A041046 -id:A041046 - cp's OEIS frontend

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127

Offset: 0

Paul Barry, Aug 16 2003

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).
The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005
a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence L_{5,n}.
Tanya Khovanova, Recursive Sequences
Wikipedia, Metallic mean
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A006497, A014448, A085447.

Cf. A086902, A000032, A052918.

I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)

{a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */

[lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.
a(n) = A100236(n) + 1.
E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007
a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008
A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010
a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010
a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020
From Peter Bala, Jul 09 2025 : (Start)
The following series telescope (Cf. A000032):
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A041018 Numerators of continued fraction convergents to sqrt(13).

Original entry on oeis.org

3, 4, 7, 11, 18, 119, 137, 256, 393, 649, 4287, 4936, 9223, 14159, 23382, 154451, 177833, 332284, 510117, 842401, 5564523, 6406924, 11971447, 18378371, 30349818, 200477279, 230827097, 431304376, 662131473

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 (0,0,0,0,36,0,0,0,0,1).

Cf. A010122 (continued fraction for sqrt(13)).

Cf. A010470, A041019 (denominators), A041046, A041090, A041150, A041226, A041318, A041426 and A041550.

a[0]:=3: a[-1]:=1: b(0):=6: b(1):=1; b(2):=1: b(3):=1: b(4):=1:
for n from 1 to 100 do  k:=n mod 5:
   a[n]:=b(k)*a[n-1]+a[n-2]:
   printf("%12d", a[n]):
end do: # Paul Weisenhorn, Aug 17 2018

Numerator[Convergents[Sqrt[13], 30]] (* Vincenzo Librandi, Oct 27 2013 *)
CoefficientList[Series[(3 + 4*x + 7*x^2 + 11*x^3 + 18*x^4 + 11*x^5 - 7*x^6 + 4*x^7 - 3*x^8 + x^9)/(1 - 36*x^5 - x^10),{x,0,50}],x] (* Stefano Spezia, Aug 31 2018 *)

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5*n) = A006497(3*n+1),
a(5*n+1) = (A006497(3*n+2)-A006497(3*n+1))/2,
a(5*n+2) = (A006497(3*n+2)+A006497(3*n+1))/2,
a(5*n+3) = A006497(3*n+2),
a(5*n+4) = A006497(3*n+3)/2.
(End)
G.f.: (3 + 4*x + 7*x^2 + 11*x^3 + 18*x^4 + 11*x^5 - 7*x^6 + 4*x^7 - 3*x^8 + x^9)/(1 - 36*x^5 - x^10). - Peter J. C. Moses, Jul 29 2013
a(n) = A010122(n)*a(n-1)+a(n-2) with a(0)=3, a(-1)=1. - Paul Weisenhorn, Aug 19 2018

A041047 Denominators of continued fraction convergents to sqrt(29).

Original entry on oeis.org

1, 2, 3, 5, 13, 135, 283, 418, 701, 1820, 18901, 39622, 58523, 98145, 254813, 2646275, 5547363, 8193638, 13741001, 35675640, 370497401, 776670442, 1147167843, 1923838285, 4994844413, 51872282415

Offset: 0

N. J. A. Sloane

The terms of this sequence can be constructed with the terms of sequence A052918.

For the terms of the periodical sequence of the continued fraction for sqrt(29) see A010128. We observe that its period is five. The decimal expansion of sqrt(29) is A010484. - Johannes W. Meijer, Jun 12 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,140,0,0,0,0,1).

Cf. A041046.

Cf. A041019, A041047, A041091, A041151, A041227, A041319, A041427, A041551.

I:=[1, 2, 3, 5, 13, 135, 283, 418, 701, 1820]; [n le 10 select I[n] else 140*Self(n-5)+Self(n-10): n in [1..50]]; // Vincenzo Librandi, Dec 10 2013

Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[29],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 18 2011 *)
Denominator[Convergents[Sqrt[29], 30]] (* Vincenzo Librandi, Dec 10 2013 *)

a(5*n) = A052918(3*n), a(5*n+1) = (A052918(3*n+1) - A052918(3*n))/2, a(5*n+2) = (A052918(3*n+1) + A052918(3*n))/2, a(5*n+3) = A052918(3*n+1) and a(5*n+4) = A052918(3*n+2)/2. - Johannes W. Meijer, Jun 12 2010
G.f.: (1 + 2*x + 3*x^2 + 5*x^3 + 13*x^4 - 5*x^5 + 3*x^6 - 2*x^7 + x^8)/(1 - 140*x^5 - x^10). - Peter J. C. Moses, Jul 29 2013
a(n) = 140*a(n-5) + a(n-10). - Vincenzo Librandi, Dec 10 2013

A041090 Numerators of continued fraction convergents to sqrt(53).

