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A041550 Numerators of continued fraction convergents to sqrt(293).

17, 137, 154, 291, 2482, 84679, 679914, 764593, 1444507, 12320649, 420346573, 3375093233, 3795439806, 7170533039, 61159704118, 2086600473051, 16753963488526, 18840563961577, 35594527450103, 303596783562401, 10357885168571737, 83166678132136297

Offset: 0

N. J. A. Sloane

From Johannes W. Meijer, Jun 12 2010: (Start)
The a(n) terms of this sequence can be constructed with the terms of sequence A090306.
For the terms of the periodical sequence of the continued fraction for sqrt(293) see A040275. We observe that its period is five. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 4964, 0, 0, 0, 0, 1).

Cf. A041018, A041046, A041090, A041150, A041226, A041318, A041426, A041550, A041551.

Numerator[Convergents[Sqrt[293], 30]] (* Vincenzo Librandi, Nov 04 2013 *)

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5n) = A090306(3n+1), a(5n+1) = (A090306(3n+2) - A090306(3n+1))/2, a(5n+2) = (A090306(3n+2) + A090306(3n+1))/2, a(5n+3) = A090306(3n+2) and a(5n+4) = A090306(3n+3)/2. (End)
G.f.: -(x^9-17*x^8+137*x^7-154*x^6+291*x^5+2482*x^4+291*x^3+154*x^2+137*x+17) / (x^10+4964*x^5-1). - Colin Barker, Nov 08 2013

More terms from Colin Barker, Nov 08 2013

A058281 Continued fraction for square root of e.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 9, 1, 1, 13, 1, 1, 17, 1, 1, 21, 1, 1, 25, 1, 1, 29, 1, 1, 33, 1, 1, 37, 1, 1, 41, 1, 1, 45, 1, 1, 49, 1, 1, 53, 1, 1, 57, 1, 1, 61, 1, 1, 65, 1, 1, 69, 1, 1, 73, 1, 1, 77, 1, 1, 81, 1, 1, 85, 1, 1, 89, 1, 1, 93, 1, 1, 97, 1, 1, 101, 1, 1, 105, 1, 1, 109, 1, 1, 113, 1, 1

Offset: 0

Robert G. Wilson v, Dec 07 2000

			sqrt(e) = 1 + 1/(1 + 1/(1 + 1/(1 + 1/(5 + ...)))). - _Harry J. Smith_, May 01 2009

Harry J. Smith, Table of n, a(n) for n = 0..20000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Keith Matthews, Finding the continued fraction of e^(l/m).
Thomas J. Osler, A proof of the continued fraction expansion of e^(1/M), Amer. Math. Monthly, 113 (No. 1, 2006), 62-66.
Gang Xiao, Contfrac.
Index entries for continued fractions for constants.
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A004766, A016813.

Cf. A019774 (decimal expansion of sqrt(e)).

ContinuedFraction[ Sqrt[E], 100]
LinearRecurrence[{0,0,2,0,0,-1},{1,1,1,1,5,1},100] (* Harvey P. Dale, Aug 05 2025 *)

PARI
```
contfrac(sqrt(exp(1)))
```

{ allocatemem(932245000); default(realprecision, 60000); x=contfrac(sqrt(exp(1))); for (n=1, 20001, write("b058281.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 01 2009

a(3k+1) = 4k+1, a(i) = 1 otherwise.
G.f.: -(x^2-x+1)*(x^3-2*x^2-2*x-1) / ((x-1)^2*(x^2+x+1)^2). - Colin Barker, Jun 24 2013
E.g.f.: exp(-x/2)*(exp(3*x/2)*(5 + 4*x) + (4 + 8*x)*cos(sqrt(3)*x/2) - 4*sqrt(3)*sin(sqrt(3)*x/2))/9. - Stefano Spezia, May 05 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi + 2*log(sqrt(2)+1)) / (4*sqrt(2)). - Amiram Eldar, May 03 2025

More terms from Jason Earls, Jul 10 2001

A169609 Period 3: repeat [1, 3, 3].

Original entry on oeis.org

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

Offset: 0

Klaus Brockhaus, Dec 03 2009

Interleaving of A000012, A010701 and A010701.
Also continued fraction expansion of (5+sqrt(65))/10 = 1.3062257748....
Also decimal expansion of 133/999.
a(n) = A144437(n) for n > 0.
Unsigned version of A154595.
Binomial transform of A168615.
Inverse binomial transform of A168673.
Essentially first differences of A047347.

