A007400 -id:A007400 - cp's OEIS frontend

A073089 a(n) = (1/2)*(4n - 3 - Sum_{k=1..n} A007400(k)).

0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1

Offset: 1

Benoit Cloitre, Aug 18 2002

From Joerg Arndt, Oct 28 2013: (Start)
Sequence is (essentially) obtained by complementing every other term of A014577.
Turns (by 90 degrees) of a curve similar to the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw).
See the linked pdf files for two renderings of the curve. (End)

			From _Paul D. Hanna_, Oct 19 2012: (Start)
Let F(x) = x + 1/x + 1/x^3 + 1/x^7 + 1/x^15 + 1/x^31 +...+ 1/x^(2^n-1) +...
then F(x) = x + 1/(x + 1/(-x + 1/(-x + 1/(-x + 1/(x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(-x + 1/(-x + 1/(-x + 1/(x +...+ 1/((-1)^a(n)*x +...)))))))))))))))))))))),
a continued fraction in which the partial quotients equal (-1)^a(n)*x.  (End)

Antti Karttunen, Table of n, a(n) for n = 1..65537
Joerg Arndt, pdf rendering of the curve described in comment, alternate rendering of the curve.
Index entries for characteristic functions

Cf. A007400, A073088 (the sum part here), A123725.

a(n)=if(n<2,0,if(n%8==1,a((n+1)/2),[1,-1,0,1,1,1,0,0,1,-1,0,1,1,0,0,0][(n%16)+1])) \\ Ralf Stephan

/* Using the Continued Fraction, Print 2^N terms of this sequence: */
{N=10;CF=contfrac(x+sum(n=1,N,1/x^(2^n-1)),2^N);for(n=1,2^N,print1((1-CF[n]/x)/2,", "))} \\ Paul D. Hanna, Oct 19 2012

a(n) = { if ( n<=1, return(0)); n-=1; my(v=2^valuation(n,2) ); return( (0==bitand(n, v<<1)) != (v%2) ); } \\ Joerg Arndt, Oct 28 2013

Recurrence: a(1) = a(4n+2) = a(8n+7) = a(16n+13) = 0, a(4n) = a(8n+3) = a(16n+5) = 1, a(8n+1) = a(4n+1).
G.f.: The following series has a simple continued fraction expansion:
x + Sum_{n>=1} 1/x^(2^n-1) = [x; x, -x, -x, -x, x, ..., (-1)^a(n)*x, ...]. - Paul D. Hanna, Oct 19 2012
a(n) = A014577(n-2) + A056594(n).  Conjecture: a(n) = (1 + (-1)^A057661(n - 1))/2 for all n > 1. - Velin Yanev, Feb 01 2021

A073097 Let x(n) denote the number of 4's among the n first elements of the continued fraction for sum k>=0 1/2^(2^k) (A007400), y(n) the number of 6's and z(n) the number of 2's. Then a(n)=x(n)-y(n)-z(n)-1.

Original entry on oeis.org

-1, -1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, -1, 0, -1

Offset: 0

Benoit Cloitre, Aug 18 2002

sign

The positive sequence has a(n) = mod(A000120(A047849(n)),2) = mod(A000120(A078008(2n)),2) - Paul Barry, Jan 13 2005

Cosh(1) in 'reflected factorial' base is 1.10101010101010101010101010101010101010101010... - see A091337 for Sinh(1) (from Robert G. Wilson v, May 04 2005)

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A000120, A005578, A007400, A008619, A047849, A078008, A091337.

up_to = 65537;
A007400(n) = if(n<3, [0, 1, 4][n+1], if(n%8==1, A007400((n+1)/2), if(n%8==2, A007400((n+2)/2), [2, 0, 0, 2, 4, 4, 6, 4, 2, 0, 0, 2, 4, 6, 4, 4][(n%16)+1]))); \\ From A007400
A073097list(up_to) = { my(v=vector(up_to), x4=0, y6=0, z2=0, k); for(n=1, up_to, k=A007400(n); if(2==k,z2++,if(4==k,x4++,if(6==k,y6++))); v[n] = (x4-y6-z2-1)); (v); };
v073097 = A073097list(up_to);
A073097(n) = if(!n,-1,v073097[n]); \\ Antti Karttunen, Jan 12 2019

It seems that a(2k+1) = 0 for k>=1.

