This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2. (Formerly M2705 N1084) 37
 3, 7, 47, 2207, 4870847, 23725150497407, 562882766124611619513723647, 316837008400094222150776738483768236006420971486980607 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS Expansion of 1/phi: 1/phi = (1-1/3)*(1-1/((3-1)*7))*(1-1/(((3-1)*7-1)*47))*(1-1/((((3-1)*7-1)*47-1)*2207))... (phi being the golden ration (1+sqrt(5))/2). - Thomas Baruchel, Nov 06 2003 An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004 Starting with 7, the terms end with 7,47,07,47,07,..., of the form 8a+7 where a = 0,1,55,121771,... Conjecture: Every a is squarefree, every other a is divisible by 55, the a's are a subset of A046194, the heptagonal triangular numbers (the first,2nd,3rd,6th,11th,?... terms). - Gerald McGarvey, Aug 08 2004 Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. - Cino Hilliard, Sep 28 2008 The subsequence of primes begins a(n) for n = 0, 1, 2, 3. - Jonathan Vos Post, Feb 26 2011 We have sum{n=0,..,N} a(n)^2 = 2*(N+1) + sum{n=1,..,N+1} a(n), sum{n=0,..,N} a(n)^4 = 5*(sum{n=1,..,N+1} a(n)) + a(N+1)^2 + 6*N -3, etc. which is very interesting with respect the fact that a(n) = Lucas(2^(n+1)) - see W. Webb's problem in Witula-Slota's paper. - Roman Witula, Nov 02 2012 From Peter Bala, Nov 11 2012: (Start) The present sequence corresponds to the case x = 3 of the following general remarks. The recurrence a(n+1) = a(n)^2 - 2 with initial condition a(0) = x > 2 has the solution a(n) = ((x + sqrt(x^2 - 4))/2)^(2^n) + ((x - sqrt(x^2 - 4))/2)^(2^n). We have the product expansion sqrt(x + 2)/sqrt(x - 2) = product {n = 0..inf} (1 + 2/a(n)) (essentially due to Euler - see Mendes-France and van der Poorten). Another expansion is sqrt(x^2 - 4)/(x + 1) = product {n = 0..inf} (1 - 1/a(n)), which follows by iterating the identity sqrt(x^2 - 4)/(x + 1) = (1 - 1/x)*sqrt(y^2 - 4)/(y + 1), where y = x^2 - 2. The sequence b(n) := a(n) - 1 satisfies b(n+1) = b(n)^2 + 2*b(n) - 2. Cases currently in the database are A145502 through A145510. The sequence c(n) := a(n)/2 satisfies c(n+1) = 2*c(n)^2 - 1. Cases currently in the database are A002812, A001601, A005828, A084764 and A084765. (End) 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. 397. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223. E. Lucas, Nouveaux theoremes d'arithmetique superieure, Comptes Rend., 83 (1876), 1286-1288. M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220. 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). R. Witula, D. Slota, delta-Fibonacci Numbers, Appl. Anal. Discrete Math., 3 (2009), 310-329. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..12 A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fib. Quart., 11 (1973), 429-437. P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68. E. Lucas, Nouveaux theoremes d'arithmetique superieure (annotated scanned copy) J. Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy] Wikipedia, Engel Expansion FORMULA a(n) = Fibonacci(2^(n+2))/Fibonacci(2^(n+1)) = A058635(n+2)/A058635(n+1). - Len Smiley, May 08 2000, and Artur Jasinski, Oct 05 2008 a(n) = ceiling(c^(2^n)) where c = (3+sqrt(5))/2 = tau^2 is the largest root of x^2-3*x+1=0. - Benoit Cloitre, Dec 03 2002 a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - Artur Jasinski, Sep 22 2008 a(n) = (G^(2^(n+1))-(1-G)^(2^(n+1)))/((G^(2^n))-(1-G)^(2^n)) = G^(2^n)+(1-G)^(2^n) = G^(2^n)+(-G)^(-2^n) where G is the golden ratio. - Artur Jasinski, Oct 05 2008 a(n) = 2*cosh(2^n*arccosh(sqrt(5)/2). - Artur Jasinski, Oct 09 2008 a(n) = Fibonacci(2^(n+1)-1)+Fibonacci(2^(n+1)+1). (3-sqrt(5))/2 = 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ... (E. Lucas).  - Philippe Deléham, Apr 21 2009 a(n)*(a(n+1)-1)/2 = A023039(2^n). - M. F. Hasler, Sep 27 2009 For n>=1, a(n) = 2+prod{i=0..n-1}(a(i)+2). - Vladimir Shevelev, Nov 28 2010 a(n) = 2*T(2^n,3/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011 Engel expansion of 1/2*(3 - sqrt(5)). Thus 1/2*(3 - sqrt(5)) = 1/3 + 1/(3*7) + 1/(3*7*47) + ... as noted above by Deleham. See Liardet and Stambul. - Peter Bala, Oct 31 2012 sqrt(5)/4 = product {n = 0..inf} (1 - 1/a(n)). sqrt(5) = product {n = 0..inf} (1 + 2/a(n)). a(n) - 1 = A145502(n+1). - Peter Bala, Nov 11 2012 a(n) == 2 (mod 9), for n>1. - Ivan N. Ianakiev, Dec 25 2013 EXAMPLE From Cino Hilliard, Sep 28 2008: (Start) Init x=1. x = (5/1+1)/2 = 3/1 x = (5/3+3)/2 = 7/3 x = (5/7/3+7/3)/2 = 47/21 x = (5/47/21+47/21)/2 = 2207/987 (2207/987)^2 = 5.000004106... (End) MAPLE a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'): seq(a(n), n=0..8); MATHEMATICA c = N[GoldenRatio, 1000]; Table[Round[c^(2^n)], {n, 1, 10}] (* Artur Jasinski, Sep 22 2008 *) c = (1 + Sqrt[5])/2; Table[Expand[c^(2^n) + (-c + 1)^(2^n)], {n, 1, 8}] (* Artur Jasinski, Oct 05 2008 *) G = (1 + Sqrt[5])/2; Table[Expand[(G^(2^(n + 1)) - (1 - G)^(2^(n + 1)))/Sqrt[5]]/Expand[((G^(2^n)) - (1 - G)^(2^n))/Sqrt[5]], {n, 1, 10}] (* Artur Jasinski, Oct 05 2008 *) Table[2*Cosh[2^n*ArcCosh[Sqrt[5]/2], {n, 1, 30}] (* Artur Jasinski, Oct 09 2008 *) NestList[#^2-2&, 3, 10] (* Harvey P. Dale, Dec 17 2014 *) PROG (PARI) {a(n) = if( n<1, 3*(n==0), a(n-1)^2 - 2)}; /* Michael Somos, Mar 14 2004 */ (PARI) g(n, p) = x=1; for(j=1, p, x=(n/x+x)/2; print1(numerator(x)", ")); g(5, 8) \\ Cino Hilliard, Sep 28 2008 (Maxima) a[0]:3\$ a[n]:=a[n-1]^2-2\$ A001566(n):=a[n]\$ makelist(A001566(n), n, 0, 7); /* Martin Ettl, Nov 12 2012 */ (PARI) {a(n) = my(w = quadgen(5)); if( n<0, 0, n++; imag( (2*w - 1) * w^2^n ))}; /* Michael Somos, Nov 30 2014 */ (PARI) {a(n) = my(y = x^2-x-1); if( n<0, 0, n++; for(i=1, n, y = polgraeffe(y)); -polcoeff(y, 1))}; /* Michael Somos, Nov 30 2014 */ CROSSREFS Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herbert S. Wilf). Cf. A000045. Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5). Cf. A058635. - Artur Jasinski, Oct 05 2008 Cf. A002812, A145502, A145274, A050614, A088334, A181393, A181419, A186750-A186751. Sequence in context: A052381 A219877 A031440 * A173771 A019039 A077559 Adjacent sequences:  A001563 A001564 A001565 * A001567 A001568 A001569 KEYWORD easy,nonn,nice AUTHOR STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified November 12 13:36 EST 2018. Contains 317109 sequences. (Running on oeis4.)