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 A002812 a(n) = 2*a(n-1)^2 - 1, starting a(0)=2. (Formerly M1817 N0720) 17
 2, 7, 97, 18817, 708158977, 1002978273411373057, 2011930833870518011412817828051050497, 8095731360557835890888779535060256832479295062749579257164654370487894017 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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..inf} (1 - 1/(2*a(n))). A second expansion is product {n = 0..inf} (1 + 1/a(n)) = sqrt((x + 1)/(x - 1)). See Mendes-France and van der Poorten. (End) Is lpf(a(p-2)) = 2^p-1 for all odd Mersenne exponents p in A000043? - Thomas Ordowski, Aug 12 2018 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. 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). LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..10 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 FORMULA a(n) = A001075(2^n). a(n) = ((2+sqrt(3))^(2^n)+(2-sqrt(3))^(2^n))/2. - Bruno Berselli, Dec 20 2011 2*sqrt(3)/5 = product {n = 0..inf} (1 - 1/(2*a(n))). sqrt(3) = product {n = 0..inf} (1 + 1/a(n)). a(n) = 1/2*A003010(n). - Peter Bala, Nov 11 2012 EXAMPLE G.f. = 2 + 7*x + 97*x^2 + 18817*x^3 + 708158977*x^4 + ... MAPLE 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 MATHEMATICA 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 *) PROG (PARI) {a(n) = if( n<1, 2 * (n==0), 2 * a(n-1)^2 - 1)}; /* Michael Somos, Mar 14 2004 */ (PARI) /* 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 (MAGMA) I:=[2]; [n le 1 select I[n] else 2*Self(n-1)^2-1: n in [1..8]]; // Vincenzo Librandi, Dec 19 2011 (GAP) a:=[2];; for n in [2..10] do a[n]:=2*a[n-1]^2-1; od; a; # Muniru A Asiru, Aug 12 2018 CROSSREFS Cf. A001075, A003010, A071579. Sequence in context: A240696 A102344 A087589 * A219280 A277434 A322642 Adjacent sequences:  A002809 A002810 A002811 * A002813 A002814 A002815 KEYWORD nonn,easy,nice AUTHOR STATUS approved

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Last modified May 27 17:53 EDT 2020. Contains 334664 sequences. (Running on oeis4.)