Original entry on oeis.org

7, 22, 29, 51, 182, 2599, 7979, 10578, 18557, 66249, 946043, 2904378, 3850421, 6754799, 24114818, 344362251, 1057201571, 1401563822, 2458765393, 8777860001, 125348805407, 384824276222, 510173081629, 894997357851, 3195165155182, 45627309530399

Offset: 0

N. J. A. Sloane

The terms of this sequence can be constructed with the terms of sequence A086902. For the terms of the periodical sequence of the continued fraction for sqrt(53) see A010139. We observe that its period is five. The decimal expansion of sqrt(53) is A010506. - Johannes W. Meijer, Jun 12 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,364,0,0,0,0,1).

Cf. A041091, A041018, A041046, A041150, A041226, A041318, A041426 and A041550.

Numerator[Convergents[Sqrt[53],30]] (* Harvey P. Dale, Sep 24 2013 *)
CoefficientList[Series[-(x^9 - 7 x^8 + 22 x^7 - 29 x^6 + 51 x^5 + 182 x^4 + 51 x^3 + 29 x^2 + 22 x + 7)/(x^10 + 364 x^5 - 1), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 27 2013 *)

a(5*n) = A086902(3*n+1), a(5*n+1) = (A086902(3*n+2)-A086902(3*n+1))/2, a(5*n+2) = (A086902(3*n+2)+A086902(3*n+1))/2, a(5*n+3) = A086902(3*n+2) and a(5*n+4) = A086902(3*n+3)/2. - Johannes W. Meijer, Jun 12 2010

G.f.: -(x^9-7*x^8+22*x^7-29*x^6+51*x^5+182*x^4+51*x^3+29*x^2+22*x+7) / (x^10+364*x^5-1). - Colin Barker, Sep 26 2013

More terms from Colin Barker, Sep 26 2013

A041150 Numerators of continued fraction convergents to sqrt(85).

Original entry on oeis.org

9, 37, 46, 83, 378, 6887, 27926, 34813, 62739, 285769, 5206581, 21112093, 26318674, 47430767, 216041742, 3936182123, 15960770234, 19896952357, 35857722591, 163327842721, 2975758891569, 12066363408997

Offset: 0

N. J. A. Sloane

From Johannes W. Meijer, Jun 17 2010: (Start)
The a(n) terms of this sequence can be constructed with the terms of sequence A087798.
For the terms of the periodic sequence of the continued fraction for sqrt(85) see A010158. We observe that its period is five. The decimal expansion of sqrt(85) is A010536. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,756,0,0,0,0,1).

Cf. A010536, A041018, A041046, A041090, A041150, A041151, A041226, A041318, A041426,

A041550.

Numerator[Convergents[Sqrt[85], 30]] (* Vincenzo Librandi, Oct 29 2013 *)

From Johannes W. Meijer, Jun 17 2010: (Start)
a(5*n) = A087798(3*n+1), a(5*n+1) = (A087798(3*n+2) - A087798(3*n+1))/2, a(5*n+2) = (A087798(3*n+2) + A087798(3*n+1))/2, a(5*n+3) = A087798(3*n+2) and a(5*n+4) = A087798(3*n+3)/2. (End)
G.f.: -(x^9-9*x^8+37*x^7-46*x^6+83*x^5+378*x^4+83*x^3+46*x^2+37*x+9) / (x^10+756*x^5-1). - Colin Barker, Nov 04 2013

A041226 Numerators of continued fraction convergents to sqrt(125).

Original entry on oeis.org

11, 56, 67, 123, 682, 15127, 76317, 91444, 167761, 930249, 20633239, 104096444, 124729683, 228826127, 1268860318, 28143753123, 141987625933, 170131379056, 312119004989, 1730726404001, 38388099893011, 193671225869056, 232059325762067, 425730551631123

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 A001946.
For the terms of the periodical sequence of the continued fraction for sqrt(125) see A010186. 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 (0,0,0,0,1364,0,0,0,0,1).

