login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

A001601
a(n) = 2*a(n-1)^2 - 1, if n>1. a(0)=1, a(1)=3.
(Formerly M3042 N1234)
22
1, 3, 17, 577, 665857, 886731088897, 1572584048032918633353217, 4946041176255201878775086487573351061418968498177
OFFSET
0,2
COMMENTS
Reduced numerators of Newton's iteration for sqrt(2). - Eric W. Weisstein
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Evaluation of the 2^n - 1 degree interpolating polynomial of 1/x at Chebyshev nodes in the interval (1,2): v = 1.0; for(i = 1, n, v *= 4*(a(i) - x*v)); v *= 2/a(n+1). - Jose Hortal, Apr 07 2012
Smallest positive integer x satisfying the Pell equation x^2 - 2^(2*n+1) * y^2 = 1, for n > 0. - A.H.M. Smeets, Sep 29 2017
REFERENCES
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. 376.
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).
LINKS
Dov Jarden, Recurring Sequences, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy]
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
Eric Weisstein's World of Mathematics, Newtons Iteration.
Eric Weisstein's World of Mathematics, Square Root.
Eric Weisstein's World of Mathematics, Pythagoras's Constant.
H. S. Wilf, D. C. B. Marsh and J. V. Whittaker, Problem E1093, Amer. Math. Monthly, 61 (1954), 424-425.
FORMULA
From Mario Catalani (mario.catalani(AT)unito.it), May 27 2003, May 30 2003: (Start)
a(n) = a(n-1)^2 + 2*A051009(n)^2 for n > 0.
a(n)^2 = 2*A051009(n+1)^2 + 1.
a(n) = Sum_{r=0..2^(n-1)} binomial(2^n, 2*r)*2^r. (End)
Expansion of 1/sqrt(2) as an infinite product: 1/sqrt(2) = Product_{k>=1} (1 - 1/(a(n)+1)). a(1)=3; a(n) = floor(1/(1-1/(sqrt(2)*Product_{k=1..n-1} 1-1/(a(k)+1)))). - Thomas Baruchel, Nov 06 2003
2*a(n+1) = A003423(n).
a(n) = (1/2)*((1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n)). - Artur Jasinski, Oct 10 2008
For n > 1: a(n) - 1 = 4^n * Product_{i=1..n-2} a(i)^2. - Jose Hortal, Apr 13 2012
From Peter Bala, Nov 11 2012: (Start)
4*sqrt(2)/7 = Product_{n>=1} (1 - 1/(2*a(n))).
sqrt(2) = Product_{n>=1} (1 + 1/a(n)).
a(n) = (1/2)*A003423(n-1). (End)
a(n) = cos(2^(n-1) * arccos(3)) = cosh(2^(n-1) * log(3 + 2*sqrt(2))) for n >= 1. - Daniel Suteu, Jul 28 2016
a(n+1) = T(2^n,3), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Feb 01 2017
a(n) = A001541(2^(n-1)). - A.H.M. Smeets, May 28 2017
MATHEMATICA
Table[Simplify[Expand[(1/2) ((1 + Sqrt[2])^(2^n) + (1 - Sqrt[2])^(2^n))]], {n, 0, 7}] (* Artur Jasinski, Oct 10 2008 *)
Join[{1}, NestList[2#^2-1&, 3, 7]] (* Harvey P. Dale, Mar 24 2011 *)
PROG
(PARI) a(n)=if(n<1, n==0, 2*a(n-1)^2-1)
CROSSREFS
a(n) = A001333(2^n).
Sequence in context: A305375 A128300 A292082 * A061119 A094133 A049985
KEYWORD
nonn,easy,nice,frac
STATUS
approved