A058635 -id:A058635 - cp's OEIS frontend

A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

3, 7, 47, 2207, 4870847, 23725150497407, 562882766124611619513723647, 316837008400094222150776738483768236006420971486980607

Offset: 0

N. J. A. Sloane

Expansion of 1/phi: 1/phi = (1-1/3)*(1-1/((3-1)*7))*(1-1/(((3-1)*7-1)*47))*(1-1/((((3-1)*7-1)*47-1)*2207))... (phi being the golden ration (1+sqrt(5))/2). - Thomas Baruchel, Nov 06 2003
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Starting with 7, the terms end with 7,47,07,47,07,..., of the form 8a+7 where a = 0,1,55,121771,... Conjecture: Every a is squarefree, every other a is divisible by 55, the a's are a subset of A046194, the heptagonal triangular numbers (the first, 2nd, 3rd, 6th, 11th, ?, ... terms). - Gerald McGarvey, Aug 08 2004
Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. - Cino Hilliard, Sep 28 2008
The subsequence of primes begins a(n) for n = 0, 1, 2, 3. - Jonathan Vos Post, Feb 26 2011
We have Sum_{n=0..N} a(n)^2 = 2*(N+1) + Sum_{n=1..N+1} a(n), Sum_{n=0..N} a(n)^4 = 5*(Sum_{n=1..N+1} a(n)) + a(N+1)^2 + 6*N -3, etc. which is very interesting with respect to the fact that a(n) = Lucas(2^(n+1)); see W. Webb's problem in Witula-Slota's paper. - Roman Witula, Nov 02 2012
From Peter Bala, Nov 11 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks.
The recurrence a(n+1) = a(n)^2 - 2 with initial condition a(0) = x > 2 has the solution a(n) = ((x + sqrt(x^2 - 4))/2)^(2^n) + ((x - sqrt(x^2 - 4))/2)^(2^n).
We have the product expansion sqrt(x + 2)/sqrt(x - 2) = Product_{n>=0} (1 + 2/a(n)) (essentially due to Euler - see Mendes-France and van der Poorten). Another expansion is sqrt(x^2 - 4)/(x + 1) = Product_{n>=0} (1 - 1/a(n)), which follows by iterating the identity sqrt(x^2 - 4)/(x + 1) = (1 - 1/x)*sqrt(y^2 - 4)/(y + 1), where y = x^2 - 2.
The sequence b(n) := a(n) - 1 satisfies b(n+1) = b(n)^2 + 2*b(n) - 2. Cases currently in the database are A145502 through A145510. The sequence c(n) := a(n)/2 satisfies c(n+1) = 2*c(n)^2 - 1. Cases currently in the database are A002812, A001601, A005828, A084764 and A084765.
(End)
E. Lucas in Section XIX of "The Theory of Simply Periodic Numerical Functions" (page 56 of English translation) equation "(127) (1-sqrt(5))/2 = -1/1 + 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ..." - Michael Somos, Oct 11 2022
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
The number of digits of a(n) is given by A094057(n+1). - Hans J. H. Tuenter, Jul 29 2025

			From _Cino Hilliard_, Sep 28 2008: (Start)
Init x=1;
x = (5/1 + 1)/2 = 3/1;
x = (5/3 + 3)/2 = 7/3;
x = ((5/7)/3 + 7/3)/2 = 47/21;
x = ((5/47)/21 + 47/21)/2 = 2207/987;
(2207/987)^2 = 5.000004106... (End)

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E.-B. Escott, Note #1741, L'Intermédiaire des Mathématiciens, 8 (1901), page 13. - N. J. A. Sloane, Mar 02 2022
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 7.
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).

