A006276 -id:A006276 - cp's OEIS frontend

A001999 a(n) = a(n-1)*(a(n-1)^2 - 3).

3, 18, 5778, 192900153618, 7177905237579946589743592924684178

Offset: 0

The next terms in the sequence contain 102 and 305 digits. - Harvey P. Dale, Jun 09 2011
From Peter Bala, Nov 13 2012: (Start)
The present sequence is the case x = 3 of the following general remarks. For other cases see A219160 (x = 4), A219161 (x = 5) and A112845 (x = 6).
Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(3^n) + (1/alpha)^(3^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^3 - 3*a(n) with the initial condition a(0) = x.
We have the following identity, valid for x > 2: sqrt((x + 2)/(x - 2)) = (1 + 2/(x-1))*sqrt((y + 2)/(y - 2)), where y = x^3 - 3*x. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n = 0..oo} (1 + 2/(a(n) - 1)), with a(0) = x and a(n+1) = a(n)^3 - 3*a(n). The rate of convergence is cubic (Fine).
For similar results to the above see A001566 and A219162. (End)
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 08 2022

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).

G. C. Greubel, Table of n, a(n) for n = 0..7
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, Vol. 44, No. 10 (1937), pp. 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.
Walther Janous, Problem B-916, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 39, No. 2 (2001), p. 181; Subscript Is Power, Solution to Problem B-916 by H.-J. Seiffert, ibid., Vol. 40, No. 1 (2002), p. 86.
Hideyuki Ohtsu, Problem B-1316, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 4 (2022), p. 365.
Eric Weisstein's World of Mathematics, Pierce Expansion.

Cf. A000032, A006276, A001566, A002814, A045529, A112845, A219160, A219161, A219162.

NestList[#(#^2-3)&,3,6] (* Harvey P. Dale, Jun 09 2011 *)
RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 3}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)

a(n)=2*fibonacci(2*3^n+1)-fibonacci(2*3^n)

a(n) = 2*F(2*3^n+1) - F(2*3^n) = ceiling(tau^(2*3^n)) where F(k) = A000045(k) is the k-th Fibonacci number and tau is the golden ratio. - Benoit Cloitre, Nov 29 2002
From Peter Bala, Nov 13 2012: (Start)
a(n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n).
Product_{n >= 0} (1 + 2/(a(n) - 1)) = sqrt(5).
a(n) = A002814(n+1) + 1. (End)
a(n) = 2*T(3^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A219161. - Peter Bala, Feb 01 2017
From Amiram Eldar, Jan 12 2022: (Start)
a(n) = A000032(2*3^n).
a(n) = A006267(n)^2 + 2.
Product_{k=0..n} (a(k)-1) = Fibonacci(3^(n+1)) = A045529(n+1) (Janous, 2001). (End)
Sum_{n>=0} arctanh(1/a(n)) = log(5)/4 (Ohtsu, 2022). - Amiram Eldar, Dec 15 2022

A006275 Pierce expansion of sqrt(2) - 1.

Original entry on oeis.org

2, 5, 7, 197, 199, 7761797, 7761799, 467613464999866416197, 467613464999866416199, 102249460387306384473056172738577521087843948916391508591105797

Offset: 0

N. J. A. Sloane

From Peter Bala, Nov 22 2012: (Start)
For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x. The present sequence is the case x = sqrt(2) - 1.
The Pierce expansion of (sqrt(2) - 1)^(3^n) is [a(0)*a(2)*...*a(2*n), a(2*n+1), a(2*n+2), ...] = [sqrt(a(2*n+1) - 1), a(2*n+1), a(2*n+2), ...]. The Pierce expansion of (sqrt(2) - 1)^(2*3^n) is [a(2*n+1), a(2*n+2), ...]. Some examples of the associated alternating series are given below.
(End)

			Let c(0)=6, c(n+1) = c(n)^3-3*c(n); then this sequence is 2, c(0)-1, c(0)+1, c(1)-1, c(1)+1, c(2)-1, c(2)+1, ...
From _Peter Bala_, Nov 22 2012: (Start)
Let x = sqrt(2) - 1. We have the alternating series expansions
x = 1/2 - 1/(2*5) + 1/(2*5*7) - 1/(2*5*7*197) + ...
x^3 = 1/14 - 1/(14*197) + 1/(14*197*199) - ...
x^9 = 1/2786 - 1/(2786*7761797) + 1/(2786*7761797*7761799) - ...,
where 2786 = 2*7*199, and also
x^2 = 1/5 - 1/(5*7) + 1/(5*7*197) - 1/(5*7*197*199) + ...
x^6 = 1/197 - 1/(197*199) + 1/(197*199*7761797) - ...
x^18 = 1/7761797 - 1/(7761797*7761799) + ....
(End)

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

G. C. Greubel, Table of n, a(n) for n = 0..14
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36, No. 10 (1929), pp. 523-525.
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
Eric Weisstein's World of Mathematics, Pierce Expansion.

