keyword:cofr - cp's OEIS frontend

A003417 Continued fraction for e.

2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, 14, 1, 1, 16, 1, 1, 18, 1, 1, 20, 1, 1, 22, 1, 1, 24, 1, 1, 26, 1, 1, 28, 1, 1, 30, 1, 1, 32, 1, 1, 34, 1, 1, 36, 1, 1, 38, 1, 1, 40, 1, 1, 42, 1, 1, 44, 1, 1, 46, 1, 1, 48, 1, 1, 50, 1, 1, 52, 1, 1, 54, 1, 1, 56, 1, 1, 58, 1, 1, 60, 1, 1, 62, 1, 1, 64, 1, 1, 66

Offset: 0

N. J. A. Sloane

This is also the Engel expansion for 3*exp(1/2)/2 - 1/2. - Gerald McGarvey, Aug 07 2004
Sorted with duplicate terms dropped, this is A004277, 1 together with the positive even numbers. - Alonso del Arte, Jan 27 2012
From Peter Bala, Nov 26 2019: (Start)
Related continued fractions expansions:
2*e = [5; 2, 3, 2, 3, 1, 2, 1, 3, 4, 3, 1, 4, 1, 3, 6, 3, 1, 6, ..., 1, 3, 2*n, 3, 1, 2*n, ...].
(1/2)*e = [1; 2, 1, 3, 1, 1, 1, 3, 3, 3, 1, 3, 1, 3, 5, 3, 1, 5, 1, 3, 7, 3, 1, 7, ..., 1, 3, 2*n + 1, 3, 1, 2*n + 1, ...].
4*e = [10, 1, 6, 1, 7, 2, 7, 2, 7, 1, 1, 1, 7, 3, 7, 1, 2, 1, 7, 4, 7, 1, 3, 1, 7, 5, 7, 1, 4, ..., 1, 7, n+1, 7, 1, n, ...].
(1/4)*e = [0, 1, 2, 8, 3, 1, 1, 1, 1, 7, 1, 1, 2, 1, 1, 1, 2, 7, 1, 2, 2, 1, 1, 1, 3, 7, 1, 3, 2, 1, 1, 1, 4, 7, 1, 4, 2, ..., 1, 1, 1, n, 7, 1, n, 2, ...]. (End)

			2.718281828459... = 2 + 1/(1 + 1/(2 + 1/(1 + 1/(1 + ...))))

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 186.
CRC Standard Mathematical Tables and Formulae, 30th ed. 1996, p. 88.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3.2.
Jay R. Goldman, The Queen of Mathematics, 1998, p. 70.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..9999
Khalil Ayadi, Chiheb Ben Bechir, and Maher Saadaoui, Continued Fractions with Predictable Patterns and Transcendental Numbers, J. Int. Seq. (2025) Vol. 28, Art. No. 25.1.4.
Thomas Baruchel and Carsten Elsner, On error sums formed by rational approximations with split denominators, arXiv preprint arXiv:1602.06445 [math.NT], 2016.
Henry Cohn, A short proof of the simple continued fraction expansion of e, Amer. Math. Monthly, 113 (No. 1, 2006), 57-62. [JSTOR] and arXiv:math/0601660 [math.NT], 2006.
Stephen Crowley, Integral Transforms of the Harmonic Sawtooth Map, The Riemann Zeta Function, and Fractal Strings, vixra:1202.0079 v2, 2012. [Wayback Machine link]
Francesco Dolce and Pierre-Adrien Tahay, Column representation of Sturmian words in cellular automata, Czech Technical University (Prague, Czechia, 2022).
W. R. Harmon, Letter to N. J. A. Sloane, Sep 1974
MathOverflow, What is the effect of adding 1/2 to a continued fraction?, 2013.
Keith Matthews, Finding the continued fraction of e^(l/m)
Sophie Morier-Genoud and Valentin Ovsienko, On q-deformed real numbers, arXiv:1908.04365 [math.QA], 2019.
C. D. Olds, The simple continued fraction expansion of e, Am. Math. Monthly 77 (9) (1970) 968-974.
Thomas J. Osler, A proof of the continued fraction expansion of e^(1/M), Amer. Math. Monthly, 113 (No. 1, 2006), 62-66.
Oskar Perron, Die Lehre von den Kettenbrüchen, 2nd ed., Teubner, Leipzig, 1929, p. 134.
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.
Ward Whitt, Weirdness in CTMC's, Notes for Course IEOR 6711: Stochastic Models I, [PDF], 2012. - From _N. J. A. Sloane_, Jan 03 2013
Eric Weisstein's World of Mathematics, e Continued Fraction.
Gang Xiao, Contfrac.
Yaohui Zhu, Kaiming Sun, Zhengdong Luo, and Lingfeng Wang, Progressive Self-Learning for Domain Adaptation on Symbolic Regression of Integer Sequences, Proc. 39th AAAI Conf. Artif. Intel. (2025) Vol. 39, No. 1, 1692-1699. See p. 1698.
Index entries for continued fractions for constants.
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A001113, A007676, A007677, A001204, A058282, A005131.
Cf. A006083, A006084, A006085, A081750.
Run lengths of A342991.