The positive sequence (assuming the pattern continues) has g.f. (1+x-x^2)/((1-x)(1-x^2)), with a(n)=(1-(1)^n)/2+0^n = mod((1+A001045(n+1))/2, 2) = mod(A005578, 2). The partial sums are A008619(n+1). - Paul Barry, Apr 28 2004

A073088 Sum of first n terms of the simple continued fraction of Sum_{k>=0} 1/2^(2^k) (cf. A007400).

Original entry on oeis.org

0, 1, 5, 7, 11, 15, 21, 25, 27, 31, 37, 39, 43, 49, 53, 57, 59, 63, 69, 71, 75, 79, 85, 89, 91, 97, 101, 103, 107, 113, 117, 121, 123, 127, 133, 135, 139, 143, 149, 153, 155, 159, 165, 167, 171, 177, 181, 185, 187, 193, 197, 199, 203, 207, 213, 217, 219, 225, 229

Offset: 1

Benoit Cloitre, Aug 18 2002

			The first 9 terms are 0, 1, 4, 2, 4, 4, 6, 4, 2, hence a(9)= 0 + 1 + 4 + 2 + 4 + 4 + 6 + 4 + 2 = 27.

Cf. A007400.

a(n) ~ 4n and a(n) < 4n.

A003285 Period of continued fraction for square root of n (or 0 if n is a square).

Original entry on oeis.org

0, 1, 2, 0, 1, 2, 4, 2, 0, 1, 2, 2, 5, 4, 2, 0, 1, 2, 6, 2, 6, 6, 4, 2, 0, 1, 2, 4, 5, 2, 8, 4, 4, 4, 2, 0, 1, 2, 2, 2, 3, 2, 10, 8, 6, 12, 4, 2, 0, 1, 2, 6, 5, 6, 4, 2, 6, 7, 6, 4, 11, 4, 2, 0, 1, 2, 10, 2, 8, 6, 8, 2, 7, 5, 4, 12, 6, 4, 4, 2, 0, 1, 2, 2, 5, 10, 2, 6, 5, 2, 8, 8, 10, 16, 4, 4, 11, 4, 2, 0, 1, 2, 12

Offset: 1

N. J. A. Sloane

Any string of five consecutive terms m^2 - 2 through m^2 + 2 for m > 2 in the sequence has the corresponding periods 4,2,0,1,2. - Lekraj Beedassy, Jul 17 2001
For m > 1, a(m^2+m) = 2 and the continued fraction is m, 2, 2*m, 2, 2*m, 2, 2*m, ... - Arran Fernandez, Aug 14 2011
Apparently the generating function of the sequence for the denominators of continued fraction convergents to sqrt(n) is always rational and of form p(x)/[1 - C*x^m + (-1)^m * x^(2m)], or equivalently, the denominators satisfy the linear recurrence b(n+2m) = C*b(n+m) - (-1)^m * b(n), where a(n) is equal to m for each nonsquare n, or 0. See A006702 for the conjecture regarding C. The same conjectures apply to the sequences of the numerators of continued fraction convergents to sqrt(n). - Ralf Stephan, Dec 12 2013
If a(n)=1, n is of form k^2+1 (A002522 except the initial term 1). See A013642 for a(n)=2, A013643 for a(n)=3, A013644 for a(n)=4, A010337 for a(n)=5, A020347 for a(n)=6, A010338 for a(n)=7, A020348 for a(n)=8, A010339 for a(n)=9, and furthermore A020349-A020439. - Ralf Stephan, Dec 12 2013
From William Krier, Dec 12 2024: (Start)
a(m^2-4) = 4 for even m>=6 since sqrt(m^2-4) = [m-1; 1, (m-4)/2, 1, 2*(m-1)].
a(m^2-4) = 6 for odd m>=5 since sqrt(m^2-4) = [m-1; 1, (m-3)/2, 2, (m-3)/2, 1, 2*(m-1)].
a(m^2+4) = 2 for even m>=2 since sqrt(m^2+4) = [m; m/2, 2*m].
a(m^2+4) = 5 for odd m>=3 since sqrt(m^2+4) = [m; (m-1)/2, 1, 1, (m-1)/2, 2*m]. (End)