Cf. A014176, A006276.

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[Sqrt[2] - 1, 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

my(r=1+quadgen(8)); for(n=1, 10, print1(floor(r), ", "); r=r/(r-floor(r)));

Let u(0)=1+sqrt(2) and u(n+1)=u(n)/frac(u(n)) where frac(x) is the fractional part of x, then a(n)=floor(u(n)). - Benoit Cloitre, Mar 09 2004
From Peter Bala, Nov 22 2012: (Start)
a(2*n+2) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n) + 1.
a(2*n+1) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n) - 1. (End)
sqrt(2) - 1 = a(0)/a(1) + (a(0)*a(2))/(a(1)*a(3)) + (a(0)*a(2)*a(4))/(a(1)*a(3)*a(5)) + ... = 2/5 + (2*7)/(5*197) + (2*7*199)/(5*197*7761797) + ... . - Peter Bala, Dec 03 2012

More terms from James Sellers, May 19 2000

A112845 Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

Original entry on oeis.org

6, 198, 7761798, 467613464999866416198, 102249460387306384473056172738577521087843948916391508591105798

Offset: 0

Eric W. Weisstein, Sep 21 2005

Identical to A006243 apart from the initial term. For some general remarks on this recurrence see A001999. - Peter Bala, Nov 13 2012

G. C. Greubel, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977, 629-630.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276.
Cf. A006243. - R. J. Mathar, Aug 15 2008
Cf. A001999, A219160, A219161.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 6}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)
NestList[#^3-3#&,6,5] (* Harvey P. Dale, Jul 23 2025 *)

a(n) = -2*cos(3^n*arccos(-3)).
From Peter Bala, Nov 13 2012: (Start)
a(n) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n).
Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(2).
(End)

A006273 Numerators of a continued fraction for (3+sqrt(13))/2.

Original entry on oeis.org

3, 10, 1297, 2186871697, 10458512317535240383929505297

Offset: 0

N. J. A. Sloane

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

Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

For denominators see A006274.

Cf. A098316, A002814, A006268, A006271, A006275, A006276.

a := proc (n) option remember; if n = 1 then 10 else a(n-1)^3 + 3*a(n-1)^2 - 3 end if; end proc:
seq(a(n), n = 1..5); # Peter Bala, Jan 19 2022

From Peter Bala, Jan 19 2022: (Start)
a(n) = (11/2 + 3/2*sqrt(13))^3^(n-1) + (11/2 - 3/2*sqrt(13))^3^(n-1) - 1.
a(1) = 10 and a(n) = a(n-1)^3 + 3*a(n-1)^2 - 3 for n >= 2.
a(1) = 10 and a(n) = 13*(Product_{k = 1..n-1} a(k))^2 - 3 for n >= 2.
a(n) = A006268(n-1)^2 + 1 for n >= 1.
13 - 9*Product_{n = 1..N} (1 + 2/a(n))^2 = 52/(a(N+1) + 3). Therefore
sqrt(13) = 3*(1 + 2/10) * (1 + 2/1297) * (1 + 2/2186871697) * ... The convergence is cubic: the first six factors of the product give sqrt(13) correct to more than 750 decimal places.
3/sqrt(13) = (1 - 2/(10+2)) * (1 - 2/(1297+2)) * (1 - 2/(2186871697+2)) * .... (End)

A006284 Pierce expansion for Euler's constant.

Original entry on oeis.org

1, 2, 6, 13, 21, 24, 225, 615, 17450, 23228, 57774, 221361, 522377, 793040, 1706305, 8664354, 19037086, 51965160, 56870701, 124645388, 784244500, 792809072, 3675221276, 42108268014, 53633289500, 56827261536, 67080647365

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

nonn

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

G. C. Greubel, Table of n, a(n) for n = 0..1000
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276, A006283.

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[EulerGamma, 7!], 25] (* G. C. Greubel, Nov 14 2016 *)

r=1/Euler;for(n=1,30,r=r/(r-floor(r));print1(floor(r),","))

If u(0) = exp(1/m), where m is an integer >=1, and u(n+1) = u(n)/frac(u(n)) then floor(u(n)) = m*n. Let u(0)=1/gamma and u(n+1) = u(n)/frac(u(n)) where frac(x) is the fractional part of x, then a(n) = floor(u(n)) - Benoit Cloitre, Mar 09 2004

A118242 Pierce expansion of 1/phi.

