A076725 -id:A076725 - cp's OEIS frontend

A092526 Decimal expansion of (2/3)cos( (1/3)arccos(29/2) ) + 1/3, the real root of x^3 - x^2 - 1.

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

Offset: 1

N. J. A. Sloane, Apr 07 2004

This is the limit x of the ratio N(n+1)/N(n) for n -> infinity of the Narayana sequence N(n) = A000930(n). The real root of x^3 - x^2 - 1. See the formula section. - Wolfdieter Lang, Apr 24 2015
This is the fourth smallest Pisot number. - Iain Fox, Oct 13 2017
Sometimes called the supergolden ratio or Narayana's cows constant, and denoted by the symbol psi. - Ed Pegg Jr, Feb 01 2019

			1.46557123187676802665673122521993910802557756847228570164318311124926...

S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.3.
Paul J. Nahin, The Logician and the Engineer, How George Boole and Claude Shannon Created the Information Age, Princeton University Press, Princeton and Oxford, 2013, Chap. 7: Some Combinational Logic Examples, Section 7.1: Channel Capacity, Shannon's Theorem, and Error-Detection Theory, page 120.

Harry J. Smith, Table of n, a(n) for n = 1..20000
Simon Baker, Exceptional digit frequencies and expansions in non-integer bases, arXiv:1711.10397 [math.DS], 2017. See the beta(2) constant pp. 3-4.
H. R. P. Ferguson, On a Generalization of the Fibonacci Numbers Useful in Memory Allocation Schema or All About the Zeroes of Z^k - Z^{k - 1} - 1, k > 0, Fibonacci Quarterly, Volume 14, Number 3, October, 1976 (see Table 2, p. 236).
Ed Pegg Jr., Images based on the supergolden ratio
Michael Penn, What is the super-golden ratio??, YouTube video, 2022.
Mario Raso, Integer Sequences in Cryptography: A New Generalized Family and its Application, Ph. D. Thesis, Sapienza University of Rome (Italy 2025). See p. 59.
Eric Weisstein's World of Mathematics, Pisot Number.
Eric Weisstein's World of Mathematics, Supergolden Ratio.
Wikipedia, Pisot number
Wikipedia, Supergolden ratio
Index entries for algebraic numbers, degree 3

Cf. A088559, A075778, A076725, A000930, A263719.
Other Pisot numbers: A060006, A086106, A228777, A293506, A293508, A293509, A293557.
Cf. A381124 (numerators of convergents).
Cf. A381125 (denominators of convergents).

RealDigits[(2 Cos[ ArcCos[ 29/2]/3] + 1)/3, 10, 111][[1]] (* Robert G. Wilson v, Apr 12 2004 *)
RealDigits[ Solve[ x^3 - x^2 - 1 == 0, x][[1, 1, 2]], 10, 111][[1]] (* Robert G. Wilson v, Oct 10 2013 *)

allocatemem(932245000); default(realprecision, 20080); x=solve(x=1, 2, x^3 - x^2 - 1); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b092526.txt", n, " ", d));  \\ Harry J. Smith, Jun 21 2009

The real root of x^3 - x^2 - 1. - Franklin T. Adams-Watters, Oct 12 2006
The only real irrational root of x^4-x^2-x-1 (-1 is also a root). [Nahim]
Equals (2/3)*cos( (1/3)*arccos(29/2) ) + 1/3.
Equals 1 + A088559.
Equals (1/6)*(116+12*sqrt(93))^(1/3) + 2/(3*(116+12*sqrt(93))^(1/3)) + 1/3. - Vaclav Kotesovec, Dec 18 2014
Equals 1/A263719. - Alois P. Heinz, Apr 15 2018
Equals (1 + 1/r + r)/3 where r = ((29 + sqrt(837))/2)^(1/3). - Peter Luschny, Apr 04 2020
Equals (1/3)*(1 + ((1/2)*(29 + (3*sqrt(93))))^(1/3) + ((1/2)*(29 - 3*sqrt(93)))^(1/3)). See A075778. - Wolfdieter Lang, Aug 17 2022

A263719 Decimal expansion of the real root r of r^3 + r - 1 = 0.