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

Cf. A000012 (all 1's sequence), A010701 (all 3's sequence), A144437 (repeat 3, 3, 1), A154595 (repeat 1, 3, 3, -1, -3, -3), A168615, A168673, A047347 (congruent to {0, 1, 4} mod 7), A010684 (repeat 1, 3).
Cf. A171419 (decimal expansion of (5+sqrt(65))/10).
Cf. A146094.

[ n mod 3 eq 0 select 1 else 3: n in [0..104] ];

&cat [[1, 3, 3]^^30]; // Wesley Ivan Hurt, Jul 02 2016

seq(op([1, 3, 3]), n=0..50); # Wesley Ivan Hurt, Jul 02 2016

PadRight[{},120,{1,3,3}] (* or *) LinearRecurrence[{0,0,1},{1,3,3},120] (* Harvey P. Dale, Apr 29 2015 *)

a(n) = a(n-3) for n > 2, with a(0) = 1, a(1) = 3, a(2) = 3.
G.f.: (1+3*x+3*x^2)/(1-x^3).
a(n) = (7/3)+(2/3)*cos((2*Pi/3)*(n+1))-(2*sqrt(3)/3)*sin((2*Pi/3)*(n+1)). - Richard Choulet, Mar 15 2010
a(n) = a(n-a(n-2)) for n>=2. Example: a(5) = a(5-a(3)) = a(5-a(3-a(1))) = a(5-a(3-3)) = a(5-a(0)) = a(5-1) = a(4) = a(4-a(2)) = a(4-3) = a(1) = 3. - Richard Choulet, Mar 15 2010; edited by Klaus Brockhaus, Nov 21 2010
a(n) = 1 + 2*sgn(n mod 3). - Wesley Ivan Hurt, Jul 02 2016
a(n) = 3/gcd(n,3). - Wesley Ivan Hurt, Jul 11 2016
E.g.f.: (7*exp(x) - 4*exp(-x/2)*cos(sqrt(3)*x/2))/3. - Stefano Spezia, Sep 05 2025

Keywords cofr, cons added by Klaus Brockhaus, Apr 20 2010

Minor edits, crossref added by Klaus Brockhaus, May 03 2010

A321091 Continued fraction expansion of the constant z that satisfies: CF(3*z, n) = CF(z, n) + 10, for n >= 0, where CF(z, n) denotes the n-th partial denominator in the continued fraction expansion of z.

Original entry on oeis.org

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

Offset: 0

Paul D. Hanna, Oct 27 2018

			The decimal expansion of this constant z begins:
z = 4.69674328597002790903797135061489969596768903498266...
The simple continued fraction expansion of z begins:
z = [4; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, ..., a(n), ...];
such that the simple continued fraction expansion of 3*z begins:
3*z = [14; 11, 12, 13, 12, 11, 13, 12, 11, 13, 11, 12, ..., a(n) + 10, ...].
RELATED CONTINUED FRACTIONS.
Continued fractions for z/2 and z/3 are also interesting:
z/2 = [2; 2, 1, 6, 1, 2, 1, 1, 2, 8, 2, 1, 1, 2, 1, 6, 1, 2, 1, 1, 5, 1, 1, 5, 1, 1, 2, 8, 2, 1, 1, 5, 1, 1, 5, 1, 1, 2, 1, 6, 1, 2, 1, 1, 2, 8, 2, ...];
z/3 = [1; 1, 1, 3, 3, 4, 1, 8, 2, 1, 3, 2, 1, 3, 3, 4, 1, 4, 3, 8, 1, 8, 2, 1, 3, 3, 4, 1, 8, 2, 1, 3, 3, 4, 1, 8, 2, 1, 3, 2, 1, 3, 3, 4, 1, 4, 3, 3, ...].
EXTENDED TERMS.
The initial 1020 terms of the continued fraction of z are
Z = [4;1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,
2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,
1,2,3,2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,
2,1,3,1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,1,2,3,2,1,3,
1,2,3,1,2,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3,2,1,3,2,1,3,1,2,3, ...].
...
The initial 1000 digits in the decimal expansion of z are
z = 4.69674328597002790903797135061489969596768903498266\
39744976065876894930180599498227467938788452668766\
73554337379179728087710868220957908662613614032798\
33388405727520089133825065104553063833471857236781\
30005222692602670820049366351838443544785254486817\
36713432024147916330132466297271088195927148747751\
61081134504127224293197514421215072180791056995531\
25282254039576642227973626045065085745006810232418\
31151864601858865871496634133860069172907530299184\
80599130262819282015136761682297696200354791299160\
72099154048193374507277756176907149849150248588944\
75702959853076891940422757945105365738782828309624\
89182729519410181321396021772327752921026193693551\
52235778358918181495624102484144418903334090227672\
68450214362152729231740655406007216125545132538964\
63321922981643984915752295515263408732183582572996\
74985150593020685391286450747231245540741605404683\
64053241113305130300450809189365250050022411738651\
75283281124343007394094179023437577924273245108697\
96469643376214173186511094645179990191104328112759...
...
GENERATING METHOD.
Start with CF = [4] and repeat (PARI code):
{M = contfracpnqn(CF + vector(#CF,i, 10));
z = (1/3)*M[1,1]/M[2,1]; CF = contfrac(z)}
This method can be illustrated as follows.
z0 = [4] = 4 ;
z1 = (1/3)*[14] = [4; 1, 2] = 14/3 ;
z2 = (1/3)*[14; 11, 12] = [4, 1, 2, 3, 2, 1, 3, 3] = 1874/399 ;
z3 = (1/3)*[14; 11, 12, 13, 12, 11, 13, 13] = [4; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 4] = 187227588/39863279 ;
z4 = (1/3)*[14; 11, 12, 13, 12, 11, 13, 12, 11, 13, 11, 12, 13, 11, 12, 13, 12, 11, 13, 11, 12, 14] = [4; 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 3, 2, 1, 4] = 254413274852180460063336/54168017999228580539039 ; ...
where this constant z equals the limit of the iterations of the above process.

