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%I M2705 N1084 #189 Mar 01 2024 09:34:51
%S 3,7,47,2207,4870847,23725150497407,562882766124611619513723647,
%T 316837008400094222150776738483768236006420971486980607
%N a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.
%C 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
%C An infinite coprime sequence defined by recursion. - _Michael Somos_, Mar 14 2004
%C 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
%C Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. - _Cino Hilliard_, Sep 28 2008
%C The subsequence of primes begins a(n) for n = 0, 1, 2, 3. - _Jonathan Vos Post_, Feb 26 2011
%C 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
%C From _Peter Bala_, Nov 11 2012: (Start)
%C The present sequence corresponds to the case x = 3 of the following general remarks.
%C 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).
%C 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.
%C 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.
%C (End)
%C 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
%C Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - _Peter Bala_, Dec 08 2022
%D 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.
%D E.-B. Escott, Note #1741, L'Intermédiaire des Mathématiciens, 8 (1901), page 13. - _N. J. A. Sloane_, Mar 02 2022
%D G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223.
%D Édouard Lucas, Nouveaux théoremes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A001566/b001566.txt">Table of n, a(n) for n = 0..12</a>
%H A. V. Aho and N. J. A. Sloane, <a href="https://www.fq.math.ca/Scanned/11-4/aho-a.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437, <a href="http://neilsloane.com/doc/doubly.html">alternative link</a>.
%H Pierre Liardet and Pierre Stambul, <a href="http://www.numdam.org/item?id=JTNB_2000__12_1_37_0">Séries d'Engel et fractions continuées</a>, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
%H Édouard Lucas, <a href="/A001566/a001566.pdf">Nouveaux théorèmes d'arithmétique supérieure</a> (annotated scanned copy)
%H Édouard Lucas, <a href="http://www.mathstat.dal.ca/FQ/Books/Complete/simply-periodic.pdf">The Theory of Simply Periodic Numerical Functions</a>, Fibonacci Association, 1969. English translation of article "Théorie des Fonctions Numériques Simplement Périodiques, I", Amer. J. Math., 1 (1878), 184-240.
%H M. Mendes France and A. J. van der Poorten, <a href="https://doi.org/10.1016/0304-3975(89)90045-5">From geometry to Euler identities</a>, Theoret. Comput. Sci., 65 (1989), 213-220.
%H Chance Sanford, <a href="https://arxiv.org/abs/1603.03765">Infinite Series Involving Fibonacci Numbers Via Apéry-Like Formulae</a>, arXiv:1603.03765 [math.NT], 2016.
%H Jeffrey Shallit, <a href="/A005248/a005248_1.pdf">An interesting continued fraction</a>, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>.
%H Roman Wituła and Damian Słota, <a href="https://doi.org/10.2298/AADM0902310W">delta-Fibonacci numbers</a>, Applicable Analysis and Discrete Mathematics, Vol. 3, No. 2 (2009), pp. 310-329.
%H <a href="/index/Aa#AHSL">Index entries for sequences of form a(n+1)=a(n)^2 + ...</a>.
%F 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
%F 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
%F a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - _Artur Jasinski_, Sep 22 2008
%F 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
%F a(n) = 2*cosh(2^n*arccosh(sqrt(5)/2). - _Artur Jasinski_, Oct 09 2008
%F 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
%F a(n)*(a(n+1)-1)/2 = A023039(2^n). - _M. F. Hasler_, Sep 27 2009
%F For n >= 1, a(n) = 2 + Product_{i=0..n-1} (a(i) + 2). - _Vladimir Shevelev_, Nov 28 2010
%F 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
%F From _Peter Bala_, Oct 31 2012: (Start)
%F 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.
%F sqrt(5)/4 = Product_{n>=0} (1 - 1/a(n)).
%F sqrt(5) = Product_{n>=0} (1 + 2/a(n)). (End)
%F a(n) - 1 = A145502(n+1). - _Peter Bala_, Nov 11 2012
%F a(n) == 2 (mod 9), for n > 1. - _Ivan N. Ianakiev_, Dec 25 2013
%F From _Amiram Eldar_, Oct 22 2020: (Start)
%F a(n) = A000032(2^(n+1)).
%F Sum_{k>=0} 1/a(k) = -1 + A338304. (End)
%F a(n) = (A000045(m+2^(n+2))+A000045(m))/A000045(m+2^(n+1)) for any m>=0. - _Alexander Burstein_, Apr 10 2021
%F a(n) = 2*cos(2^n*arccos(3/2)). - _Peter Luschny_, Oct 12 2022
%F a(n) == -1 ( mod 2^(n+2) ). - _Peter Bala_, Nov 07 2022
%F a(n) = 5*Fibonacci(2^n)^2+2 = 5*A058635(n)^2+2, for n>0. - _Jianglin Luo_, Sep 21 2023
%F Sum_{n>=0} a(n)/Fibonacci(2^(n+2)) = A094874 (Sanford, 2016). - _Amiram Eldar_, Mar 01 2024
%e From _Cino Hilliard_, Sep 28 2008: (Start)
%e Init x=1;
%e x = (5/1 + 1)/2 = 3/1;
%e x = (5/3 + 3)/2 = 7/3;
%e x = ((5/7)/3 + 7/3)/2 = 47/21;
%e x = ((5/47)/21 + 47/21)/2 = 2207/987;
%e (2207/987)^2 = 5.000004106... (End)
%p a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'):
%p seq(a(n), n=0..8);
%t c = N[GoldenRatio, 1000]; Table[Round[c^(2^n)], {n, 1, 10}] (* _Artur Jasinski_, Sep 22 2008 *)
%t c = (1 + Sqrt[5])/2; Table[Expand[c^(2^n) + (-c + 1)^(2^n)], {n, 1, 8}] (* _Artur Jasinski_, Oct 05 2008 *)
%t 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 *)
%t Table[2*Cosh[2^n*ArcCosh[Sqrt[5]/2],{n,1,30}] (* _Artur Jasinski_, Oct 09 2008 *)
%t NestList[#^2-2&,3,10] (* _Harvey P. Dale_, Dec 17 2014 *)
%t Table[LucasL[2^n], {n, 1, 8}] (* _Amiram Eldar_, Oct 22 2020 *)
%o (PARI) {a(n) = if( n<1, 3*(n==0), a(n-1)^2 - 2)}; /* _Michael Somos_, Mar 14 2004 */
%o (PARI) g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","));
%o g(5,8) \\ _Cino Hilliard_, Sep 28 2008
%o (PARI) {a(n) = my(w = quadgen(5)); if( n<0, 0, n++; imag( (2*w - 1) * w^2^n ))}; /* _Michael Somos_, Nov 30 2014 */
%o (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 */
%o (Maxima)
%o a[0]:3$
%o a[n]:=a[n-1]^2-2$
%o A001566(n):=a[n]$
%o makelist(A001566(n),n,0,7); /* _Martin Ettl_, Nov 12 2012 */
%Y Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herbert S. Wilf). Cf. A000045.
%Y Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5).
%Y Cf. A058635. - _Artur Jasinski_, Oct 05 2008
%Y Cf. A002812, A094874, A145502, A145274, A050614, A088334, A181393, A181419, A186750, A186751, A338304.
%K easy,nonn,nice
%O 0,1
%A _N. J. A. Sloane_
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