Vincenzo Librandi, Table of n, a(n) for n = 0..12
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!)
Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
Édouard Lucas, The Theory of Simply Periodic Numerical Functions, Fibonacci Association, 1969. English translation of article "Théorie des Fonctions Numériques Simplement Périodiques, I", Amer. J. Math., 1 (1878), 184-240.
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
E. L. Roettger and H. C. Williams, Some Remarks Concerning the Lucas-Lehmer Primality Test, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.5. See p. 2.
Chance Sanford, Infinite Series Involving Fibonacci Numbers Via Apéry-Like Formulae, arXiv:1603.03765 [math.NT], 2016.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Wikipedia, Engel Expansion.
Roman Wituła and Damian Słota, delta-Fibonacci numbers, Applicable Analysis and Discrete Mathematics, Vol. 3, No. 2 (2009), pp. 310-329.
Index entries for sequences of form a(n+1)=a(n)^2 + ....

Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herbert S. Wilf). Cf. A000045.
Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5).
Cf. A058635. - Artur Jasinski, Oct 05 2008
Cf. A002812, A094874, A145502, A145274, A050614, A088334, A181393, A181419, A186750, A186751, A338304.

a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'):
seq(a(n), n=0..8);

NestList[#^2-2&,3,10] (* Harvey P. Dale, Dec 17 2014 *)
Table[LucasL[2^n], {n, 1, 8}] (* Amiram Eldar, Oct 22 2020 *)

a[0]:3$
a[n]:=a[n-1]^2-2$
A001566(n):=a[n]$
makelist(A001566(n),n,0,7); /* Martin Ettl, Nov 12 2012 */

{a(n) = if( n<1, 3*(n==0), a(n-1)^2 - 2)}; /* Michael Somos, Mar 14 2004 */

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","));
g(5,8) \\ Cino Hilliard, Sep 28 2008

{a(n) = my(w = quadgen(5)); if( n<0, 0, n++; imag( (2*w - 1) * w^2^n ))}; /* Michael Somos, Nov 30 2014 */

{a(n) = my(y = x^2-x-1); if( n<0, 0, n++; for(i=1, n, y = polgraeffe(y)); -polcoeff(y, 1))}; /* Michael Somos, Nov 30 2014 */

a(n) = Fibonacci(2^(n+2))/Fibonacci(2^(n+1)) = A058635(n+2)/A058635(n+1). - Len Smiley, May 08 2000, and Artur Jasinski, Oct 05 2008
a(n) = ceiling(c^(2^n)) where c = (3+sqrt(5))/2 = tau^2 is the largest root of x^2-3*x+1=0. - Benoit Cloitre, Dec 03 2002
a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - Artur Jasinski, Sep 22 2008
a(n) = (G^(2^(n+1))-(1-G)^(2^(n+1)))/((G^(2^n))-(1-G)^(2^n)) = G^(2^n)+(1-G)^(2^n) = G^(2^n)+(-G)^(-2^n) where G is the golden ratio. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(2^(n+1)*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
a(n) = Fibonacci(2^(n+1)-1) + Fibonacci(2^(n+1)+1). (3-sqrt(5))/2 = 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ... (E. Lucas). - Philippe Deléham, Apr 21 2009
a(n)*(a(n+1)-1)/2 = A023039(2^n). - M. F. Hasler, Sep 27 2009
For n >= 1, a(n) = 2 + Product_{i=0..n-1} (a(i) + 2). - Vladimir Shevelev, Nov 28 2010
a(n) = 2*T(2^n,3/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
From Peter Bala, Oct 31 2012: (Start)
Engel expansion of 1/2*(3 - sqrt(5)). Thus 1/2*(3 - sqrt(5)) = 1/3 + 1/(3*7) + 1/(3*7*47) + ... as noted above by Deleham. See Liardet and Stambul.
sqrt(5)/4 = Product_{n>=0} (1 - 1/a(n)).
sqrt(5) = Product_{n>=0} (1 + 2/a(n)). (End)
a(n) - 1 = A145502(n+1). - Peter Bala, Nov 11 2012
a(n) == 2 (mod 9), for n > 1. - Ivan N. Ianakiev, Dec 25 2013
From Amiram Eldar, Oct 22 2020: (Start)
a(n) = A000032(2^(n+1)).
Sum_{k>=0} 1/a(k) = -1 + A338304. (End)
a(n) = (A000045(m+2^(n+2))+A000045(m))/A000045(m+2^(n+1)) for any m>=0. - Alexander Burstein, Apr 10 2021
a(n) = 2*cos(2^n*arccos(3/2)). - Peter Luschny, Oct 12 2022
a(n) == -1 ( mod 2^(n+2) ). - Peter Bala, Nov 07 2022
a(n) = 5*Fibonacci(2^n)^2+2 = 5*A058635(n)^2+2, for n>0. - Jianglin Luo, Sep 21 2023
Sum_{n>=0} a(n)/Fibonacci(2^(n+2)) = A094874 (Sanford, 2016). - Amiram Eldar, Mar 01 2024