Original entry on oeis.org

1, 2, 4, 17, 19, 5777, 5779, 192900153617, 192900153619, 7177905237579946589743592924684177, 7177905237579946589743592924684179

Offset: 1

Eric W. Weisstein, Apr 17 2006

Differs from A006276 in the first term only.

G. C. Greubel, Table of n, a(n) for n = 1..17
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006276. A002813, A002814.

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[(Sqrt[5] - 1)/2, 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

A118242(n) = A006276(n-1) for n>1.
From Peter Bala, Dec 03 2012: (Start)
Odd-indexed terms give A002813; even-indexed terms give A002814.
The Pierce series expansion is the alternating series 1/phi = 1/2*(sqrt(5) - 1) = 1/1 - 1/(1*2) + 1/(1*2*4) - 1/(1*2*4*17) + 1/(1*2*4*17*19) - ....
Another series expansion is
1/phi = a(1)/a(2) + (a(1)*a(3))/(a(2)*a(4)) + (a(1)*a(3)*a(5))/(a(2)*a(4)*a(6)) + ... = 1/2 + (1*4)/(2*17) + (1*4*19)/(2*17*5777) + ....
(End)

A006271 Numerators of a continued fraction for 1 + sqrt(2).

Original entry on oeis.org

2, 5, 197, 7761797, 467613464999866416197, 102249460387306384473056172738577521087843948916391508591105797

Offset: 0

N. J. A. Sloane

With b(n) = floor((1+sqrt(2))^n) (cf. A080039) the terms appear to be b(2*3^n). - Joerg Arndt, Apr 29 2013

Note that 1 + sqrt(2) = (c + sqrt(c^2+4))/2 and has regular continued fraction [c, c, ...] with c = 2. With b(n) = A006266(n), it can be expanded into an irregular continued fraction f(1) = b(1) and f(n) = (b[n-1]^2+1)/(b[n]-b[n-1]), and numerator(f(n)) = a(n) (cf. Shallit). - Michel Marcus, Apr 29 2013

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

F. L. Bauer, Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.
Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

For denominators see A006272. Cf. A002814, A006266, A006273, A006275, A006276.

a := proc (n) option remember; if n = 1 then 5 else a(n-1)^3 + 3*a(n-1)^2 - 3 end if; end proc:
seq(a(n), n = 1 .. 5); # Peter Bala, Jan 19 2022

From Peter Bala, Jan 18 2022: (Start)
a(n) = (3 + 2*sqrt(2))^3^(n-1) + (3 - 2*sqrt(2))^3^(n-1) - 1 for n >= 1.
a(n) = A006266(n)^2 + 1 for n >= 1.
a(1) = 5 and a(n) = a(n-1)^3 + 3*a(n-1)^2 - 3 for n >= 2.
a(1) = 5 and a(n) = 8*(Product_{k = 1..n-1} a(k))^2 - 3 for n >= 2.
2 - Product_{n = 1..N} (1 + 2/a(n))^2 = 8/(a(N+1) + 3). Therefore
sqrt(2) = (1 + 2/5) * (1 + 2/197) * (1 + 2/7761797) * (1 + 2/ 467613464999866416197) * ... - see Bauer.
The convergence is cubic - see Fine. The first six factors of the product give sqrt(2) correct to more than 500 decimal places. (End)

Previous values for a(3) and a(4) were 776 and 1797. They have been merged into 7761797 to reflect the 2nd continued fraction on page 6 of Shallit paper by Michel Marcus, Apr 29 2013

A091831 Pierce expansion of 1/sqrt(2).

Original entry on oeis.org

1, 3, 8, 33, 35, 39201, 39203, 60245508192801, 60245508192803, 218662352649181293830957829984632156775201, 218662352649181293830957829984632156775203

Offset: 0

Benoit Cloitre, Mar 09 2004

nonn

If u(0)=exp(1/m) m integer>1 and u(n+1)=u(n)/frac(u(n)) then floor(u(n))=m*n.

G. C. Greubel, Table of n, a(n) for n = 0..14
P. Erdős and Jeffrey Shallit, New bounds on the length of finite Pierce and Engel series, Sem. Théor. Nombres Bordeaux (2) 3 (1991), no. 1, 43-53.
Vlado Keselj, Length of Finite Pierce Series: Theoretical Analysis and Numerical Computations .
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
Pelegrí Viader, Lluís Bibiloni, Jaume Paradís, On a problem of Alfred Renyi, Economics Working Paper No. 340.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276, A006283.

Cf. A006784 (Pierce expansion definition), A028254

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[2^(-1/2), 7!], 17] (* G. C. Greubel, Nov 13 2016 *)

r=sqrt(2);for(n=1,10,r=r/(r-floor(r));print1(floor(r),","))

Let u(0)=sqrt(2) and u(n+1)=u(n)/frac(u(n)) where frac(x) is the fractional part of x, then a(n)=floor(u(n)).
1/sqrt(2)= 1/a(1) - 1/a(1)/a(2) + 1/a(1)/a(2)/a(3) - 1/a(1)/a(2)/a(3)/a(4)...
limit n -> infinity a(n)^(1/n) = e.

