A145504 -id:A145504 - cp's OEIS frontend

A145502 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=2.

2, 6, 46, 2206, 4870846, 23725150497406, 562882766124611619513723646, 316837008400094222150776738483768236006420971486980606

Offset: 1

Artur Jasinski, Oct 11 2008

General formula for a(n+1) = a(n)^2+2*a(n)-2 and a(1) = k+1 is a(n) = floor(((k + sqrt(k^2 + 4))/2)^(2^((n+1) - 1))).
From Peter Bala, Nov 12 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks. Sequences A145503 through A145510 correspond to the cases x = 4 through x = 11 respectively.
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^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1. Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^2 + 2*a(n) - 2 with the initial condition a(1) = x - 1.
A second recurrence is a(n) = (a(1) + 2)*{Product_{k = 1..n-1} a(k)} - 2.
The following algebraic identity is valid for x > 2:
(x + 1)/sqrt(x^2 - 4) = (1 + 1/(x - 1))*(y + 1)/sqrt(y^2 - 4), where y - 1 = (x - 1)^2 + 2*(x - 1) - 2. Iterating the identity yields the product expansion (x + 1)/sqrt(x^2 - 4) = Product_{n >= 1} (1 + 1/a(n)).
A second expansion is Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt((x + 2)/(x - 2)). For an alternative approach to these identities see the Bala link.
(End)

Peter Bala, Notes on A145502-A145510.
Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, Journal of Mathematical Sciences-University of Tokyo, Vol. 8, No. 2 (2001), pp. 275-316.
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.

Cf. A145503, A145504, A145505, A145506, A145507, A145508, A145509, A145510.

Cf. A000324, A001566.

aa = {}; k = 2; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k = 1; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}]
NestList[#^2+2#-2&,2,10] (* Harvey P. Dale, Dec 14 2021 *)

From Peter Bala, Nov 12 2012: (Start)
a(n) = phi^(2^n) + (1/phi)^(2^n) - 1, where phi := (1 + sqrt(5))/2 is the golden ratio.
a(n) = A001566(n-1) - 1.
Recurrence: a(n) = 4*(Product_{k = 1..n-1} a(k)) - 2 with a(1) = 2.
Product_{n >= 1} (1 + 1/a(n)) = 4/sqrt(5).
Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt(5). (End)
From Amiram Eldar, Sep 10 2022: (Start)
a(n) = A000324(n) - 3.
Sum_{n>=1} (-2)^n/a(n) = -1/2 (Duverney, 2001). (End)
Product_{n>=1} (1 + 3/a(n)) = 4 (Duverney and Kurosawa, 2022). - Amiram Eldar, Jan 07 2023

A003487 a(n) = a(n-1)^2 - 2, with a(0) = 5.

Original entry on oeis.org

5, 23, 527, 277727, 77132286527, 5949389624883225721727, 35395236908668169265765137996816180039862527, 1252822795820745419377249396736955608088527968701950139470082687906021780162741058825727

Offset: 0

N. J. A. Sloane

The next term has 175 digits. - Harvey P. Dale, Feb 19 2015

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

Seiichi Manyama, Table of n, a(n) for n = 0..10
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
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566 (starting with 3), A003010 (starting with 4), A003423 (starting with 6). A001601, A145504.

a:= n-> simplify(2*ChebyshevT(2^n, 1/2*5), 'ChebyshevT'):
seq(a(n), n=0..7);

NestList[#^2-2&,5,10] (* Harvey P. Dale, Feb 19 2015 *)
a[ n_] := If[ n < 0, 0, 2 ChebyshevT[2^n, 5/2]]; (* Michael Somos, Dec 06 2016 *)

{a(n) = if( n<0, 0, polchebyshev(2^n, 1, 5/2) * 2)}; /* Michael Somos, Dec 06 2016 */

a(n) = ceiling(c^(2^n)) where c=(5+sqrt(21))/2 is the largest root of x^2-5x+1=0. - Benoit Cloitre, Dec 03 2002
a(n) = 2*T(2^n,5/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
Engel expansion of 1/2*(5 - sqrt(21)). Thus 1/2*(5 - sqrt(21)) = 1/5 + 1/(5*23) + 1/(5*23*527) + .... See Liardet and Stambul. Cf. A001566, A003010 and A003423. - Peter Bala, Oct 31 2012
From Peter Bala, Nov 11 2012: (Start)
a(n) = ((5 + sqrt(21))/2)^(2^n) + ((5 - sqrt(21))/2)^(2^n).
sqrt(21)/6 = Product_{n = 0..oo} (1 - 1/a(n)).
sqrt(7/3) = Product_{n = 0..oo} (1 + 2/a(n)).
a(n) - 1 = A145504(n+1). (End)
a(n) = A003501(2^n). - Michael Somos, Dec 06 2016
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2 + 3*Product_{k = 0 ..n-1} (a(k) + 2) for n >= 1.
Let b(n) = a(n) - 5. 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. (End)

One more term from Harvey P. Dale, Feb 19 2015

A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10.

Original entry on oeis.org

10, 118, 14158, 200477278, 40191139395243838, 1615327685887921300502934267457918, 2609283532796026943395592527806764363779539144932833602430435810558

Offset: 1

Artur Jasinski, Oct 11 2008

General formula for a(n+1)=a(n)^2+2*a(n)-2 and a(1)=k+1 is a(n)=Floor[((k + Sqrt[k^2 + 4])/2)^(2^((n+1) - 1))

Cf. A145502, A145503, A145504, A145505, A145506, A145507, A145508, A145509.

aa = {}; k = 10; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k =9; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 1/2*(11 + sqrt(117)). a(n) = 1 (mod 9).
Recurrence: a(n) = 12*{Product_{k = 1..n-1} a(k)} - 2 with a(1) = 10.
Product {n = 1..inf} (1 + 1/a(n)) = 12/sqrt(117).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(13/9).
(End)

cp's OEIS Frontend

A145502 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A003487 a(n) = a(n-1)^2 - 2, with a(0) = 5.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Formula