A071579 -id:A071579 - cp's OEIS frontend

A058635 a(n) = Fibonacci(2^n).

1, 1, 3, 21, 987, 2178309, 10610209857723, 251728825683549488150424261, 141693817714056513234709965875411919657707794958199867

Offset: 0

Robert G. Wilson v, Jan 16 2001

nonn

The next term has 107 digits.
From Peter Bala, Oct 30 2013: (Start)
Apart from giving the numerators in the Engel series representation of (1/2)*(7 - sqrt(5)), as noted below by Cloitre, this sequence (excluding the initial term) is also a generalized Pierce expansion defined as follows. Let x and b be positive real numbers. We define a Pierce expansion of x to the base b to be a nondecreasing sequence [a(1), a(2), a(3), ...] of positive integers such that we have an alternating series representation x = b/a(1) - b^2/(a(1)*a(2)) + b^3/(a(1)*a(2)*a(3)) - ....
The present sequence, apart from the initial term, is a Pierce expansion of the real number x := (1/2)*(3 - sqrt(5)) to the base b := 1/sqrt(5). The associated series representation begins (1/2)*(3 - sqrt(5)) = b/1 - b^2/(1*3) + b^3/(1*3*21) - b^4/(1*3*21*987) + .... Cf. A071579 and A230338.
More generally, for n >= 0, the sequence [a(n+1), a(n+2), a(n+3), ...] gives a Pierce expansion of ( (1/2)*(3 - sqrt(5)) )^(2^n) to the base b = 1/sqrt(5). Some examples are given below. (End)
a(n) is the denominator of the n-th iterate when Newton's method is applied to the function x^2 - x - 1 with initial guess x = 1. The n-th iterate is A192222(n)/a(n). - Jason Zimba, Jan 20 2023

			Let b = 1/sqrt(5) and x = (1/2)*(3 - sqrt(5)). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*3) + b^3/(1*3*21) - b^4/(1*3*21*987) + ....
x^2 = b/3 - b^2/(3*21) + b^3/(3*21*987) - b^4/(3*21*987*2178309) + ....
x^4 = b/21 - b^2/(21*987) + b^3/(21*987*2178309) - ....
x^8 = b/987 - b^2/(987*2178309) + .... - _Peter Bala_, Oct 30 2013

Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, p. 446.

Vincenzo Librandi, Table of n, a(n) for n = 0..12
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
S. B. Ekhad and D. Zeilberger, How To Generate As Many Somos-Like Miracles as You Wish, arXiv preprint arXiv:1303.5306 [math.CO], 2013.
John Gill and Matthew Miller, Newton's Method and Ratios of Fibonacci Numbers, Fibonacci Quarterly, 19(1):1-3, February 1981.
H. Hu, Z.-W. Sun and J.-X. Liu, Reciprocal sums of second order recurrent sequences, Fib. Quart. 39(2001), no. 3, 214-220.
Hideyuki Ohtsuka, Problem H-767, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 53, No. 1 (2015), p. 88; Nested Radicals and Fibonacci Numbers, Solution to Problem H-767 by the proposer, ibid., Vol. 55, No. 1 (2017), pp. 90-91.

Cf. A000032, A000045, A000079, A054783, A001566, A071579, A230338.

[Fibonacci(2^n): n in [0..10]]; // Vincenzo Librandi, Mar 25 2014

a:= n-> (<<0|1>, <1|1>>^(2^n))[1,2]:
seq(a(n), n=0..10);  # Alois P. Heinz, Nov 21 2014

Table[ Fibonacci[ 2^n ], {n, 0, 9} ]
G = (1 + Sqrt[5])/2; Table[Expand[(G^(2^n) - (1 - G)^(2^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[2^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 10}] (* Artur Jasinski, Oct 05 2008 *)