numtheory[cfrac](exp(1),100,'quotients'); # Jani Melik, May 25 2006
A003417:=(2+z+2*z**2-3*z**3-z**4+z**6)/(z-1)**2/(z**2+z+1)**2; # Simon Plouffe in his 1992 dissertation

ContinuedFraction[E, 100] (* Stefan Steinerberger, Apr 07 2006 *)
a[n_] := KroneckerDelta[1, n] + 2 n/3 - (2 n - 3)/3 DirichletCharacter[3, 1, n]; Table[a[n], {n, 1, 20}] (* Enrique Pérez Herrero, Feb 23 2013 *)
Table[Piecewise[{{2, n == 0}, {2 (n + 1)/3, Mod[n, 3] == 2}}, 1], {n, 0, 120}] (* Eric W. Weisstein, Jan 05 2019 *)
Join[{2}, LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 2, 1, 1, 4, 1}, 120]] (* Eric W. Weisstein, Jan 05 2019 *)
Join[{2}, Table[(2 (n + 4) + (1 - 2 n) Cos[2 n Pi/3] + Sqrt[3] (1 - 2 n) Sin[2 n Pi/3])/9, {n, 120}]] (* Eric W. Weisstein, Jan 05 2019 *)
Join[{2}, Flatten[Table[{1, 2n, 1}, {n, 40}]]] (* Harvey P. Dale, Jan 21 2020 *)

contfrac(exp(1)) \\ Alexander R. Povolotsky, Feb 23 2008

{ allocatemem(932245000); default(realprecision, 25000); x=contfrac(exp(1)); for (n=1, 10000, write("b003417.txt", n-1, " ", x[n])); } \\ Harry J. Smith, Apr 14 2009

A003417(n)=if(n%3<>2,1+(n==0),(n+1)/3*2) \\ M. F. Hasler, May 01 2013

def A003417(n): return 2 if n == 0 else 1 if n % 3 != 2 else (n+1)//3<<1 # Chai Wah Wu, Jul 27 2022

def eContFracTrio(n: Int): List[Int] = List(1, 2 * n, 1)
2 +: ((1 to 40).map(eContFracTrio).flatten) // Alonso del Arte, Nov 22 2020, with thanks to Harvey P. Dale