Paul D. Hanna, Table of n, a(n) for n = 0..10000

Cf. A321090, A321092, A321093, A321094, A321095, A321096, A321097, A321098.

/* Generate over 5000 terms */
{CF=[4]; for(i=1,8, M = contfracpnqn( CF + vector(#CF,i,10) ); z = (1/3)*M[1,1]/M[2,1]; CF = contfrac(z) )}
for(n=0,200,print1(CF[n+1],", "))

Formula for terms:
(1) a(0) = 4,
(2) a(3*n) = 3 for n >= 1,
(3) a(3*n+2) = 3 - a(3*n+1) for n >= 0,
(4) a(9*n+1) = 1 for n >= 0,
(5) a(9*n+7) = 2 for n >= 0,
(6) a(9*n+4) = 3 - a(3*n+1) for n >= 0.
a(3*n+1) = A189706(n+1) + 1 for n >= 0.
a(n) = A321090(n) + 1 for n >= 1, with a(0) = 4.
a(n) = A321093(n) - A321090(n), for n >= 0.
a(n) = A321095(n) - 2*A321090(n), for n >= 0.
a(n) = A321097(n) - 3*A321090(n), for n >= 0.

A001148 Final digit of 3^n.

Original entry on oeis.org

1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1

Offset: 0

N. J. A. Sloane

Let G = {1,3,7,9}, and let the binary operator o be defined as: X o Y = least significant digit of the product XY, where X,Y belong to G. Then (G,o) is an Abelian group and 3 is a generator of this group. - K.V.Iyer, Apr 19 2009
3^n mod 10 and 3^n mod 20. - Zerinvary Lajos, Nov 25 2009
Continued fraction expansion of (243+17*sqrt(285))/4020 = 0.13183906... (see A178148). - Klaus Brockhaus, Apr 17 2011

[3^n mod 10: n in [0..150]]; // Vincenzo Librandi, Apr 12 2011

Table[PowerMod[3, n, 10], {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)

a(n)=[1, 3, 9, 7][n%4+1] \\ Charles R Greathouse IV, Dec 27 2012

[power_mod(3, n, 10) for n in range(0, 81)] # Zerinvary Lajos, Nov 24 2009

Periodic with period 4.
From R. J. Mathar, Apr 13 2010: (Start)
a(n) = a(n-1) - a(n-2) + a(n-3).
G.f.: (1+2*x+7*x^2)/ ((1-x) * (1+x^2)). (End)
a(n) = 5 - (2+i)*(-i)^n - (2-i)*i^n, where i is the imaginary unit. Also a(n) = A001903(A159966(n)). - Bruno Berselli, Feb 08 2011
a(0)=1, a(1)=3, a(n) = 10 - a(n-2). - Vincenzo Librandi, Feb 08 2011

A010122 Continued fraction for sqrt(13).

Original entry on oeis.org

3, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6, 1, 1, 1, 1, 6

Offset: 0

N. J. A. Sloane

Eventual period is (1, 1, 1, 1, 6). - Zak Seidov, Mar 05 2011

			3.605551275463989293119221267... = 3 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 02 2009

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 96 at p. 264.
Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 428.

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

Cf. A010470 (decimal expansion).