A219506 Pierce expansion of 2 - sqrt(3).

Original entry on oeis.org

3, 5, 51, 53, 140451, 140453, 2770663499604051, 2770663499604053, 21269209556953516583554114034636483645584976451, 21269209556953516583554114034636483645584976453

Offset: 0

Peter Bala, Nov 22 2012

For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x.
The present sequence is the case x = 2 - sqrt(3).
Shallit has shown that the Pierce expansion of the quadratic irrational (c - sqrt(c^2 - 4))/2 has the form [c(0) - 1, c(0) + 1, c(1) - 1, c(1) + 1, c(2) - 1, c(2) + 1, ...], where c(0) = c and c(n+1) = c(n)^3 - 3*c(n). This is the case c = 4. For other cases see A006276 (c = 3), A219507 (c = 5) and A006275 (essentially c = 6 apart from the initial term).
The Pierce expansion of ((c - sqrt(c^2 - 4))/2)^(3^n) is [c(n) - 1, c(n) + 1, c(n+1) - 1, c(n+1) + 1, c(n+2) - 1, c(n+2) + 1, ...].

			We have the alternating series expansions
2 - sqrt(3) = 1/3 - 1/(3*5) + 1/(3*5*51) - 1/(3*5*51*53) + ...
(2 - sqrt(3))^3 = 1/51 - 1/(51*53) + 1/(51*53*140451) - ...
(2 - sqrt(3))^9 = 1/140451 - 1/(140451*140453) + ....

G. C. Greubel, Table of n, a(n) for n = 0..13
F. L. Bauer, Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276, A112845, A219160, A219507, A219508.

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[2 - Sqrt[3] , 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

a(2*n) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) - 1.
a(2*n + 1) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) + 1.
From Peter Bala, Jan 18 2022: (Start)
a(2*n+2) = a(2*n)^3 + 3*a(2*n)^2 - 3; a(2*n+1) = a(2*n-1)^3 - 3*a(2*n-1)^2 + 3.
a(2*n) = 6*(Product_{k = 1..n-1} a(2*k))^2 - 3, with a(0) = 1;
a(2*n+1) = 2*(Product_{k = 0..n-1} a(2*k+1))^2 + 3, with a(1) = 5.
sqrt(3) = (1 + 2/3)*(1 + 2/51)*(1 + 2/140451)*(1 + 2/2770663499604051)* ....  See Bauer.
1/sqrt(3) = (1 - 2/5)*(1 - 2/53)*(1 - 2/140453)*(1 - 2/2770663499604053)* .... (End)

A219507 Pierce expansion of (5 - sqrt(21))/2.

Original entry on oeis.org

4, 6, 109, 111, 1330669, 1330671, 2356194280407770989, 2356194280407770991, 13080769480548649962914459850235688797656360638877986029, 13080769480548649962914459850235688797656360638877986031

Offset: 0

Peter Bala, Nov 22 2012

For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x.
The present sequence is the case x = 1/2*(5 - sqrt(21)).
Jeffrey Shallit has shown that the Pierce expansion of the quadratic irrational (c - sqrt(c^2 - 4))/2 has the form [c(0) - 1, c(0) + 1, c(1) - 1, c(1) + 1, c(2) - 1, c(2) + 1, ...], where c(0) = c and c(n+1) = c(n)^3 - 3*c(n). This is the case c = 5. For other cases see A006276 (c = 3), A219506 (c = 4) and A006275 (essentially c = 6 apart from the initial term).
The Pierce expansion of ((c - sqrt(c^2 - 4))/2)^(3^n) is [c(n) - 1, c(n) + 1, c(n+1) - 1, c(n+1) + 1, c(n+2) - 1, c(n+2) + 1, ...].

			Let x = 1/2*(5 - sqrt(21)). We have the alternating series expansions
x = 1/4 - 1/(4*6) + 1/(4*6*109) - 1/(4*6*109*111) + ...
x^3 = 1/109 - 1/(109*111) + 1/(109*111*1330669) - ...
x^9 = 1/1330669 - 1/(1330669*1330671) + ....

G. C. Greubel, Table of n, a(n) for n = 0..13
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276, A219160, A219506, A219508.

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[(5 - Sqrt[21])/2 , 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

a(2*n) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n) - 1.

a(2*n+1) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n) + 1.

cp's OEIS Frontend

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A006273 Numerators of a continued fraction for (3+sqrt(13))/2.

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A006271 Numerators of a continued fraction for 1 + sqrt(2).

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