From Paul Barry, Jun 27 2006: (Start)
G.f.: (2 + x + 2*x^2 - 3*x^3 - x^4 + x^6)/(1 - 2*x^3 + x^6).
a(n) = 0^n + Sum{k = 0..n} 2*sin(2*Pi*(k - 1)/3)*floor((2*k - 1)/3)/sqrt(3). [Corrected and simplified by Jianing Song, Jan 05 2019] (End)
a(n) = 2*a(n-3) - a(n-6), n >= 7. - Philippe Deléham, Feb 10 2009
G.f.: 1 + U(0) where U(k)= 1 + x/(1 - x*(2*k + 1)/(1 + x*(2*k + 1) - 1/((2*k + 1) + 1 - (2*k + 1)*x/(x + 1/U(k+1))))); (continued fraction, 5-step). - Sergei N. Gladkovskii, Oct 07 2012
a(3*n-1) = 2*n, a(0) = 2, a(n) = 1 otherwise (i.e., for n+1 > 1, not a multiple of 3). - M. F. Hasler, May 01 2013
E.g.f.: First derivative of (2/9)*exp(x)*(x + 3) + (2/9)*exp(-x/2)*(2*x*cos((sqrt(3)/2)*x+2*Pi/3) - 3*cos((sqrt(3)/2)*x)) + x. - Jianing Song, Jan 05 2019
a(n) = floor(1/(n+1))-(floor(n/3)-floor((n+1)/3))*(2*n-1)/3+1. - Aaron J Grech, Sep 06 2024
Sum_{n>=1} (-1)^(n+1)/a(n) = 1 - log(2)/2. - Amiram Eldar, May 03 2025

Offset changed by Andrew Howroyd, Aug 07 2024

A013631 Continued fraction for zeta(3).

Original entry on oeis.org

1, 4, 1, 18, 1, 1, 1, 4, 1, 9, 9, 2, 1, 1, 1, 2, 7, 1, 1, 7, 11, 1, 1, 1, 3, 1, 6, 1, 30, 1, 4, 1, 1, 4, 1, 3, 1, 2, 7, 1, 3, 1, 2, 2, 1, 16, 1, 1, 3, 3, 1, 2, 2, 1, 6, 1, 1, 1, 6, 1, 1, 4, 428, 5, 1, 1, 3, 1, 1, 11, 2, 4, 4, 5, 4, 1, 5, 14, 1, 3, 1, 2, 19, 1, 2, 5, 1, 7, 1, 1, 1, 1, 1, 57, 3, 2, 14, 2

Offset: 0

N. J. A. Sloane, John Morrison (John.Morrison(AT)armltd.co.uk)

			zeta(3) = 1.2020569031595942... = 1 + 1/(4 + 1/(1 + 1/(18 + 1/(1 + ...)))). - _Harry J. Smith_, Apr 20 2009

Harry J. Smith, Table of n, a(n) for n = 0..19999
Eric Weisstein's World of Mathematics, Apery's Constant.
G. Xiao, Contfrac.
Index entries for continued fractions for constants.
Index entries for zeta function.

Cf. A002117 (decimal expansion), A078984, A078985 (convergents).

Cf. continued fractions for zeta(2)-zeta(20): A013679, A013680-A013696.

Mathematica
```
ContinuedFraction[ Zeta[3], 100]
```

{ allocatemem(932245000); default(realprecision, 21000); x=contfrac(zeta(3)); for (n=1, 20000, write("b013631.txt", n-1, " ", x[n])); } \\ Harry J. Smith, Apr 20 2009

Offset changed by Andrew Howroyd, Jul 10 2024

A174500 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003500(n)) ), where A003500(n) = (2+sqrt(3))^n + (2-sqrt(3))^n.

Original entry on oeis.org

1, 2, 1, 12, 1, 50, 1, 192, 1, 722, 1, 2700, 1, 10082, 1, 37632, 1, 140450, 1, 524172, 1, 1956242, 1, 7300800, 1, 27246962, 1, 101687052, 1, 379501250, 1, 1416317952, 1, 5285770562, 1, 19726764300, 1, 73621286642, 1, 274758382272, 1