ContinuedFraction[Sqrt[13],300] (* Vladimir Joseph Stephan Orlovsky, Mar 05 2011 *)

{ allocatemem(932245000); default(realprecision, 13000); x=contfrac(sqrt(13)); for (n=0, 20000, write("b010122.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 02 2009

From Amiram Eldar, Nov 12 2023: (Start)
Multiplicative with a(5^e) = 6, and a(p^e) = 1 for p != 5.
Dirichlet g.f.: zeta(s) * (1 + 1/5^(s-1)). (End)
G.f.: (3 + x + x^2 + x^3 + x^4 + 3*x^5)/(1 - x^5). - Stefano Spezia, Aug 17 2024

A019433 Continued fraction for tan(1/10).

Original entry on oeis.org

0, 9, 1, 28, 1, 48, 1, 68, 1, 88, 1, 108, 1, 128, 1, 148, 1, 168, 1, 188, 1, 208, 1, 228, 1, 248, 1, 268, 1, 288, 1, 308, 1, 328, 1, 348, 1, 368, 1, 388, 1, 408, 1, 428, 1, 448, 1, 468, 1, 488, 1, 508, 1, 528, 1, 548, 1, 568, 1, 588, 1, 608, 1, 628, 1, 648, 1, 668, 1, 688, 1, 708, 1, 728

Offset: 0

David W. Wilson

From Peter Bala, Oct 04 2023: (Start)
Related simple continued fractions expansions (see my comments in A019425):
tan(1/(10*k)) = [0; 10*k - 1, 1, 30*k - 2, 1, 50*k - 2, 1, 70*k - 2, 1, 90*k - 2, 1, ...] for k >= 1.
If d is a divisor of 10 with d*d' = 10 then the simple continued fraction expansion of d*tan(1/10) begins [0; d' - 1, 1, 30*d - 2, 1, 5*d' - 2, 1, 70*d - 2, 1, 9*d' - 2, 1, 110*d - 2, 1, 13*d' - 2, ...], while the simple continued fraction expansion of (1/d)*tan(1/10) begins [ 0; 10*d - 1, 1, 3*d'- 2, 1, 50*d - 2, 1, 7*d' - 2, 1, 90*d - 2, 1, 11*d' - 2, 1, 130*d - 2, ...]. (End)

			0.10033467208545054505808004... = 0 + 1/(9 + 1/(1 + 1/(28 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009

Harry J. Smith, Table of n, a(n) for n = 0..20000
G. Xiao, Contfrac
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A161019 (decimal expansion), A019425 through A019432.

LinearRecurrence[{0,2,0,-1},{0,9,1,28,1,48},80] (* or *) Join[{0,9},Riffle[NestList[20+#&,28,40],1,{1,-1,2}]] (* Harvey P. Dale, Jul 23 2023 *)

{ allocatemem(932245000); default(realprecision, 99000); x=contfrac(tan(1/10)); for (n=0, 20000, write("b019433.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 14 2009

Vec(x*(x^4-x^3+10*x^2+x+9)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Sep 08 2013

From Colin Barker, Sep 08 2013: (Start)
a(n) = -1/2+(3*(-1)^n)/2+5*n-5*(-1)^n*n for n>1.
a(n) = 2*a(n-2)-a(n-4) for n>5.
G.f.: x*(x^4-x^3+10*x^2+x+9) / ((x-1)^2*(x+1)^2). (End)

A040005 Continued fraction for sqrt(8).

Original entry on oeis.org

2, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4

Offset: 0

N. J. A. Sloane

			2.828427124746190097603377448... = 2 + 1/(1 + 1/(4 + 1/(1 + 1/(4 + ...)))). - _Harry J. Smith_, Jun 02 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.
Index entries for continued fractions for constants.
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A010466 (decimal expansion).

Essentially the same as A010685.

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

ContinuedFraction[Sqrt[8],300] (* Vladimir Joseph Stephan Orlovsky, Mar 04 2011 *)

{ allocatemem(932245000); default(realprecision, 16000); x=contfrac(sqrt(8)); for (n=0, 20000, write("b040005.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 02 2009

From Amiram Eldar, Nov 12 2023: (Start)
Multiplicative with a(2^e) = 4, and a(p^e) = 1 for an odd prime p.
Dirichlet g.f.: zeta(s) * (1 + 3/2^s). (End)
G.f.: (2 + x + 2*x^2)/(1 - x^2). - Stefano Spezia, Jul 26 2025

A041014 Numerators of continued fraction convergents to sqrt(11).