Original entry on oeis.org

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

Offset: 0

Paul D. Hanna, Oct 24 2015

Constant from Narayana's cows sequence: Limit A000930(n)/A000930(n+1) = r.

Reciprocal of constant described by A092526.

			0.682327803828019327369483739711048256891188581897998577803728606639896...

G. C. Greubel, Table of n, a(n) for n = 0..5000
Cyril Banderier and Michael Wallner, Lattice paths with catastrophes, arXiv:1707.01931 [math.CO], 2017. See Corollary 4.22. on p. 24.
Index entries for algebraic numbers, degree 3

Cf. A000930, A025157, A076725, A092526, A274112.

RealDigits[ ((Sqrt[93] + 9)/18)^(1/3) - ((Sqrt[93] - 9)/18)^(1/3), 10, 100][[1]] (* G. C. Greubel, May 01 2017 *)

a(n) = my(r = (sqrt(93)/18 + 1/2)^(1/3) - (sqrt(93)/18 - 1/2)^(1/3)); floor(r*10^(n+1))%10
for(n=0,120,print1(a(n),", "))

solve(r=0, 1,  r^3 + r - 1 ) \\ Michel Marcus, Oct 25 2015

r = (sqrt(93)/18 + 1/2)^(1/3) - (sqrt(93)/18 - 1/2)^(1/3).
Constant r satisfies:
(1) 1/(1 - r*i) = (r + r^2*i) where i^2 = -1.
(2) r = real( 1/(1 - r*i) ).
(3) r = norm( 1/(1 - r*i) ).
(4) r = r^2 + r^4.
Equals 1/A092526. - Vaclav Kotesovec, Nov 27 2017

A005154 a(0) = 1, a(1) = 2; thereafter a(n) = 3a(n-1)^2 - 2a(n-2)^4.

Original entry on oeis.org

1, 2, 10, 268, 195472, 104310534400, 29722161121961969778688, 2413441860555924454205324333893477339897004032, 15913289476042091181119569948276231488639535067163704670852319029791565485433738366445158400

Offset: 0

N. J. A. Sloane

A lower bound for maximal number of stable matchings (or marriages) possible with 2^n men and 2^n women for suitable preference ordering.

D. Gusfield and R. W. Irving, The Stable Marriage Problem: Structure and Algorithms. MIT Press, 1989, p. 25.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. W. Irving and P. Leather, The complexity of counting stable marriages, SIAM J. Computing 15 (1986), 655-667. [The sequence is v_n =g(2^n), where g(n) appears on page p. 657.]
Anna R. Karlin, Shayan Oveis Gharan, and Robbie Weber, A Simply Exponential Upper Bound on the Maximum Number of Stable Matchings, arXiv:1711.01032 [cs.DM], 2017.
Anna R. Karlin, Shayan Oveis Gharan, and Robbie Weber, A Simply Exponential Upper Bound on the Maximum Number of Stable Matchings, STOC 2018: Proceedings of the 50th Annual ACM SIGACT Symposium on Theory of Computing, June 2018, pp. 920-925.
D. E. Knuth, Mariages Stables, Presses Univ. de Montréal, 1976 (gives 10 matchings illustrating a(2)).
J. C. Lagarias, J. H. Spencer and J. P. Vinson, Counting dyadic equipartitions of the unit square, Discrete Math. 257 (2002), 481-499.
Clayton Thomas, A recurrence giving a lower bound for stable matchings (analysis of the asymptotic behavior of a_n, with proof due to Peter Shor)
E. G. Thurber, Concerning the maximum number of stable matchings in the stable marriage problem, Discrete Math., 248 (2002), 195-219 (see Eq. (1)).

Recurrence is similar to A076725.