Offset: 1

Paul D. Hanna, Mar 20 2010

			Let L = Sum_{n>=1} 1/(n*A003500(n)) or, more explicitly,
L = 1/4 + 1/(2*14) + 1/(3*52) + 1/(4*194) + 1/(5*724) + 1/(6*2702) +...
so that L = 0.2937696594138291094177057532058145970820225289928...
then exp(L) = 1.3414748719687236691269115428250035920032300984596...
equals the continued fraction expansion given by this sequence:
exp(L) = [1;2,1,12,1,50,1,192,1,722,1,2700,1,10082,1,...]; i.e.,
exp(L) = 1 + 1/(2 + 1/(1 + 1/(12 + 1/(1 + 1/(50 + 1/(1 +...)))))).
Compare these partial quotients to A003500(n), n=1,2,3,...:
[4,14,52,194,724,2702,10084,37634,140452,524174,1956244,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-5,0,1).

Cf. A003500, A174501.

a[n_?OddQ] = 1; a[n_?EvenQ] := a[n] = 4*a[n-2] - a[n-4] + 4;  a[2] = 2; a[4] = 12; Table[a[n], {n, 1, 41}] (* Jean-François Alcover, May 15 2014, after the first conjecture *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((2+sqrt(3))^m+(2-sqrt(3))^m))));contfrac(exp(L))[n]}

a(2n-1) = 1, a(2n) = A003500(n) - 2, for n>=1 [conjecture].
From Peter Bala, Jan 04 2013: (Start)
The above conjectures are correct. The real number exp( Sum_{n>=1} 1/(n*A003500(n)) ) is equal to the infinite product F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = 2 - sqrt(3). Ramanujan has given a continued fraction expansion for F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(N - sqrt(N^2 - 4))), N an integer greater than 3. The present case is when N = 4. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F((2 - sqrt(3))^k), k = 1, 2, 3, ...: if [1; c(1), 1, c(2), 1, c(3), 1, ...] denotes the present sequence then the simple continued fraction expansion of F((2 - sqrt(3))^k) is given by [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. (End)
a(n) = 5*a(n-2)-5*a(n-4)+a(n-6). G.f.: -x*(x^4+2*x^3-4*x^2+2*x+1) / ((x-1)*(x+1)*(x^4-4*x^2+1)). [Colin Barker, Jan 20 2013]

A007400 Continued fraction for Sum_{n>=0} 1/2^(2^n) = 0.8164215090218931...

Original entry on oeis.org

0, 1, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 6, 4

Offset: 0

Simon Plouffe, Jeffrey Shallit

			0.816421509021893143708079737... = 0 + 1/(1 + 1/(4 + 1/(2 + 1/(4 + ...))))

M. Kmošek, Rozwinieçie Niektórych Liczb Niewymiernych na Ulamki Lancuchowe (Continued Fraction Expansion of Some Irrational Numbers), Master's thesis, Uniwersytet Warszawski, 1979.

Harry J. Smith, Table of n, a(n) for n = 0..20000
Henry Cohn, Symmetry and specializability in continued fractions, Acta Arithmetica, volume 75, number 4, 1996, pages 297-320 (PDF). Also arXiv:math/0008221 [math.NT].
W. F. Lunnon, Q-D Tables and Zero-Squares, Manuscript, Jan. 1974. (Annotated scanned copy)
R. M. MacGregor, Generalizing the notion of a periodic sequence, American Math. Monthly 87 (1980), 90-102. (Annotated scanned copy)
B. Salvy, Email to N. J. A. Sloane, May 1994
Ratpoly info, May 1994
Jeffrey Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), no. 2, 209-217.
Jeffrey O. Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), no. 2, 209-217.
A. J. van der Poorten, An introduction to continued fractions, Unpublished.
A. J. van der Poorten, An introduction to continued fractions, Unpublished [Cached copy]
G. Xiao, Contfrac
Index entries for continued fractions for constants

Cf. A007404 (decimal), A073088 (partial sums), A073414/A073415 (convergents), A088431 (half), A089267, A092910.