Original entry on oeis.org

3, 10, 63, 199, 1257, 3970, 25077, 79201, 500283, 1580050, 9980583, 31521799, 199111377, 628855930, 3972246957, 12545596801, 79245827763, 250283080090, 1580944308303, 4993116004999, 31539640338297

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

Cf. A010468, A041015 (denominators).

Analog for other sqrt(m): A001333 (m=2), A002531 (m=3), A001077 (m=5), A041006 (m=6), A041008 (m=7), A041010 (m=8), A005667 (m=10), A041016 (m=12), ..., A042936 (m=1000).

Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[11],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2011 *)
Numerator[Convergents[Sqrt[11], 30]] (* Vincenzo Librandi, Oct 28 2013 *)

A041014=contfracpnqn(c=contfrac(sqrt(11)), #c)[1,][^-1] \\ Discard last element which may be incorrect. Use e.g. \p999 to get more terms, or extend as follows:
{A041014_upto(N,A=Vec(A041014,N))=for(n=#A041014+1,N, A[n]=20*A[n-2]-A[n-4]); A041014=A} \\ M. F. Hasler, Nov 01 2019

G.f.: (3 + 10*x + 3*x^2 - x^3)/(1 - 20*x^2 + x^4).

A041017 Denominators of continued fraction convergents to sqrt(12).

Original entry on oeis.org

1, 2, 13, 28, 181, 390, 2521, 5432, 35113, 75658, 489061, 1053780, 6811741, 14677262, 94875313, 204427888, 1321442641, 2847313170, 18405321661, 39657956492, 256353060613, 552364077718, 3570537526921

Offset: 0

N. J. A. Sloane

a(2n+1)/a(2n) tends to 1/(sqrt(12) - 3) = 2.154700538...; e.g., a(7)/a(6) = 5432/2521 = 2.1547005...; but a(2n)/a(2n - 1) tends to 6.464101615... = sqrt(12) + 3; e.g., a(8)/a(7) = 35113/5432 = 6.46101620... - Gary W. Adamson, Mar 28 2004
The constant sqrt(12) + 3 = 6.464101615... is the "curvature" (reciprocal of the radius) of the inner or 4th circle in the Descartes circle equation; given 3 mutually tangent circles of radius 1, the radius of the innermost tangential circle = 0.1547005383... = 1/(sqrt(12) + 3). The Descartes circle equation states that given 4 mutually tangent circles (i.e., 3 tangential plus the innermost circle) with curvatures a,b,c,d (curvature = 1/r), then (a^2 + b^2 + c^2 + d^2) = 1/2(a + b + c + d)^2. - Gary W. Adamson, Mar 28 2004
Sequence also gives numerators in convergents to barover[6,2] = CF: [6,2,6,2,6,2,...] = 0.1547005... = 1/(sqrt(12) + 3), the first few convergents being 1/6, 2/13, 13/84, 28/181, 181/1170, 390/2521... with 390/2521 = 0.154700515... - Gary W. Adamson, Mar 28 2004
Sqrt(12) = 3 + continued fraction [2, 6, 2, 6, 2, 6, ...] = 6/2 + 6/13 + 6/(13*181) + 6/(181*2521) + ... - Gary W. Adamson, Dec 21 2007
Also, values i where A227790(i)/i reaches a new maximum (conjectured). - Ralf Stephan, Sep 23 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A010469, A040008, A041016 (numerators).

with (numtheory): seq( nthdenom(cfrac(sin(Pi/6)*tan(Pi/3),25),i)-nthnumer(cfrac(sin(Pi/6)*tan(Pi/3),25),i), i=2..24 ); # Zerinvary Lajos, Feb 10 2007

Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[12],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2011 *)
Denominator[Convergents[Sqrt[12],50]] (* Harvey P. Dale, Feb 18 2012 *)
a0[n_] := ((7-4*Sqrt[3])^n*(2+Sqrt[3])-(-2+Sqrt[3])*(7+4*Sqrt[3])^n)/4 // Simplify
a1[n_] := 2*Sum[a0[i], {i, 1, n}]
Flatten[MapIndexed[{a0[#],a1[#]}&,Range[11]]] (* Gerry Martens, Jul 10 2015 *)

G.f.: (1+2*x-x^2)/(1-14*x^2+x^4). - Colin Barker, Jan 01 2012
From Gerry Martens, Jul 11 2015: (Start)
Interspersion of 2 sequences [a0(n),a1(n)]:
a0(n) = ((7-4*sqrt(3))^n*(2+sqrt(3)) - (-2+sqrt(3))*(7+4*sqrt(3))^n)/4.
a1(n) = 2*Sum_{i=1..n} a0(i). (End)

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A001148 Final digit of 3^n.

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