A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.

Original entry on oeis.org

1, 2, 34, 196418, 37889062373143906, 271964099255182923543922814194423915162591622175362

Offset: 0

Jose Eduardo Blazek, Dec 11 1999

The next term, a(6), has 153 digits. - Harvey P. Dale, Oct 24 2011

Seiichi Manyama, Table of n, a(n) for n = 0..7
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.
Zalman Usiskin, Problem B-265, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 333; Fibonacci Numbers for Powers of 3, Solution to Problem B-265 by Ralph Garfield and David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), this sequence (k=3), A145231 (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

Cf. A000032, A000045, A001622, A001999, A002814, A006267, A094214.

a := proc(n) option remember; if n = 0 then 1 else 5*a(n-1)^3 - 3*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(3^n) - (1 - G)^(3^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[3^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 4}] (* Artur Jasinski, Oct 05 2008 *)
RecurrenceTable[{a[0]==1,a[n]==5a[n-1]^3-3a[n-1]},a[n],{n,6}] (* Harvey P. Dale, Oct 24 2011 *)
NestList[5#^3-3#&,1,5] (* Harvey P. Dale, Dec 21 2014 *)

A045529(n):=fib(3^n)$
makelist(A045529(n),n,0,10); /* Martin Ettl, Nov 12 2012 */

The first example I know in which a(n) can be expressed as (4/5)^(1/2)*cosh(3^n*arccosh((5/4)^(1/2))).
a(n) = Fibonacci(3^n). - Leroy Quet, Mar 17 2002
a(n+1) = a(n)*A002814(n+1). - Lekraj Beedassy, Jun 16 2003
a(n) = (phi^(3^n) - (1 - phi)^(3^n))/sqrt(5), where phi is the golden ratio (A001622). - Artur Jasinski, Oct 05 2008
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) - 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 24 2022: (Start)
a(2*n+2) == a(2*n) (mod 3^(2*n+1)); a(2*n+3) == a(2*n+1) (mod 3^(2*n+2));
a(2*n+1) + a(2*n) == 0 (mod 3^(2*n+1)).
a(2*n) == 1 (mod 3) and a(2*n+1) == 2 (mod 3).
5*a(n)^2 == 2 (mod 3^(n+1)).
In the ring of 3-adic integers, the sequences {a(2*n)} and {a(2*n+1)} are both Cauchy sequences and converge to the pair of 3-adic roots of the quadratic equation 5*x^2 - 2 = 0. (End)
From Amiram Eldar, Jan 07 2023: (Start)
Product_{n>=1} (1 + 2/(sqrt(5)*a(n)-1)) = phi (A001622).
Product_{n>=1} (1 - 2/(sqrt(5)*a(n)+1)) = 1/phi (A094214).
Both formulas are from Duverney and Kurosawa (2022). (End)

A020762 Decimal expansion of 1/sqrt(5).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

This number is the cosine of the central angle of a regular icosahedron; see A105199 for the angle itself. - Clark Kimberling, Feb 10 2009

Largest radius of ten circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013

			0.447213595499957939281834733746255247088123671922305144854179449082104...