a:= proc(n) option remember; local n8, n16;
    n8:= n mod 8;
    if n8 = 0 or n8 = 3 then return 2
    elif n8 = 4 or n8 = 7 then return 4
    elif n8 = 1 then return procname((n+1)/2)
    elif n8 = 2 then return procname((n+2)/2)
    fi;
    n16:= n mod 16;
    if n16 = 5 or n16 = 14 then return 4
    elif n16 = 6 or n16 = 13 then return 6
    fi
end proc:
a(0):= 0: a(1):= 1: a(2):= 4:
map(a, [$0..1000]); # Robert Israel, Jun 14 2016

a[n_] := a[n] = Which[n < 3, {0, 1, 4}[[n+1]], Mod[n, 8] == 1, a[(n+1)/2], Mod[n, 8] == 2, a[(n+2)/2], True, {2, 0, 0, 2, 4, 4, 6, 4, 2, 0, 0, 2, 4, 6, 4, 4}[[Mod[n, 16]+1]]]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, Nov 29 2013, after Ralf Stephan *)

a(n)=if(n<3,[0,1,4][n+1],if(n%8==1,a((n+1)/2),if(n%8==2,a((n+2)/2),[2,0,0,2,4,4,6,4,2,0,0,2,4,6,4,4][(n%16)+1]))) /* Ralf Stephan */

a(n)=contfrac(suminf(n=0,1/2^(2^n)))[n+1]

{ allocatemem(932245000); default(realprecision, 26000); x=suminf(n=0, 1/2^(2^n)); x=contfrac(x); for (n=1, 20001, write("b007400.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 07 2009

(define (A007400 n) (cond ((<= n 1) n) ((= 2 n) 4) (else (case (modulo n 8) ((0 3) 2) ((4 7) 4) ((1) (A007400 (/ (+ 1 n) 2))) ((2) (A007400 (/ (+ 2 n) 2))) (else (case (modulo n 16) ((5 14) 4) ((6 13) 6))))))) ;; (After Ralf Stephan's recurrence) - Antti Karttunen, Aug 12 2017

From Ralf Stephan, May 17 2005: (Start)
a(0)=0, a(1)=1, a(2)=4; for n > 2:
a(8k)    = a(8k+3)   = 2;
a(8k+4)  = a(8k+7)   = a(16k+5) = a(16k+14) = 4;
a(16k+6) = a(16k+13) = 6;
a(8k+1)  = a(4k+1);
a(8k+2)  = a(4k+2). (End)

A013679 Continued fraction for zeta(2) = Pi^2/6.

Original entry on oeis.org

1, 1, 1, 1, 4, 2, 4, 7, 1, 4, 2, 3, 4, 10, 1, 2, 1, 1, 1, 15, 1, 3, 6, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 3, 1, 1, 5, 1, 2, 2, 1, 1, 6, 27, 20, 3, 97, 105, 1, 1, 1, 1, 1, 45, 2, 8, 19, 1, 4, 1, 1, 3, 1, 2, 1, 1, 1, 5, 1, 1, 2, 3, 6, 1, 1, 1, 2, 1, 5, 1, 1, 2, 9, 5, 3, 2, 1, 1, 1

Offset: 0

N. J. A. Sloane

			1.644934066848226436472415166... = 1 + 1/(1 + 1/(1 + 1/(1 + 1/(4 + ...))))

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.
David Wells, "The Penguin Dictionary of Curious and Interesting Numbers," Revised Edition, Penguin Books, London, England, 1997, page 23.

T. D. Noe, Table of n, a(n) for n = 0..9999
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
G. Xiao, Contfrac
Index entries for continued fractions for constants
Index entries for zeta function.

Cf. A013661 (decimal expansion).

Cf. continued fractions for zeta(3)-zeta(20): A013631, A013680-A013696.

Mathematica
```
ContinuedFraction[ Pi^2/6, 100]
```

{ allocatemem(932245000); default(realprecision, 21000); x=contfrac(Pi^2/6); for (n=1, 20000, write("b013679.txt", n-1, " ", x[n])); } \\ Harry J. Smith, Apr 29 2009

Offset changed by Andrew Howroyd, Jul 10 2024

A040002 Continued fraction for sqrt(5).