A. Brousseau, Number Theory Tables. Fibonacci Association, San Jose, CA, 1973, p. 197.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 5000 terms from T. D. Noe)
Marius Beceanu, Period of the Continued Fraction of sqrt(n), (Feb 05 2003)
Leon Bernstein, Fundamental units and cycles in the period of real quadratic number fields, I. Pacific J. Math 63 (1976): 37-61.
Ron Knott, All square-root continued fractions eventually repeat
R. Luczak, Patterns in the period lengths of simple periodic continued fractional representations of square roots of integers near perfect squares, QED: Chicago's Youth Math Research Symposium (April 2013).
R. Mestrovic, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012-2018. - From _N. J. A. Sloane_, Jun 13 2012
Justin T. Miller, Families of Continued Fractions, translated (2000) from a document of Nikos Drakos, Computer Based Learning Unit, University of Leeds.
C. D. Patterson and H. C. Williams, Some Periodic Continued Fractions with Long Periods, Mathematics of Computation, Vol. 44 (1985), No. 170, pp. 523-532.
A. M. Rockett and P. Szuesz, On the lengths of the periods of the continued fractions of square-roots of integers, Forum Mathematicum, 2 (1990), 119-123.
R. G. Stanton, C. Sudler, Jr. and H. C. Williams, An Upper Bound for the Period of the Simple Continued Fraction for Sqrt(D), Pacific Journal of Math., Vol. 67 (1976), No. 2, pp. 525-536.
Hanna Uscka-Wehlou, Continued Fractions and Digital Lines with Irrational Slopes, in Discrete Geometry for Computer Imagery, Lecture Notes in Computer Science, Volume 4992/2008, Springer-Verlag.
A. J. van der Poorten, Fractional Parts of the Period of the Continued Fraction Expansion of Quadratic Integers [Refined and revised text of a talk given at the 2nd Conference of the Canadian Number Theory Association, Vancouver, 1989]
A. J. van der Poorten, An introduction to continued fractions, Unpublished.
A. J. van der Poorten, An introduction to continued fractions, Unpublished [Cached copy]
H. C. Williams, A Numerical Investigation Into the Length of the Period of the Continued Fraction Expansion of Sqrt(D), Mathematics of Computation, Vol. 36 (1981), No. 154, pp. 593-601.

Cf. A013943, A035015, A054269, A061490, A065938, A067280, A097853.

f:= n ->  if issqr(n) then 0
   else nops(numtheory:-cfrac(sqrt(n),'periodic','quotients')[2]) fi:
map(f, [$1..100]); # Robert Israel, Sep 02 2015

a[n_] := ContinuedFraction[Sqrt[n]] // If[Length[ # ] == 1, 0, Length[Last[ # ]]]&
pcf[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],0,Length[ContinuedFraction[s][[2]]]]]; Array[pcf,110] (* Harvey P. Dale, Jul 15 2017 *)

a(n)=if(issquare(n),return(0));my(s=sqrt(n),x=s,f=floor(s),P=[0],Q=[1],k);while(1,k=#P;P=concat(P,f*Q[k]-P[k]);Q=concat(Q,(n-P[k+1]^2)/Q[k]);k++;for(i=1,k-1,if(P[i]==P[k]&&Q[i]==Q[k],return(k-i)));x=(P[k]+s)/Q[k];f=floor(x)) \\ Charles R Greathouse IV, Jul 31 2011

isok(n, p) = {localprec(p); my(cf = contfrac(sqrt(n))); setsearch(Set(cf), 2*cf[1]);}
a(n) = {if (issquare(n), 0, my(p=100); while (! isok(n, p), p+=100); localprec(p); my(cf = contfrac(sqrt(n))); for (k=2, #cf, if (cf[k] == 2*cf[1], return (k-1))););} \\ Michel Marcus, Jul 07 2021

from sympy.ntheory.continued_fraction import continued_fraction_periodic
def a(n):
    cfp = continued_fraction_periodic(0, 1, d=n)
    return 0 if len(cfp) == 1 else len(cfp[1])
print([a(n) for n in range(1, 104)]) # Michael S. Branicky, Aug 22 2021

A007404 Decimal expansion of Sum_{n>=0} 1/2^(2^n).