I:=[1,2]; [m le 2 select I[m] else 3*Self(m-1)^2-2*Self(m-2)^4: m in [1..9]]; // Marius A. Burtea, Aug 09 2019

A005154 := proc(n) option remember; if n <= 1 then n+1 else 3*A005154(n-1)^2-2*A005154(n-2)^4; fi; end;

RecurrenceTable[{a[0]==1,a[1]==2,a[n]==3a[n-1]^2-2a[n-2]^4},a,{n,8}] (* Harvey P. Dale, Mar 19 2012 *)

a(n) ~ r*s^(2^n), where r = (sqrt(3)-1)/2 = 0.366025... and s = 2.28014... . - Clayton Thomas, Aug 09 2019

The Karlin, Gharan, Weber upper bound is C^(2^n) for a large C. - Domotor Palvolgyi, Feb 09 2020

Formula and comment swapped by N. J. A. Sloane, Mar 01 2020

A338293 Number of matchings in the complete binary tree of n rows.

Original entry on oeis.org

1, 1, 3, 15, 495, 467775, 448046589375, 396822986774382287109375, 316789536051348354157896789538095888519287109375

Offset: 0

Kevin Ryde, Oct 21 2020

nonn

A recurrence is formed by considering the root vertex matched or unmatched and a(n-1) or a(n-2) matchings in the subtrees below.
    unmatched         matched            matched
     /    \           /   \\             //   \
  any     any      any   matched     matched  any
                         /   \        /   \
                       any   any    any   any
  so:
    a(n-1)^2     +       2 * a(n-1)*a(n-2)^2     = a(n)
The Jacobsthal product formula (below) follows from this recurrence by induction by substituting the products for a(n-1) and a(n-2) and using J(n+1) = J(n) + 2*J(n-1) (its recurrence in A001045).
The Jacobsthal product terms, with multiplicity, in a(n) are a subset of the terms in any bigger a(m), so a(n) divides any bigger a(m) and so in particular this is a divisibility sequence.
Asymptotically, a(n) ~ (1/2)*C^(2^n) where C = 1.537176.. = A338294.  For growth power C, let c(n) = (2*a(n))^(1/2^n) so that C = lim_{n->oo} c(n).  The Jacobsthal products formula gives log(c(n)) = log(2)/2^n + log(J(n+1))/2^n + Sum_{k=1..n} log(J(k))/2^k.  Then discarding log(J(1)) = log(J(2)) = 0, and log(2)/2^n -> 0, and log(J(n+1))/2^n -> 0, leaves the terms of A242049 so that log(C) = A242049.
The asymptotic factor F = 1/2 is found by letting f(n) = a(n)/a(n-1)^2, so f(n) = J(n+1) / J(n) by the products formula, and f(n) = 2 + (-1)^n/J(n) -> 2 = 1/F.  This factor makes no difference to the growth power C, since any F^(1/2^n) -> 1, but it brings the approximation closer to a(n) sooner.

			n=0 rows is the empty tree and n=1 row is a single vertex.  Both have only the empty matching so a(0) = a(1) = 1.
n=2 rows is a path-3 and has 3 matchings: first two vertices, last two, or the empty matching, so a(2) = 3.
Jacobsthal products formula:
  a(4) = J(5) * J(4) * J(3)^2 * J(2)^4 * J(1)^8
       =  11  *   5  *    3^2 *    1^4 *    1^8 = 495.

Kevin Ryde, Table of n, a(n) for n = 0..12
Kevin Ryde, vpar examples/complete-binary-matchings.gp calculations and code in PARI/GP.
Index entries for sequences of form a(n+1)=a(n)^2 + ...
Index to divisibility sequences

Cf. A001045 (Jacobsthal numbers), A338294 (growth power), A242049 (log of growth power).