Cf. A041227, A041018, A041046, A041090, A041150, A041226, A041318, A041426, A041550.

Numerator[Convergents[Sqrt[125], 30]] (* Vincenzo Librandi, Oct 31 2013 *)

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5n)   = A001946(3n+1),
a(5n+1) = (A001946(3n+2) - A001946(3n+1))/2,
a(5n+2) = (A001946(3n+2) + A001946(3n+1))/2,
a(5n+3) = A001946(3n+2),
a(5n+4) = A001946(3n+3)/2. (End)
G.f.: -(x^9 -11*x^8 +56*x^7 -67*x^6 +123*x^5 +682*x^4 +123*x^3 +67*x^2 +56*x +11) / ((x^2 +4*x -1)*(x^4 -7*x^3 +19*x^2 -3*x +1)*(x^4 +3*x^3 +19*x^2 +7*x +1)). - Colin Barker, Nov 08 2013

More terms from Colin Barker, Nov 08 2013

A041318 Numerators of continued fraction convergents to sqrt(173).

Original entry on oeis.org

13, 79, 92, 171, 1118, 29239, 176552, 205791, 382343, 2499849, 65378417, 394770351, 460148768, 854919119, 5589663482, 146186169651, 882706681388, 1028892851039, 1911599532427, 12498490045601, 326872340718053, 1973732534353919, 2300604875071972

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, 0, 0, 0, 2236, 0, 0, 0, 0, 1).

Cf. A041319, A041018, A041046, A041090, A041150, A041226, A041318, A041426, A041550.

Cf. A010217 (continued fraction).

Numerator[Convergents[Sqrt[173], 30]] (* Vincenzo Librandi, Nov 01 2013 *)
LinearRecurrence[{0,0,0,0,2236,0,0,0,0,1},{13,79,92,171,1118,29239,176552,205791,382343,2499849},30] (* Harvey P. Dale, Jul 28 2018 *)

a(5*n) = A088316(3*n+1), a(5*n+1) = (A088316(3*n+2) - A088316(3*n+1))/2, a(5*n+2) = (A088316(3*n+2)+A088316(3*n+1))/2, a(5*n+3) = A088316(3*n+2) and a(5*n+4) = A088316(3*n+3)/2. [Johannes W. Meijer, Jun 12 2010]

G.f.: -(x^9-13*x^8+79*x^7-92*x^6+171*x^5+1118*x^4+171*x^3+92*x^2+79*x+13) / (x^10+2236*x^5-1). - Colin Barker, Nov 08 2013

More terms from Colin Barker, Nov 08 2013

A041426 Numerators of continued fraction convergents to sqrt(229).

Original entry on oeis.org

15, 106, 121, 227, 1710, 51527, 362399, 413926, 776325, 5848201, 176222355, 1239404686, 1415627041, 2655031727, 20000849130, 602680505627, 4238764388519, 4841444894146, 9080209282665, 68402909872801, 2061167505466695, 14496575448139666, 16557742953606361

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 A090301.
For the terms of the periodical sequence of the continued fraction for sqrt(229) see A040213. 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, 3420, 0, 0, 0, 0, 1).

Cf. A041427, A041018, A041046, A041090, A041150, A041226, A041318, A041426, A041550.

Numerator[Convergents[Sqrt[229], 30]] (* Vincenzo Librandi, Nov 01 2013 *)
LinearRecurrence[{0,0,0,0,3420,0,0,0,0,1},{15,106,121,227,1710,51527,362399,413926,776325,5848201},30] (* Harvey P. Dale, Dec 19 2016 *)

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5n) = A090301(3n+1), a(5n+1) = (A090301(3n+2) - A090301(3n+1))/2, a(5n+2) = (A090301(3n+2) + A090301(3n+1))/2, a(5n+3) = A090301(3n+2) and a(5n+4) = A090301(3n+3)/2. (End)
G.f.: -(x^9-15*x^8+106*x^7-121*x^6+227*x^5+1710*x^4+227*x^3+121*x^2+106*x+15) / (x^10+3420*x^5-1). - Colin Barker, Nov 08 2013

More terms from Colin Barker, Nov 08 2013

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

Original entry on oeis.org

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

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Original entry on oeis.org

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

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