a(n)=fibonacci(2^n) \\ Charles R Greathouse IV, Oct 03 2016

a(n) = a(n-1)*A001566(n-2). - Joe Keane (jgk(AT)jgk.org), May 31 2002
Sum_{n>=0} 1/a(n) = (1/2)*(7-sqrt(5)). - Benoit Cloitre, Jan 26 2003
1/phi^2 = (0.6180339...)^2 = 2/(3+sqrt(5)) = Sum_{n>=2} 1/a(n) = 1/3 + 1/21 + 1/987 + 1/2178309 + ... - Gary W. Adamson, Jun 12 2003
From Artur Jasinski, Oct 05 2008: (Start)
a(n) = (G^(2^n) - (1 - G)^(2^n))/sqrt(5) where G = GoldenRatio = (1 + sqrt(5))/2.
a(n) = sqrt(4/5)*cosh((2^n)*arccosh(sqrt(5/4))). (End)
a(n) = (a(n-1)^3 / a(n-2)^2 + 5 * a(n-1) * a(n-2)^2) / 2, for n > 1. - Lee A. Newberg, Jul 20 2010
Recurrence equations from Peter Bala, Oct 30 2013: (Start)
a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 3.
a(n)/a(n-1) = 5*a(n-2)^2 + 2 for n >= 3.
a(n) = a(n-1)*sqrt(5*a(n-1)^2 + 4) for n >= 2. (End)
0 = a(n)^2 * ( a(n+3) - 2*a(n+2) ) - a(n+1)*a(n+2) * ( a(n+2) - 2*a(n+1)) if n > 0. - Michael Somos, Mar 24 2014
From Amiram Eldar, Dec 02 2021: (Start)
a(n) = A000045(A000079(n)).
Limit_{n->oo} sqrt(a(1)^2 + sqrt(a(2)^2 + sqrt(a(3)^2 + ... + sqrt(a(n))))) = 3 (Ohtsuka, 2015). (End)
a(n) = Product_{k=0..n-1} L(2^k), for n >= 1, where L(k) is the k-th Lucas number (A000032). - Amiram Eldar, Mar 30 2023

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

Original entry on oeis.org

2, 7, 97, 18817, 708158977, 1002978273411373057, 2011930833870518011412817828051050497, 8095731360557835890888779535060256832479295062749579257164654370487894017

Offset: 0

N. J. A. Sloane

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
2^p-1 is prime iff it divides a(p-2), since a(n) = A003010(n)/2, where A003010 is the Lucas-Lehmer sequence used for Mersenne number primality testing. - M. F. Hasler, Mar 09 2007
From Cino Hilliard, Sep 28 2008: (Start)
Also numerators of the convergents to the square root of 3 using the following recursion for initial x = 1: x1=x, x=3/x, x=(x+x1)/2.
This recursion was derived by experimenting with polynomial recursions of the form x = -a(0)/(a(n-1)x^(n-1) + ... + a(1)) in an effort to find a root for the polynomial a(n)x^n + a(n-1)x^(n-1) + ... + a(0). The process was hit-and-miss until I introduced the averaging step described above. This method is equivalent to Newton's Method although derived somewhat differently. (End)
The sequence satisfies the Pell equation a(n)^2 - 3*A071579(n)^2 = 1. - Vincenzo Librandi, Dec 19 2011
From Peter Bala, Nov 2012: (Start)
The present sequence corresponds to the case x = 2 of the following general remarks. Let x > 1 and let alpha := {x + sqrt(x^2 - 1)}. Define a sequence a(n) (which depends on x) by setting a(n) = (1/2)*(alpha^(2^n) + (1/alpha)^(2^n)) for n >= 0. It is easy to verify that a(n) is a solution to the recurrence equation a(n+1) = 2*a(n)^2 - 1 with the initial condition a(0) = x.
The following algebraic identity is valid for x > 1:
sqrt(4*x^2 - 4)/(2*x + 1) = (1 - 1/(2*x))*sqrt(4*y^2 - 4)/(2*y + 1), where y = 2*x^2 - 1. Iterating the identity yields the product expansion sqrt(4*x^2 - 4)/(2*x + 1) = Product_{n >= 0} (1 - 1/(2*a(n))). A second expansion is Product_{n >= 0} (1 + 1/a(n)) = sqrt((x + 1)/(x - 1)). See Mendes-France and van der Poorten. (End)

			G.f. = 2 + 7*x + 97*x^2 + 18817*x^3 + 708158977*x^4 + ...