Cf. A076725 (independent sets), A158681 (Wiener index), A000975 (independence number and matching number).

a(n) = my(x=1,y=1); for(i=2,n, [x,y] = [x^2 + 2*x*y^2, x]); x;

a(n) = a(n-1)^2 + 2*a(n-1)*a(n-2)^2 starting a(0)=1 and a(1)=1.
a(n) = J(n+1) * J(n) * J(n-1)^2 * J(n-2)^4 * ... * J(1)^(2^(n-1)) where J(n) = (2^n - (-1)^n)/3 = A001045(n) is the Jacobsthal numbers.
A228607(n) = a(n+1)^2*(a(n+1)+3*a(n)^2). - R. J. Mathar, Jul 22 2022

A128384 a(n) = numerator of r(n): r(n) is such that the continued fraction (of rational terms) [r(1);r(2),...,r(n)] = b(n) for every positive integer n, where b(1) = 1 and b(n+1) = 1 + 1/b(n)^2 for.every positive integer n.

Original entry on oeis.org

1, 1, 1, 9, 91, 1110627, 2340346013551, 1153990596161360976412397283, 4046008078695782327150934410204261346038863974869654053

Offset: 1

Leroy Quet, Feb 28 2007

b(n) = A076725(n)/A076725(n-1)^2. The limit, as n -> infinity, of r(n)*r(n+1) = (2 /x^3) + (x^3 /2) - 2, where x is the real root of x^3 -x^2 -1 = 0. (This limit result needs some checking.)

a(10) has 113 digits. - Michel Marcus, Jan 13 2014

			{r(n)}: 1, 1, 1/3, 9/13, 91/289,...
b(4) = 41/25 = 1 + 1/(1 + 1/(1/3 + 13/9)).
And b(5) = 2306/1681 = 1 + 1/(1 + 1/(1/3 + 1/(9/13 + 289/91))).

Cf. A128385, A076725.

lista(nn) = {kill(n); kill(m); nn1 = nn + 1; v = vector(nn1); v[1] = 1; v[2] = 1; for (i=3, nn1, v[i] = v[i-1]^2 + v[i-2]^4;); b = vector(nn1, i, if (i==1, 1, v[i]/v[i-1]^2)); r = vector(nn); r[1] = b[1]; for (i=2, nn, pol = (n/m); forstep (k = i-1, 1, -1, pol = 1/pol + r[k];); mat = matrix(2, 2); mat[1, 1] = polcoeff( polcoeff( numerator(pol), 1, n), 0, m); mat[1, 2] = polcoeff( polcoeff( numerator(pol), 1, m), 0, n); mat[2, 1] = polcoeff( polcoeff( denominator(pol), 1, n), 0, m); mat[2, 2] = polcoeff( polcoeff( denominator(pol), 1, m), 0, n); col = vector(2); col[1] = numerator(b[i+1]); col[2] = denominator(b[i+1]); vres = matsolve(mat, col~); r[i] = vres[1]/vres[2];); r;} \\ Michel Marcus, Jan 12 2014

More terms from Michel Marcus, Jan 12 2014

A128385 a(n) = denominator of r(n): r(n) is such that the continued fraction (of rational terms) [r(1);r(2),...,r(n)] = b(n) for every positive integer n, where b(1) = 1 and b(n+1) = 1 + 1/b(n)^2 for.every positive integer n.

Original entry on oeis.org

1, 1, 3, 13, 289, 1645423, 7499988983197, 1716234423353399580977511919, 12985299047930678223817284541389710796223289877600061663

Offset: 1

Leroy Quet, Feb 28 2007

b(n) = A076725(n)/A076725(n-1)^2. The limit, as n -> infinity, of r(n)*r(n+1) = (2 /x^3) + (x^3 /2) - 2, where x is the real root of x^3 -x^2 -1 = 0. (This limit result needs some checking.)

a(10) has 113 digits. - Michel Marcus, Jan 13 2014

			{r(n)}: 1, 1, 1/3, 9/13, 91/289,...
b(4) = 41/25 = 1 + 1/(1 + 1/(1/3 + 13/9)).
And b(5) = 2306/1681 = 1 + 1/(1 + 1/(1/3 + 1/(9/13 + 289/91))).