Original entry on oeis.org

2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 0

N. J. A. Sloane

Decimal expansion of 11/45. - Natan Arie Consigli, Jan 19 2016

			2.236067977499789696409173668... = 2 + 1/(4 + 1/(4 + 1/(4 + 1/(4 + ...)))). - _Harry J. Smith_, Jun 01 2009

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 186.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 276.

Harry J. Smith, Table of n, a(n) for n = 0..20000
G. Xiao, Contfrac
Index entries for continued fractions for constants
Index of eventually constant sequences
Index of divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A002163 (decimal expansion), A001077/A001076 (convergents), A248235 (Egyptian fraction).
Cf. Continued fraction for sqrt(a^2+1) = (a, 2a, 2a, 2a....): A040000 (contfrac(sqrt(2)) = (1,2,2,...)), A040002, A040006, A040012, A040020, A040030, A040042, A040056, A040072, A040090, A040110 (contfrac(sqrt(122)) = (11,22,22,...)), A040132, A040156, A040182, A040210, A040240, A040272, A040306, A040342, A040380, A040420 (contfrac(sqrt(442)) = (21,42,42,...)), A040462, A040506, A040552, A040600, A040650, A040702, A040756, A040812, A040870, A040930 (contfrac(sqrt(962)) = (31,62,62,...)).
Essentially the same as A010709.

Digits := 100: convert(evalf(sqrt(N)),confrac,90,'cvgts'):

ContinuedFraction[Sqrt[5],300] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011 *)
PadRight[{2},120,{4}] (* Harvey P. Dale, Jul 06 2019 *)

{ allocatemem(932245000); default(realprecision, 26000); x=contfrac(sqrt(5)); for (n=0, 20000, write("b040002.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 01 2009

a(0) = 2, a(n) = 4 n>0. - Natan Arie Consigli, Jan 19 2016
From Elmo R. Oliveira, Feb 16 2024: (Start)
G.f.: 2*(1+x)/(1-x).
E.g.f.: 4*exp(x) - 2.
a(n) = 2*A040000(n). (End)

A041025 Denominators of continued fraction convergents to sqrt(17).

Original entry on oeis.org

1, 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, 151693352, 1232221121, 10009462320, 81307919681, 660472819768, 5365090477825, 43581196642368, 354014663616769, 2875698505576520, 23359602708228929, 189752520171407952, 1541379764079492545

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A041024(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = +1, a(2*n) with b(2*n) := A041024(2*n), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = -1 (cf. Emerson reference).
Bisection: a(2*n) = T(2*n+1,sqrt(17))/sqrt(17) = A078988(n), n >= 0 and a(2*n+1) = 8*S(n-1,66), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003
Sqrt(17) = 8/2 + 8/65 + 8/(65*4289) + 8/(4289*283009) + ... . - Gary W. Adamson, Dec 26 2007
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 8's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
De Moivre's formula: a(n) = (r^n - s^n)/(r-s), for r > s, gives sequences with integers if r and s are conjugates. With r=4+sqrt(17) and s=4-sqrt(17), a(n+1)/a(n) converges to r=4+sqrt(17). - Sture Sjöstedt, Nov 11 2011
a(n) equals the number of words of length n on alphabet {0,1,...,8} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 8-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 8 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle, Chaos, Solitons & Fractals 2007; 33(1): 38-49.
S. Falcón and Á. Plaza, On k-Fibonacci sequences and polynomials and their derivatives, Chaos, Solitons & Fractals (2007).
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,1).

Cf. A041024, A040012.
Cf. A000045, A000129, A006190, A001076, A052918, A005668, A054413, A243399.
Row n=8 of A073133, A172236 and A352361.
Cf. A099369 (squares).