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Hideyuki Ohtsuka, Problem B-1148, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 52, No. 2 (2014), p. 179; The Exact Value of an Infinite Series, Solution to B-1148, ibid., Vol. 53, No. 2 (2015), pp. 183-184.
Index entries for algebraic numbers, degree 2.

Cf. A000032, A000045, A010476, A001566, A058635, A105199.

RealDigits[5^(-1/2), 10, 150] (* Stefan Steinerberger, Apr 08 2006 *)
Circs[n_] := With[{r = Sin[Pi/n]/(1 - Sin[Pi/n])}, Graphics[Append[
  Table[Circle[(r + 1) {Sin[2 Pi k/n], Cos[2 Pi k/n]}, r], {k, n}],
    {Blue, Circle[{0, 0}, 1]}]]]
Circs[10] (* Charles R Greathouse IV, Jan 14 2013 *)

1/sqrt(5) \\ Charles R Greathouse IV, Jan 14 2013

Equals cos(arctan(2)). - Clark Kimberling, Feb 10 2009
Equals lim_{n -> infinity} A000045(n)/A000032(n). - Bruno Berselli, Jan 22 2018
From Christian Katzmann, Mar 19 2018: (Start)
Equals Sum_{n>=0} (2*n)!/(n!^2*3^(2*n+1)).
Equals Sum_{n>=0} 5*(2*n+1)!/(n!^2*3^(2*n+3)). (End)
Equals A010476/10. - R. J. Mathar, Jan 14 2021
Equals Sum_{k>=1} F(2^(k-1))/(L(2^k)+1) = Sum_{k>=0} A058635(k)/(A001566(k)+1), where F(k) = A000045(k) is the k-th Fibonacci number and L(k) = A000032(k) is the k-th Lucas number (Ohtsuka, 2014). - Amiram Eldar, Dec 09 2021

More terms from Stefan Steinerberger, Apr 08 2006

A079585 Decimal expansion of c = (7-sqrt(5))/2.

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Jan 26 2003

c is an integer in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Jan 08 2018
From Amiram Eldar, Jul 16 2021: (Start)
Sum_{k>=0} 1/F(2^k) is sometimes called "Millin series" after D. A. Millin, a high school student at Annville, Pennsylvania, who posed in 1974 the problem of proving that it equals (7-sqrt(5))/2. This identity was in fact already known to Lucas in 1878.
Mahler (1975) provided a false proof that this sum is transcendental. The mistake was corrected in Mahler (1976). (End)
The name "Millin" was a misprint of "Miller", the author of the problem was Dale A. Miller. His name was corrected in the solution to the problem (1976). - Amiram Eldar, Feb 29 2024

			c = 2.3819660112501051517954131656343618822796908201942371378645513772947...

Jean-Paul Allouche and Jeffrey Shallit, Automatic sequences, Cambridge University Press, 2003, p. 65.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.2, p. 7.
Ross Honsberger, Mathematical Gems III, Washington, DC: Math. Assoc. Amer., 1985, pp. 135-137.
Alfred S. Posamentier and Ingmar Lehmann, [Phi], The Glorious Golden Ratio, Prometheus Books, 2011, page 75.