Cf. A128384, A076725.

PARI
```
see A128384.
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More terms from Michel Marcus, Jan 12 2014

A144229 The numerators of the convergents to the recursion x=1/(x^2+1).

Original entry on oeis.org

1, 1, 4, 25, 1681, 5317636, 66314914699609, 8947678119828215014722891025, 178098260698995011212395018312912894502905113202338936836

Offset: 0

Cino Hilliard, Sep 15 2008

The recursion converges to the real root of 1/(x^2+1) - x = 0, 0.682327803...
An interesting consequence of this result occurs if we multiply by x^2+1 to get 1-x-x^3=0. These different equations intersect at the same root 0.682327803... Note also that a(n) is a square. The square roots form sequence A076725.
a(n) is the number of (0,1)-labeled perfect binary trees of height n such that no adjacent nodes have 1 as the label and the root is labeled 0. - Ran Pan, May 22 2015

Ran Pan, Exercise R, Project P.

Cf. A076725, A088559.

f[n_]:=(n+1/n)/n;Prepend[Denominator[NestList[f,2,7]],1] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
RecurrenceTable[{a[n]==(a[n-2]^2 + a[n-1])^2, a[0]==1, a[1]==1},a,{n,0,10}] (* Vaclav Kotesovec, May 22 2015 after Ran Pan *)

x=0;for(j=1,10,x=1/(x^2+1);print1((numerator(x))","))

a(n+2) = (a(n)^2 + a(n+1))^2. - Ran Pan, May 22 2015

a(n) ~ c * d^(2^n), where c = A088559 = 0.465571231876768... is the root of the equation c*(1+c)^2 = 1, d = 1.6634583970724267140029... . - Vaclav Kotesovec, May 22 2015

A252584 a(n) = a(n-1)^3 + a(n-2)^9, a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 1, 2, 9, 1241, 2298661010, 19127219051011953293860241761, 8789440239853164630485833302292601093162389737995133605845884014903267053091248194081

Offset: 0

Vaclav Kotesovec, Dec 18 2014

nonn

In general, if a(n) = a(n-1)^k + a(n-2)^(k^2), k > 1, then a(n) ~ d * c^(k^n), where c is a constant (dependent only on k, a(0) and a(1)), d is the root of the equation d^(k-1) * (1 + d^(k*(k-1))) = 1.
From Massimo Galasi, Jul 26 2018: (Start)
The recursion a(n)=a(n-1)^k+a(n-2)^(k^2) with a(0)=1, and a(1)=2, gives the number of independent sets of the k-ary tree with n levels. (For n=0 one has the empty tree.)
Thus this sequence, starting with 1, 2, 9, 1241, ... is the case k=3. Sequence A076725 is the case k=2. For k=1 the tree becomes a path and a(n) is Fibonacci(n+1). (End)

Cf. A000278, A076725.

a:=proc(n) option remember; if n=0 then 0 else if n=1 then 1 else a(n-1)^3+a(n-2)^9 fi fi end: seq(a(n), n = 0..9);

RecurrenceTable[{a[0]==0, a[1]==1, a[n] == a[n-1]^3 + a[n-2]^9}, a, {n, 0, 10}]

a(n) ~ d * c^(3^n), where c = 1.03028886637346769106..., d = 0.85117093406701547... is the root of the equation d^2 + d^8 = 1.

A242117 The number of independent sets in a complete binary tree with n levels (i.e., with 2^n-1 vertices) in which every vertex has degree 3.

Original entry on oeis.org

0, 0, 3, 24, 1680, 5317635, 66314914699608, 8947678119828215014722891024, 178098260698995011212395018312912894502905113202338936835

Offset: 1

Steven Kuipers and Bradford M. Morris, May 04 2014

nonn