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. 399.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
W. Sierpiński, Sur les développements systématiques des nombres en produits infinis, in Œuvres choisies, tome 1, PWN Editions scientifiques de Pologne, 1974.
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..10
Georg Cantor, Zwei Sätze über eine gewisse Zerlegung der Zahlen in unendliche Producte, Zeitschrift für Mathematik und Physik. Band 14, 1869, S. 152-158. (See page 155, II, error in the fourth term.)
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
Jeffrey Shallit, Rational numbers with non-terminating, non-periodic modified Engel-type expansions, Fib. Quart., 31 (1993), 37-40.
Eric Weisstein's World of Mathematics, Newton's Iteration
Index entries for sequences related to Engel expansions

Cf. A001075, A002531, A003010, A071579.

Cf. A177879 (lpf).

a:=[2];; for n in [2..10] do a[n]:=2*a[n-1]^2-1; od; a; # Muniru A Asiru, Aug 12 2018

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

a:=n->((2+sqrt(3))^(2^n)+(2-sqrt(3))^(2^n))/2: seq(floor(a(n)),n=0..10); # Muniru A Asiru, Aug 12 2018

Table[((2 + Sqrt[3])^2^n + (2 - Sqrt[3])^2^n)/2, {n, 0, 7}] (* Bruno Berselli, Dec 20 2011 *)
NestList[2#^2-1&,2,10] (* Harvey P. Dale, May 04 2013 *)
a[ n_] := If[ n < 0, 0, ChebyshevT[2^n, 2]]; (* Michael Somos, Dec 06 2016 *)

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

/* Roots by recursion. Find first root of ax^2 + b^x + c */
rroot2(a,b,c,p) = { local(x=1,x1=1,j); for(j=1,p, x1=x; x=-c/(a*x+b); x=(x1+x)/2; /* Let x be the average of the last 2 values */ print1(numerator(x)","); ); } \\ Cino Hilliard, Sep 28 2008

a(n) = A001075(2^n).
a(n) = ((2+sqrt(3))^(2^n) + (2-sqrt(3))^(2^n))/2. - Bruno Berselli, Dec 20 2011
From Peter Bala, Nov 11 2012: (Start)
2*sqrt(3)/5 = Product_{n >= 0} (1 - 1/(2*a(n))).
sqrt(3) = Product_{n >= 0} (1 + 1/a(n)).
a(n) = (1/2)*A003010(n). (End)
a(n) = cos(2^n*arccos(2)). - Peter Luschny, Oct 12 2022
a(n) = A002531(2^(n+1)). - Robert FERREOL, Apr 13 2023

A244012 Numerators of rational approximations to sqrt(7) obtained from Newton's method.

Original entry on oeis.org

2, 11, 233, 108497, 23543191457, 1108563727961872518977, 2457827077905448997994482872789298261401217, 12081827889770476116093110581355561229584727594431650162181251776430351279198649072897

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 11/4, 233/88, 108497/41008, 23543191457/8898489952, ...

R. Parimala, A Hasse principle for quadratic forms over function fields, Bull. Amer. Math. Soc. (N.S.) 51 (2014), no. 3, 447--461. MR3196794.

Cf. A244013 (denominators).

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

N:=7;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A244013 Denominators of rational approximations to sqrt(7) obtained from Newton's method.

Original entry on oeis.org

1, 4, 88, 41008, 8898489952, 418997705236253480128, 928971316248341903257187589777603944778112, 4566501711345281867283814391125123371716411674583075407993026856131137508750543524608

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 11/4, 233/88, 108497/41008, 23543191457/8898489952, ...

R. Parimala, A Hasse principle for quadratic forms over function fields, Bull. Amer. Math. Soc. (N.S.) 51 (2014), no. 3, 447--461. MR3196794.

Cf. A244012 (numerators).

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

N:=7;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A244014 Numerators of rational approximations to sqrt(6) obtained from Newton's method.

Original entry on oeis.org

2, 5, 49, 4801, 46099201, 4250272665676801, 36129635465198759610694779187201, 2610701117696295981568349760414651575095962187244375364404428801

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 5/2, 49/20, 4801/1960, 46099201/18819920, ...

Cf. A244015, A084765.

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

N:=6;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A244015 Denominators of rational approximations to sqrt(6) obtained from Newton's method.

Original entry on oeis.org

1, 2, 20, 1960, 18819920, 1735166549767840, 14749861913749949808286047759680, 1065814268211609269094400465471990022332221793124358274759711360

Offset: 0

N. J. A. Sloane, Jun 18 2014

			2, 5/2, 49/20, 4801/1960, 46099201/18819920, ...