I:=[1, 8]; [n le 2 select I[n] else 8*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013

CoefficientList[Series[1/(-z^2 - 8 z + 1), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 23 2011 *)
Denominator[Convergents[Sqrt[17],30]] (* Harvey P. Dale, Aug 15 2011 *)
LinearRecurrence[{8,1}, {1,8}, 50] (* Sture Sjöstedt, Nov 11 2011 *)

Vec(1/(1-8*x-x^2)+O(x^99)) \\ Charles R Greathouse IV, Dec 09 2014

[lucas_number1(n,8,-1) for n in range(1, 20)] # Zerinvary Lajos, Apr 25 2009

G.f.: 1/(1 - 8*x - x^2).
a(n) = ((-i)^n)*S(n, 8*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind and i^2 = -1. See A049310.
a(n) = F(n, 8), the n-th Fibonacci polynomial evaluated at x=8. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = ((4 + sqrt(17))^n - (4 - sqrt(17))^n)/(2*sqrt(17));
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*8^(n-1-2i). (End)
Let T be the 2 X 2 matrix [0, 1; 1, 8]. Then T^n * [1, 0] = [a(n-2), a(n-1)]. - Gary W. Adamson, Dec 26 2007
a(n) = 8*a(n-1) + a(n-2), n > 1; a(0)=1, a(1)=8. - Philippe Deléham, Nov 20 2008
a(p-1) == 68^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009 [Corrected by Jason Yuen, Apr 05 2025. See A087475 for more info about this congruence.]
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = sqrt(17) - 4. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 8*x - x^2) = Sum_{n >= 0} x^n *( Product_{k = 1..n} (m*k + 8 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A122553 a(0)=1, a(n)=3 for n > 0.

Original entry on oeis.org

1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 0

Philippe Deléham, Sep 20 2006

Continued fraction for (sqrt(13) - 1)/2 = A223139.
Decimal expansion of 4/30. - Alonso del Arte, Aug 16 2012
4/3 is the volume of the regular octahedron inscribed in the unit-radius sphere. - Amiram Eldar, Jun 02 2023

Calvin C. Clawson, Mathematical Mysteries, The Beauty and Magic of Numbers, Springer, 2013, pp. 95-96, 224.

Jun Yan, Results on pattern avoidance in parking functions, arXiv:2404.07958 [math.CO], 2024. See p. 4.
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A118273 (cube), A339259 (regular icosahedron), A363437 (regular tetrahedron), A363438 (regular dodecahedron).

Cf. A223139.

RealDigits[4/3, 10, 105][[1]] (* Alonso del Arte, Aug 16 2012 *)
PadRight[{1},120,3] (* Harvey P. Dale, Jul 21 2023 *)

a(n)=(n>=0)+2*(n>0) \\ Jaume Oliver Lafont, Mar 26 2009

a(n) = 3 - 2*0^n.
G.f.: (1 + 2*x)/(1 - x).
Sum_{n >= 0} a(n)*10^(-n) = 4/3.
From Amiram Eldar, Jun 05 2021: (Start)
4/3 = Product_{k>=1} (1 + 1/2^(2^k)).
4/3 = Sum_{k>=0} binomial(2*k,k)/((k+2)*4^k). (End)
Sum_{k>0} 3*k/4^k = 4/3 [Nicole Oresme]. - Stefano Spezia, Jun 27 2024
K_{n>=3} n/(n-2) = 4/3 (see Clawson at p. 224). - Stefano Spezia, Jul 01 2024
E.g.f.: 3*exp(x) - 2. - Elmo R. Oliveira, Aug 05 2024

A013680 Continued fraction for zeta(4).

