|
%I #37 Mar 08 2023 11:50:37
%S 1,4,5,19,24,91,115,436,551,2089,2640,10009,12649,47956,60605,229771,
%T 290376,1100899,1391275,5274724,6665999,25272721,31938720,121088881,
%U 153027601,580171684,733199285,2779769539,3512968824
%N Denominators in continued fraction [0; 1, 3, 1, 3, 1, 3, ...].
%C A136210(n)/A136211(n) tends to 0.791287847... = [0; 1, 3, 1, 3, 1, 3, ...] = (sqrt(21) - 3)/2 = the inradius of a right triangle with hypotenuse 3, legs 2 and sqrt(21).
%C The number 0.791287847... = (sqrt(21) - 3)/2 arises in finding a number which is 5 less than its square; the result is: 2.791287847... because (2.791287847...)^2 = 7.791287847... In general the quadratic equation for finding such numbers is x^2 - x = N, so x = (1 + sqrt(1 + 4N))/2. - _Alexander R. Povolotsky_, Dec 23 2007
%C Prepending a 1 to the sequence gives [1, 1, 4, 5, 19, 24, ...]. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) with the parameters R = 3 and Q = -1. It is a strong divisibility sequence, that is, GCD(a(n),a(m)) = a(GCD(n,m)) for all natural numbers n and m. - _Peter Bala_, May 14 2014
%H G. C. Greubel, <a href="/A136211/b136211.txt">Table of n, a(n) for n = 1..1000</a>
%H P. Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry4/barry64.html">Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays</a>, JIS 12 (2009) 09.8.6
%H J. L. Ramirez, F. Sirvent, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Ramirez/ramirez9.html">A q-Analogue of the Bi-Periodic Fibonacci Sequence</a>, J. Int. Seq. 19 (2016) # 16.4.6, t_n at a=1, b=3.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LehmerNumber.html">Lehmer Number</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 5, 0, -1).
%F a(1) = 1, a(2) = 4, then for n>2, a(2n) = 3*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Let T = the 2 X 2 matrix [1, 3; 1, 4]. Then T^n = [A136210(2n-1), A136210(2n); a(2n-1), a(2n)].
%F From _R. J. Mathar_, May 18 2008: (Start)
%F O.g.f.: x*(1+4*x-x^3)/(1-5*x^2+x^4).
%F a(2*n) = A004253(n+1).
%F a(2*n+1) = A004254(n). (End)
%F a(n)*a(n+1) = A099025(n). - _R. K. Guy_, May 18 2008
%F {-a(n) + 5 a(n + 2) - a(n + 4), a(0) = 1, a(1) = 4, a(2) = 5, a(3) = 19}. - _Robert Israel_, May 14 2008
%e a(4) = 19 = 3*a(3) + a(2) = 3*5 + 4.
%e a(5) = 24 = a(4) + a(3) = 19 + 5.
%e T^3 = [19, 72; 24, 91], where the bottom row [24, 91] = [a(5), a(6)].
%t Denominator[NestList[(3/(3+#))&,0,60]] (* _Vladimir Joseph Stephan Orlovsky_, Apr 13 2010 *)
%t a[n_] := FromContinuedFraction[ Join[{0}, 3 - 2*Array[Mod[#, 2]&, n]]] // Denominator; Table[a[n], {n, 1, 30}] (* _Jean-François Alcover_, May 15 2014 *)
%o (PARI) x='x + O('x^25); Vec(x*(1+4*x-x^3)/(1-5*x^2+x^4)) \\ _G. C. Greubel_, Feb 18 2017
%Y Cf. A136210.
%K nonn,frac,easy
%O 1,2
%A _Gary W. Adamson_, Dec 21 2007
|