Original entry on oeis.org

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

Offset: 0

Simon Plouffe

Kempner shows that numbers of a general form (which includes this constant) are transcendental. - Charles R Greathouse IV, Nov 07 2017

			0.81642150902189314370....

M. J. Knight, An "oceans of zeros" proof that a certain non-Liouville number is transcendental, The American Mathematical Monthly, Vol. 98, No. 10 (1991), pp. 947-949.

Harry J. Smith, Table of n, a(n) for n = 0..20000
Boris Adamczewski, The Many Faces of the Kempner Number, Journal of Integer Sequences, Vol. 16 (2013), #13.2.15.
David H. Bailey, Jonathan M. Borwein, Richard E. Crandall, and Carl Pomerance, On the Binary Expansions of Algebraic Numbers, Journal de Théorie des Nombres de Bordeaux, volume 16, number 3, 2004, pages 487-518. Also LBNL-53854 2003, and authors' copies one, four.
D. H. Bailey and H. R. P. Ferguson, Numerical results on relations between fundamental constants using a new algorithm, Mathematics of Computation, Vol.53 No. 188 (1989), 649-656. (Annotated scanned copy)
F. R. Bernhart & N. J. A. Sloane, Emails, April-May 1994
Aubrey J. Kempner, On transcendental numbers, Transactions of the American Mathematical Society 17 (1916), pp. 476-482.
Simon Plouffe, Plouffe's Inverter, sum(1/2^(2^n), n=0..infinity); to 20000 digits
Simon Plouffe, sum(1/2^(2^n), n=0..infinity) to 1024 digits
Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.
Index entries for transcendental numbers

Cf. A007400, A078885, A078585, A078886, A078887, A078888, A078889, A078890, A036987.

RealDigits[ N[ Sum[1/2^(2^n), {n, 0, Infinity}], 110]] [[1]]

default(realprecision, 20080); x=suminf(n=0, 1/2^(2^n)); x*=10; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b007404.txt", n, " ", d)); \\ Harry J. Smith, May 07 2009

suminf(k = 0, 1/(2^(2^k))) \\ Michel Marcus, Mar 26 2017

suminf(k=0,1.>>2^k) \\ Charles R Greathouse IV, Nov 07 2017

Equals -Sum_{k>=1} mu(2*k)/(2^k - 1) = Sum_{k>=1, k odd} mu(k)/(2^k - 1). - Amiram Eldar, Jun 22 2020

Edited by Robert G. Wilson v, Dec 11 2002

Deleted old PARI program Harry J. Smith, May 20 2009

A004200 Continued fraction for Sum_{k>=0} 1/3^(2^k).

Original entry on oeis.org

0, 2, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 3, 1, 5, 3, 1, 3, 3, 5, 3, 1, 5, 3, 3, 1, 3, 5, 1, 3, 5, 3, 1, 3, 3, 5, 1, 3, 5, 3, 3, 1, 3, 5, 3

Offset: 0

N. J. A. Sloane

			0.456942562477639661115491826... = 0 + 1/(2 + 1/(5 + 1/(3 + 1/(3 + ...)))).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Harry J. Smith, Table of n, a(n) for n = 0..20000
Jeffrey O. Shallit, Letter to N. J. A. Sloane with attachment, Aug. 1979
Jeffrey O. 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.
Jeffrey O. Shallit, Simple Continued Fractions for Some Irrational Numbers II, J. Number Theory 14 (1982), 228-231.