Chai Wah Wu, Table of n, a(n) for n = 1..10001
I. J. Good, A Reciprocal Series of Fibonacci Numbers, Fib. Quart., Vol. 12, No. 4 (1974), p. 346.
History of Science and Mathematics StackExchange, Who was D.A. Millin, the eponym of the Millin Series?, 2022.
Edouard Lucas, Théorie des Fonctions Numériques Simplement Périodiques. [Continued], American Journal of Mathematics, Vol. 1, No. 3 (1878), pp. 197-240. See p. 225, equations 125 and 127.
Kurt Mahler, On the transcendency of the solutions of special class of functional equations, Bull. Austral. Math. Soc., Vol. 13, No. 3 (1975), pp. 389-410.
Kurt Mahler, On the transcendency of the solutions of a special class of functional equations: Corrigendum, Bull. Austral. Math. Soc., Vol. 14, No. 3 (1976), pp. 477-478.
Dale Miller, Publications.
D. A. Millin, Problem H-237, The Fibonacci Quarterly, Vol. 12, No. 3 (1974), p. 309; Sum Reciprocal!, Solution to Problem H-237 by A. G. Shannon, ibid., Vol. 14, No. 2 (1976), pp. 186-187.
Michael Penn, The Millin Series (A nice Fibonacci sum), YouTube video, 2020.
Proofwiki, Definition:Millin Series.
Stanley Rabinowitz, A note on the sum 1/w_{k2^n}, Missouri J. Math. Sci., Vol. 10, No. 3 (1998), pp. 141-146.
Eric Weisstein's World of Mathematics, Millin Series.
Index entries for algebraic numbers, degree 2

Cf. A001622, A020837, A058635.

RealDigits[4 - GoldenRatio, 10, 111][[1]] (* Robert G. Wilson v, Jan 31 2012 *)

(7 - sqrt(5))/2 \\ Michel Marcus, Sep 05 2017

c = (7-sqrt(5))/2 = 4 - phi, with phi from A001622.
c = 7/2 - 10*A020837.
c = Sum_{k>=0} 1/F(2^k), where F(k) denotes the k-th Fibonacci number; c = Sum_{k>=0} 1/A058635(k).
Periodic continued fraction representation is [2, 2, 1, 1, 1, 1, ....]. - R. J. Mathar, Mar 24 2011
Minimal polynomial: 11 - 7*x + x^2. - Stefano Spezia, Oct 16 2024

A250486 A(n,k) is the n^k-th Fibonacci number; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 3, 2, 1, 0, 1, 21, 34, 3, 1, 0, 1, 987, 196418, 987, 5, 1, 0, 1, 2178309, 37889062373143906, 10610209857723, 75025, 8, 1

Offset: 0

Alois P. Heinz, Nov 24 2014

			Square array A(n,k) begins:
  1,  0,  0,      0,       0,    0,  0,  0,   ...
  1,  1,  1,      1,       1,    1,  1,  1,   ...
  1,  1,  3,      21,      987,  2178309,     ...
  1,  2,  34,     196418,  37889062373143906, ...
  1,  3,  987,    10610209857723,             ...
  1,  5,  75025,  59425114757512643212875125, ...
  1,  8,  14930352,                           ...
  1,  13, 7778742049,                         ...

Alois P. Heinz, Antidiagonals n = 0..10, flattened
Wikipedia, Fibonacci number

Columns k=0-8 give: A000012, A000045, A054783, A182149, A250490, A250491, A250492, A250493, A250494.
Rows n=0-10 give: A000007, A000012, A058635, A045529, A145231, A145232, A145233, A145234, A250487, A250488, A250489.
Main diagonal gives A250495.
Cf. A000045.

A:= (n, k)-> (<<0|1>, <1|1>>^(n^k))[1, 2]:
seq(seq(A(n, d-n), n=0..d), d=0..8);

A[n_, k_] := MatrixPower[{{0, 1}, {1, 1}}, n^k][[1, 2]]; A[0, 0] = 1;
Table[A[n, d-n], {d, 0, 8}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A(n,k) = [0, 1; 1, 1]^(n^k)[1,2].

A145231 a(n) = Fibonacci(4^n).

