is a convergent to an infinitely repeating palindromic continued fraction.
Conversely, if tan(A) is such a convergent, then tan(2A) must be rational.
The conjecture applies to any palindromic form such as [a,a,a,...], [a,b,a,a,b,a,...], [a,b,b,a,a,b,b,a,...], [a,b,c,c,b,a,...] and so on.
Conjecture proved by the author, Jul 05 2010. - Gary W. Adamson, Mar 25 2014
Tan(2) corrected to tan(A), which is the half angle of a right triangle with sides (a, b) and c = sqrt(a^2 + b^2), not an integer. Example: Given a right triangle with sides a = 3, b = 5, c = sqrt(34), we calculate the two acute angles 2A and divide each by two, giving us two angles A; then perform tan(A) on each. These have an algebraic equivalency of two constants: 3/(5 + sqrt(34)) = 0.276983964... and 5/(3 + sqrt(34)) = 0.566190378... The continued fraction representations of these constants are respectively barover[3,1,1,1,1,3] and barover[1,1,3,3,1,1], both palindromic. The conjecture and proof don't apply to Pythagorean triangles, in which case the tangents of the half angles would be rational. - Gary W. Adamson, Apr 07 2014

Examples

			The first few convergents given
[1,...2,...1,...1,...2,...1,... =
.1....2....3....5...13,..18,...
...
a(5) = 13 = 2*a(4) + a(3) = 2*10 + 3 since 5 == 2 mod 3
a(6) = 18 = a(5) + a(4) = 13 + 5 since 6 == 0 mod 3.
...
Examples relating to the conjecture: Since A179238 has two recursive multipliers, we can extract the sequence A179237: (1, 2, 13, 80, 493, 3038,...) = convergents of A179238 such that a(n+1) = 2*a(n) + a(n-1). The convergent of this sequence = 1/(sqrt(10) - 3) = 6.16227766, ... = a, where 1/a = 0.1622776, ... the convergent to the palindromic continued fraction [6,6,6,...]. Then letting 1/a = tan(A), then tan(2A) = 1/3, rational. Similarly, taking a = 6.16227766, ... = tan(A), then tan(2A) = -1/3, rational, with a and 1/a irrational.

Links

Crossrefs

Cf. A179237.

Formula

a(1) = 1; a(n+1) = a(n) + a(n-1) if n == (0,1) mod 3.
a(n+1) = 2*a(n) + a(n-1) if n == 2 mod 3.
a(n) = +6*a(n-3) +a(n-6). G.f.: x*(1+2*x+3*x^2-x^3+x^4)/(1-6*x^3-x^6). - R. J. Mathar, Jul 06 2010
a(3n+2) = A179237(n+1). - R. J. Mathar, Feb 14 2024
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