Cf. A007400, A078885 (decimal expansion).

u := 3: v := 7: Buv := [u,1,[0,u-1,u+1]]: for k from 2 to v do n := nops(Buv[3]): Buv := [u,Buv[2]+1,[seq(Buv[3][i],i=1..n-1),Buv[3][n]+1,Buv[3][n]-1,seq(Buv[3][n-i],i=1..n-2)]] od: seq(Buv[3][i],i=1..2^v);# first 2^v terms of A004200 # Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Dec 02 2002

ContinuedFraction[ NSum[1/3^(2^n), {n, 0, Infinity}, WorkingPrecision -> 105], 105] (* Jean-François Alcover, Jul 18 2011 *)

{ allocatemem(932245000); default(realprecision, 20000); x=suminf(n=0, 1/3^(2^n)); x=contfrac(x); for (n=1, 20001, write("b004200.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 10 2009

Recurrence: a(0)=0, a(1)=2, a(2)=5, a(16n+5)=a(16n+12)=a(32n+9)=a(32n+24)=1, a(8n+3)=a(8n+6)=a(16n+4)=a(16n+13)=a(32n+8)=a(32n+25)=3, a(8n+2)=a(8n+7)=5, a(16n)=a(8n), a(16n+1)=a(8n+1). - Ralf Stephan, May 17 2005

Better description and more terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 19 2001

A088431 Half of the (n+1)-st component of the continued fraction expansion of Sum_{k>=0} 1/2^(2^k).

Original entry on oeis.org

2, 1, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 3, 2, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 2, 2, 1, 3, 2, 1, 2, 2, 3, 2, 1, 3, 2, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 3, 2, 1, 2, 3, 2, 2, 1, 3, 2, 1, 2, 2, 3, 2, 1, 2, 3

Offset: 1

Benoit Cloitre, Nov 08 2003

nonn

To construct the sequence use the rule: a(1)=2, then a(a(1) + a(2) + ... + a(n) + 1) = 2 and fill in any undefined places with the sequence 1,3,1,3,1,3,1,3,1,3,1,3,....
This sequence appears to be the sequence of run lengths of the regular paperfolding sequence A014577, i.e., the latter starts as follows: 2 zeros, 1 one, 2 zeros, 2 ones, etc. - Dimitri Hendriks, May 06 2010
Hendriks' conjecture is proved in Bunder, Bates, and Arnold (2024).  Also see Shallit (2024). - Jeffrey Shallit, Mar 10 2025

			Example to illustrate the comment: a(a(1)+1)=a(3)=2 and a(2) is undefined. The rule requires a(2)=1. Next, a(a(1)+a(2)+1)=a(4)=2, a(a(1)+a(2)+a(3)+1)=a(6)=2 and a(5) is undefined. The rule now requires a(5)=3.

Antti Karttunen, Table of n, a(n) for n = 1..8192
Martin Bunder, Bruce Bates, and Stephen Arnold, The summed paperfolding sequence, Bull. Austral. Math. Soc. 110 (2024), 189-198.
Kevin Ryde, Iterations of the Dragon Curve, see index "TurnRun", with a(n) = TurnRun(n-1).
Jeffrey Shallit, Runs in Paperfolding Sequences, arXiv:2412.17930 [math.CO], 2024. See p. 10.

Cf. A007400, A014577, A088435, A092910.

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]]]; Array[a[# + 1]/2 &, 98] (* after Jean-François Alcover at A007400 *)

(define (A088431 n) (* 1/2 (A007400 (+ 1 n)))) ;; Code for A007400 given under that entry. - Antti Karttunen, Aug 12 2017

a(n) = (1/2)*A007400(n+1); a(a(1) + a(2) + ... + a(n) + 1) = 2.

A010145 Continued fraction for sqrt(61).

Original entry on oeis.org

7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14

Offset: 0

N. J. A. Sloane

			7.810249675906654394129722735... = 7 + 1/(1 + 1/(4 + 1/(3 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 07 2009

Harry J. Smith, Table of n, a(n) for n = 0..20000
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
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Cf. A010514 Decimal expansion. - Harry J. Smith, Jun 07 2009

ContinuedFraction[Sqrt[61],300] (* Vladimir Joseph Stephan Orlovsky, Mar 08 2011 *)
PadRight[{7},120,{14,1,4,3,1,2,2,1,3,4,1}] (* Harvey P. Dale, Mar 27 2013 *)

{ allocatemem(932245000); default(realprecision, 18000); x=contfrac(sqrt(61)); for (n=0, 20000, write("b010145.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 07 2009

A056469 Number of elements in the continued fraction for Sum_{k=0..n} 1/2^2^k.