For example, when n=3, there are two degree 3 vertices which do not share an edge. There are then three degree 3 (regular) independent subsets so a(3)=3. a(10) has 113 digits and is too large to include. The sequence is related to A076725, the number of independent sets in a complete binary tree.

The independent sets sought are those in the subgraph induced by the degree-3 vertices. This subgraph is a forest comprising two complete binary trees with n-2 levels each. These trees have A076725(n-2+1) independent sets each and the empty set (empty in both) is excluded here so a(n) = A076725(n-1)^2 - 1. - Kevin Ryde, Mar 10 2020

			a(3) = (a(1) + 1)^4 + 2(a(2)+1)(a(1) + 1)^2 + (a(2) + 1)^2 - 1 = (0+1)^4+2(0+1)(0+1)^2+(0+1)^2-1 = 3.
a(4) = (a(2) + 1)^4 + 2(a(3)+1)(a(2) + 1)^2 + (a(3) + 1)^2 - 1 = (0+1)^4+2(3+1)(0+1)^2+(3+1)^2-1 = 24.
a(5) = (a(3) + 1)^4 + 2(a(4)+1)(a(3) + 1)^2 + (a(4) + 1)^2 - 1 = (3+1)^4+2(24+1)(3+1)^2+(24+1)^2-1 = 1680.

Cf. A076725.

a(n) = my(x=1,y=1); for(i=3,n, [x,y] = [(x + y^2)^2, x]); x-1; \\ Kevin Ryde, Mar 10 2020

a(n) = (a(n-2) + 1)^4 + 2(a(n-1)+1)(a(n-2) + 1)^2 + (a(n-1) + 1)^2 - 1, with a(1) = a(2) = 0.

A252583 a(n) = a(n-1)^3 + a(n-2)^6, a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 1, 2, 9, 793, 499208698, 124407462491481058208408441, 1925481259009966865844002992692350885969632796673886663638811207308483184039785

Offset: 0

Vaclav Kotesovec, Dec 18 2014

nonn

Cf. A000278, A076725, A252584.

RecurrenceTable[{a[0]==0, a[1]==1, a[n] == a[n-1]^3 + a[n-2]^6}, a, {n, 0, 10}]
nxt[{a_,b_}]:={b,b^3+a^6}; NestList[nxt,{0,1},10][[All,1]] (* Harvey P. Dale, Jul 22 2022 *)

a(n) ~ c^(3^n), where c = 1.0278548747403916038597317519446646178420852580529382519451043895427806227... . - Vaclav Kotesovec, Dec 18 2014

cp's OEIS Frontend

A092526 Decimal expansion of (2/3)*cos( (1/3)*arccos(29/2) ) + 1/3, the real root of x^3 - x^2 - 1.

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A263719 Decimal expansion of the real root r of r^3 + r - 1 = 0.

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A005154 a(0) = 1, a(1) = 2; thereafter a(n) = 3*a(n-1)^2 - 2*a(n-2)^4.

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A338293 Number of matchings in the complete binary tree of n rows.

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A128384 a(n) = numerator of r(n): r(n) is such that the continued fraction (of rational terms) [r(1);r(2),...,r(n)] = b(n) for every positive integer n, where b(1) = 1 and b(n+1) = 1 + 1/b(n)^2 for.every positive integer n.

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A128385 a(n) = denominator of r(n): r(n) is such that the continued fraction (of rational terms) [r(1);r(2),...,r(n)] = b(n) for every positive integer n, where b(1) = 1 and b(n+1) = 1 + 1/b(n)^2 for.every positive integer n.

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A144229 The numerators of the convergents to the recursion x=1/(x^2+1).

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A092526 Decimal expansion of (2/3)cos( (1/3)arccos(29/2) ) + 1/3, the real root of x^3 - x^2 - 1.

A005154 a(0) = 1, a(1) = 2; thereafter a(n) = 3a(n-1)^2 - 2a(n-2)^4.