Cf. A244014 (numerators).

The analogs for sqrt(k), k=2,3,5,6,7 are: A001601/A051009, A002812/A071579, A081459/A081460, A244014/A244015, A244012/A244013.

m:=9; f:=[n eq 1 select 2 else (Self(n-1)+6/Self(n-1))/2: n in [1..m]]; [Denominator(f[n]): n in [1..m]]; // Vincenzo Librandi, Jan 12 2016

N:=6;
s:=[floor(sqrt(N))];
M:=8;
for n from 1 to M do
x:=s[n];
h:=(N-x^2)/(2*x);
s:=[op(s),x+h]; od:
lprint(s);
s1:=map(numer,s);
s2:=map(denom,s);

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)sqrt(21a(n-1)^2 + 4) for n >= 1.

Original entry on oeis.org

1, 5, 115, 60605, 16831644835, 1298263252133919638045, 7723873922612696850892381990249713732303715, 273388347343560518533856033712658350781293745092679040607342582493129736504927611387805

Offset: 0

Peter Bala, Oct 30 2013

For integer N, the recurrence equation a(n) = a(n-1)*sqrt((N^2 - 4)*a(n-1)^2 + 4) for n >= 1, with starting value a(0) = 1, produces an integer sequence. The present sequence is the case N = 5. Cf. A000079 (case N = 2), A058635 (case N = 3) and A071579 (case N = 4).
Sequence of numerators in the Engel series representation of 1/2*(7 - sqrt(21)) = 1 + 1/5 + 1 /115 + 1/60605 + .... The corresponding Engel expansion is A003487.
The sequence also has a description as a Pierce expansion of the quadratic irrational 1/2*(5 - sqrt(21)) to the base b := 1/sqrt(21) (see A058635 for a definition of this term).
The associated Pierce series representation of 1/2*(5 - sqrt(21)) to the base b begins 1/2*(5 - sqrt(21)) = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + ....
More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of ( 1/2*(5 - sqrt(21)) )^(2^n) to the base b = 1/sqrt(21). Some examples are given below.

			Let b = 1/sqrt(21) and x = 1/2*(5 - sqrt(21)). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + b^5/(1*5*115*60605*16831644835) - ....
x^2 = b/5 - b^2/(5*115) + b^3/(5*115*60605) - b^4/(5*115*60605*16831644835) + ....
x^4 = b/115 - b^2/(115*60605) + b^3/(115*60605*16831644835) - ....
x^8 = b/60605 - b^2/(60605*16831644835) + ....

Seiichi Manyama, Table of n, a(n) for n = 0..10

Cf. A003487, A004254, A058635, A071579.

a[n_] := a[n - 1]*Sqrt[21 a[n - 1]^2 + 4]; a[0] = 1; Array[a, 8, 0] (* Robert G. Wilson v, Mar 19 2014 *)
a[ n_] := If[ n < 0, 0, ChebyshevU[2^n - 1, 5/2]]; (* Michael Somos, Dec 06 2016 *)

a(n) = if( n<0, 0, imag( (5 + quadgen(84))^2^n) / 2^(2^n - 1)); /* Michael Somos, Dec 06 2016 */

a(n) = 1/sqrt(21)*( alpha^(2^n) - (1/alpha)^(2^n) ), where alpha = 1/2*(5 + sqrt(21)).
a(n) = product {k = 0..n-1} A003487(k).
Defining recurrence equation:
a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
Other recurrence equations:
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = 21*a(n-2)^2 + 2 for n >= 2.
a(n) = A004254(2^n). - Michael Somos, Dec 06 2016

cp's OEIS Frontend

A058635 a(n) = Fibonacci(2^n).

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

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A244012 Numerators of rational approximations to sqrt(7) obtained from Newton's method.

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Maple

A244013 Denominators of rational approximations to sqrt(7) obtained from Newton's method.

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Maple

A244014 Numerators of rational approximations to sqrt(6) obtained from Newton's method.

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Maple

A244015 Denominators of rational approximations to sqrt(6) obtained from Newton's method.

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Magma

Maple

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.

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A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)sqrt(21a(n-1)^2 + 4) for n >= 1.