Original entry on oeis.org

1, 12, 6, 1, 3, 1, 4, 183, 1, 1, 2, 1, 3, 1, 1, 5, 4, 2, 7, 23, 1, 1, 1, 1, 3, 2, 4, 2, 2, 22, 1, 13, 5, 1, 4, 2, 1, 3, 1, 1, 1, 6, 11, 40, 1, 7, 5, 2, 4, 1, 2, 3, 14, 9, 1, 33, 78, 1, 12, 4, 1, 2, 551, 1, 1, 1, 1, 1, 1, 2, 1, 9, 2, 7, 3, 1, 3, 2, 15, 1, 1, 2, 2

Offset: 0

N. J. A. Sloane

			zeta(4) = 1 + 1/(12 + 1/(6 + 1/(1 + 1/(3 + ...)))). - _Harry J. Smith_, Apr 29 2009

T. D. Noe, Table of n, a(n) for n = 0..9999
G. Xiao, Contfrac
Index entries for continued fractions for constants
Index entries for zeta function

Cf. A013662 (zeta(4)). - Harry J. Smith, Apr 29 2009

Cf. continued fractions for zeta(2)-zeta(20): A013679, A013631, A013681-A013696.

ContinuedFraction[Zeta[4],80] (* Harvey P. Dale, Oct 13 2013 *)

{ allocatemem(932245000); default(realprecision, 21000); x=contfrac(Pi^4/90); for (n=1, 20000, write("b013680.txt", n-1, " ", x[n])); } \\ Harry J. Smith, Apr 29 2009

Offset changed by Andrew Howroyd, Jul 09 2024

A176059 Periodic sequence: Repeat 3, 2.

Original entry on oeis.org

3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3

Offset: 0

Klaus Brockhaus, Apr 07 2010

Interleaving of A010701 and A007395.
Also continued fraction expansion of (3+sqrt(15))/2.
Also decimal expansion of 32/99.
a(n) = A010693(n+1).
Essentially first differences of A047218.
Binomial transform of 3 followed by -A122803.
Inverse binomial transform of 3 followed by A020714.
Second inverse binomial transform of A057198 without initial term 1.

Index entries for linear recurrences with constant coefficients, signature (0, 1).

Cf. A010701 (all 3's sequence), A007395 (all 2's sequence), A176058 (decimal expansion of (3+sqrt(15))/2), A010693 (repeat 2, 3), A047218 (congruent to {0, 3} mod 5), A122803 (powers of -2), A020714 (5*2^n), A057198 ((5*3^(n-1)+1)/2, n > 0).

Cf. A026532 (partial products).

a176059 = (3 -) . (`mod` 2) -- Reinhard Zumkeller, Nov 27 2012

a176059_list = cycle [3,2]  -- Reinhard Zumkeller, Apr 04 2012

&cat[ [3, 2]: n in [0..52] ];
[ (5+(-1)^n)/2: n in [0..104] ];

A176059:=n->(5+(-1)^n)/2; seq(A176059(n), n=0..100); # Wesley Ivan Hurt, Feb 26 2014

a[n_] := {3, 2}[[Mod[n, 2] + 1]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Jul 19 2013 *)
PadRight[{},120,{3,2}] (* Harvey P. Dale, Oct 06 2019 *)

a(n)=3-n%2 \\ Charles R Greathouse IV, Jul 13 2016

a(n) = (5+(-1)^n)/2.
a(n) = a(n-2) for n > 1; a(0) = 3, a(1) = 2.
a(n) = -a(n-1)+5 for n > 0; a(0) = 3.
a(n) = 3*((n+1) mod 2)+2*(n mod 2).
G.f.: (3+2*x)/((1-x)*(1+x)).
E.g.f.: 3*cosh(x) + 2*sinh(x). - Stefano Spezia, Aug 04 2025

cp's OEIS Frontend

A003417 Continued fraction for e.

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A013631 Continued fraction for zeta(3).

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A174500 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003500(n)) ), where A003500(n) = (2+sqrt(3))^n + (2-sqrt(3))^n.

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A007400 Continued fraction for Sum_{n>=0} 1/2^(2^n) = 0.8164215090218931...

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A013679 Continued fraction for zeta(2) = Pi^2/6.

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A040002 Continued fraction for sqrt(5).

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