Original entry on oeis.org

1, 3, 987, 10610209857723, 141693817714056513234709965875411919657707794958199867

Offset: 0

Artur Jasinski, Oct 05 2008

This sequence has the property that a(n+1) is divisible by a(n). Conjecture: each prime divisor can occur only once (i.e., all terms are squarefree). - Artur Jasinski, Oct 05 2008

Seiichi Manyama, Table of n, a(n) for n = 0..6

Cf. A000045, A341601, A341603.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), this sequence (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

a := proc(n) option remember; if n = 1 then 3 else a(n-1)*(5*a(n-1)^2 + 2)*sqrt(5*a(n-1)^2 + 4) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 14 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 6}]
Table[Round[(4/5)^(1/2)*Cosh[4^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 7}]
Fibonacci[4^Range[5]] (* Harvey P. Dale, Mar 28 2012 *)

a(n) = (G^(4^n) - (1-G)^(4^n) )/sqrt(5) where G = (1 + sqrt 5)/2 = A001622.
a(n) = round( sqrt(4/5) *cosh( 4^n*arccosh (sqrt(5/4)) )).
a(n)= A000045(A000302(n)). - Michel Marcus, Nov 07 2013
From Peter Bala, Nov 11 2022: (Start)
a(n+1) = a(n)*(5*a(n)^2 + 2)*sqrt(5*a(n)^2 + 4) for n >= 1.
a(n) == 3 (mod 4) for n >= 1.
a(n+1) == a(n) (mod 2^(2*n+1)).
A341601(n) == a(n) (mod 2^n) for n >= 2.
In the ring of 2-adic integers, the sequence {Fibonacci(4^n)} converges to the 2-adic integer A341603. (End)

A145232 a(n) = Fibonacci(5^n).

Original entry on oeis.org

1, 5, 75025, 59425114757512643212875125, 18526362353047317310282957646406309593963452838196423660508102562977229905562196608078556292556795045922591488273554788881298750625

Offset: 0

Artur Jasinski, Oct 05 2008

Seiichi Manyama, Table of n, a(n) for n = 0..5
Robert Frontczak, Problem B-1341, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 1 (2024), p. 84.
Thomas Koshy and Zhenguang Gao, Polynomial Extensions of a Diminnie Delight, Fibonacci Quart. 55 (2017), no. 1, 13-20.
Achilleas Sinefakopoulos, Solution to Problem 1909, Crux Mathematicorum, 20 (1994), 295-296.

Cf. A000045.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), A045529 (k=3), A145231 (k=4), this sequence (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

a := proc(n) option remember; if n = 0 then 1 else 25*a(n-1)^5 - 25*a(n-1)^3 + 5*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(5^n) - (1 - G)^(5^n))/Sqrt[5]], {n, 1, 6}]
Table[Round[N[(2/Sqrt[5])*Cosh[5^n*ArcCosh[Sqrt[5]/2]], 1000]], {n, 1, 4}]
Fibonacci[5^Range[0,4]] (* Harvey P. Dale, Nov 29 2018 *)

a(n) = (G^(5^n) - (1 - G)^(5^n))/sqrt(5) where G = (1 + sqrt(5))/2.
a(n) = (2/sqrt(5))*cosh((2*k+1)^n*arccosh(sqrt(5)/2)).
a(n) = (2/sqrt(5))*cosh(5^n*arccosh(sqrt(5)/2)).
a(n) = (5^n)*A128935(n). - R. J. Mathar, Nov 04 2010
a(n) = A000045(A000351(n)). - Michel Marcus, Nov 07 2013
a(n+1) = 25*a(n)^5 - 25*a(n)^3 + 5*a(n) with a(0) = 1. - Peter Bala, Nov 24 2022
a(n) = 5^n * Product_{k=0..n-1} (5*a(k)^4 - 5*a(k)^2 + 1) (Frontczak, 2024). - Amiram Eldar, Feb 29 2024

A071579 a(n) = 2a(n-1)A002812(n-1), starting a(0)=1.

