A001601 a(n) = 2*a(n-1)^2 - 1, if n>1. a(0)=1, a(1)=3.
1, 3, 17, 577, 665857, 886731088897, 1572584048032918633353217, 4946041176255201878775086487573351061418968498177
Offset: 0
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
- Dennis Martin, Table of n, a(n) for n = 0..11
- 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.
- 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, 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.
- Index entries for sequences related to Engel expansions
Programs
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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 *)
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PARI
a(n)=if(n<1,n==0,2*a(n-1)^2-1)
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
Comments