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 A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2. (Formerly M2705 N1084) 41
 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 to 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} (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} (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) E. Lucas in Section XIX of "The Theory of Simply Periodic Numerical Functions" (page 56 of English translation) equation "(127) (1-sqrt(5))/2 = -1/1 + 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ..." - Michael Somos, Oct 11 2022 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. E.-B. Escott, Note #1741, L'Intermédiaire des Mathématiciens, 8 (1901), page 13. - N. J. A. Sloane, Mar 02 2022 G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223. Édouard Lucas, Nouveaux théoremes d'arithmétique supérieure, 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). LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..12 A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437, alternative link. Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68. Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy) Édouard Lucas, The Theory of Simply Periodic Numerical Functions, Fibonacci Association, 1969. English translation of article "Théorie des Fonctions Numériques Simplement Périodiques, I", Amer. J. Math., 1 (1878), 184-240. Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy] Wikipedia, Engel Expansion. Roman Wituła and Damian Słota, delta-Fibonacci numbers, Applicable Analysis and Discrete Mathematics, Vol. 3, No. 2 (2009), pp. 310-329. 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 + Product_{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 From Peter Bala, Oct 31 2012: (Start) 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. sqrt(5)/4 = Product_{n>=0} (1 - 1/a(n)). sqrt(5) = Product_{n>=0} (1 + 2/a(n)). (End) 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 From Amiram Eldar, Oct 22 2020: (Start) a(n) = A000032(2^(n+1)). Sum_{k>=0} 1/a(k) = -1 + A338304. (End) a(n) = (A000045(m+2^(n+2))+A000045(m))/A000045(m+2^(n+1)) for any m>=0. - Alexander Burstein, Apr 10 2021 a(n) = 2*cos(2^n*arccos(3/2)). - Peter Luschny, Oct 12 2022 a(n) == -1 ( mod 2^(n+2) ). - Peter Bala, Nov 07 2022 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 *) Table[LucasL[2^n], {n, 1, 8}] (* Amiram Eldar, Oct 22 2020 *) 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 (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 */ (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 */ 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, A338304. Sequence in context: A219877 A031440 A331038 * A173771 A019039 A077559 Adjacent sequences: A001563 A001564 A001565 * A001567 A001568 A001569 KEYWORD easy,nonn,nice,changed AUTHOR STATUS approved

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Last modified November 30 02:32 EST 2022. Contains 358431 sequences. (Running on oeis4.)