Original entry on oeis.org

2, 3, 4, 6, 10, 18, 34, 66, 130, 258, 514, 1026, 2050, 4098, 8194, 16386, 32770, 65538, 131074, 262146, 524290, 1048578, 2097154, 4194306, 8388610, 16777218, 33554434, 67108866, 134217730, 268435458, 536870914, 1073741826, 2147483650

Offset: 0

Benoit Cloitre, Dec 07 2002

Let f_1(x) := 1 - sqrt(1 - x^2) = 2*x^2 + 2*x^4 + 4*x^6 + ... and for n>1 let f_n(x) := f_{n-1}(f_1(x)) = x^(2^n)*(2 + 2^n*x^2 + 2^n*a(n-1)*x^4 + ...). - Michael Somos, Jun 29 2023

			G.f. = 2 + 3*x + 4*x^2 + 6*x^3 + 10*x^4 + 18*x^5 + 34*x^6 + ... - _Michael Somos_, Jun 29 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A007400. Apart from initial term, same as A052548. See also A089985.

[Floor(2^(n-1)+2): n in [0..60]]; // Vincenzo Librandi, Sep 21 2011

LinearRecurrence[{3,-2},{2,3,4},40] (* Harvey P. Dale, Apr 23 2015 *)
a[ n_] := If[n < 0, 0, Floor[2^n/2] + 2]; (* Michael Somos, Jun 29 2023 *)

{a(n) = if(n<0, 0, 2^n\2 + 2)}; /* Michael Somos, Jun 29 2023 */

[floor(gaussian_binomial(n,1,2)+3) for n in range(-1,32)] # Zerinvary Lajos, May 31 2009

a(0)=2; for n > 0, a(n) = 2^(n-1) + 2 = A052548(n-1) + 2.
a(n) = floor(2^(n-1) + 2). - Vincenzo Librandi, Sep 21 2011
From Colin Barker, Mar 22 2013: (Start)
a(n) = 3*a(n-1) - 2*a(n-2) for n > 2.
G.f.: -(x^2+3*x-2) / ((x-1)*(2*x-1)). (End)
E.g.f.: exp(x)*(2 + sinh(x)). - Stefano Spezia, Oct 19 2023

A010124 Continued fraction for sqrt(19).

Original entry on oeis.org

4, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1

Offset: 0

N. J. A. Sloane

			4.358898943540673552236981983... = 4 + 1/(2 + 1/(1 + 1/(3 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 03 2009

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
A. J. van der Poorten, An introduction to continued fractions, Unpublished.
A. J. van der Poorten, An introduction to continued fractions, Unpublished [Cached copy]
Index entries for continued fractions for constants
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. A041028/A041029 (convergents).

Cf. A010475 (decimal expansion).

ContinuedFraction[Sqrt[19],300] (* Vladimir Joseph Stephan Orlovsky, Mar 05 2011 *)

{ allocatemem(932245000); default(realprecision, 17000); x=contfrac(sqrt(19)); for (n=0, 20000, write("b010124.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 03 2009

G.f.: (4 + 2*x + x^2 + 3*x^3 + x^4 + 2*x^5 + 4*x^6)/(1 - x^6). - Stefano Spezia, Jul 26 2025

cp's OEIS Frontend

A073089 a(n) = (1/2)*(4n - 3 - Sum_{k=1..n} A007400(k)).

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A073097 Let x(n) denote the number of 4's among the n first elements of the continued fraction for sum k>=0 1/2^(2^k) (A007400), y(n) the number of 6's and z(n) the number of 2's. Then a(n)=x(n)-y(n)-z(n)-1.

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A073088 Sum of first n terms of the simple continued fraction of Sum_{k>=0} 1/2^(2^k) (cf. A007400).

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A003285 Period of continued fraction for square root of n (or 0 if n is a square).

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A007404 Decimal expansion of Sum_{n>=0} 1/2^(2^n).

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A004200 Continued fraction for Sum_{k>=0} 1/3^(2^k).

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A088431 Half of the (n+1)-st component of the continued fraction expansion of Sum_{k>=0} 1/2^(2^k).

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