Original entry on oeis.org

1, 4, 56, 10864, 408855776, 579069776145402304, 1161588808526051807570761628582646656, 4674072680304961790168962360144614650442718636276775741658113370728376064

Offset: 0

Joe Keane (jgk(AT)jgk.org), May 31 2002

Also the denominators of the convergents to sqrt(3) using Newton's recursion x = (3/x+x)/2. - Cino Hilliard, Sep 28 2008
For n>1, Egyptian fraction of 2-sqrt(3): 2-sqrt(3) = 1/4 + 1/56 + 1/10864 + 1/408855776 + ... - Simon Plouffe, Feb 20 2011
The sequence satisfies the Pell equation A002812(n)^2-3*a(n)^2 = 1. - Vincenzo Librandi, Dec 19 2011
From Peter Bala, Oct 30 2013: (Start)
Apart from giving the numerators in the Engel series representation of 2 - sqrt(3), as stated above by Plouffe, this sequence is also a Pierce expansion of the real number x = 2 - sqrt(3) to the base b := 1/sqrt(12) (see A058635 for a definition of this term).
The associated series representation begins 2 - sqrt(3) = b/1 - b^2/(1*4) + b^3/(1*4*56) - b^4/(1*4*56*10864) + .... Cf. A230338.
More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of (2 - sqrt(3))^(2^n) to the base b = 1/sqrt(12). Some examples are given below. (End)

			Let b = 1/sqrt(12) and x = 2 - sqrt(3). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*4) + b^3/(1*4*56) - b^4/(1*4*56*10864) + b^5/(1*4*56*10864*408855776) - ....
x^2 = b/4 - b^2/(4*56) + b^3/(4*56*10864) - b^4/(4*56*10864*408855776) + ....
x^4 = b/56 - b^2/(56*10864) + b^3/(56*10864*408855776) - ....
x^8 = b/10864 - b^2/(10864*408855776) + .... - _Peter Bala_, Oct 30 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..10
Eric Weisstein's World of Mathematics, Newton's Iteration.
Doron Zeilberger, Another Book of Somos-Like Miracles, Prop. Number, 2; Local copy.

Cf. A002812, A001353. A058635, A230338.

I:=[1,4]; [n le 2 select I[n] else 2*Self(n-1)*(6*Self(n-2)^2+1): n in [1..8]]; // Vincenzo Librandi, Dec 19 2011

a[ n_] := If[n<0, 0, Coefficient[PolynomialMod[x^2^n, x^2 - 4*x + 1], x]]; (* Michael Somos, Jul 29 2024 *)
a[ n_] := If[n<1, Boole[n==0], a[n] = a[n-1]*Sqrt[12*a[n-1]^2 + 4] ]; (* Michael Somos, Jul 29 2024 *)

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(denominator(x)","))

{a(n) = if(n<0, 0, imag((2 + quadgen(12))^2^n))}; /* Michael Somos, Jul 29 2024 */

{a(n) = if(n<0, 0, polcoef(lift(Mod(x^2^n, x^2 - 4*x + 1)), 1))}; /* Michael Somos, Jul 29 2024 */
g(3,8) \\ Cino Hilliard, Sep 28 2008

a(n) = 1/sqrt(12)*( (2 + sqrt(3))^2^n - (2 - sqrt(3))^2^n ) = A001353(2^n).
a(n) = 2*a(n-1)*(6*a(n-2)^2+1). - Max Alekseyev, Apr 19 2006
Recurrence equations:
a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(n) = a(n-1)*sqrt(12*a(n-1)^2 + 4) for n >= 1. - Peter Bala, Oct 30 2013
0 = 6*a(n)^2*a(n+2) - 6*a(n+1)^3 - 2*a(n+1) + a(n+2) for n>=1. - Michael Somos, Dec 05 2016
0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 for n>=1. - Michael Somos, Dec 05 2016
a(n) = A001353(2^n). - Michael Somos, Jul 29 2024

cp's OEIS Frontend

A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

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A045529 a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1.

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A020762 Decimal expansion of 1/sqrt(5).

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A079585 Decimal expansion of c = (7-sqrt(5))/2.

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A250486 A(n,k) is the n^k-th Fibonacci number; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A145231 a(n) = Fibonacci(4^n).

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A145232 a(n) = Fibonacci(5^n).

A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.

A071579 a(n) = 2a(n-1)A002812(n-1), starting a(0)=1.