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A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.

%I A230338 #40 Jun 20 2022 05:15:27
%S A230338 1,5,115,60605,16831644835,1298263252133919638045,
%T A230338 7723873922612696850892381990249713732303715,
%U A230338 273388347343560518533856033712658350781293745092679040607342582493129736504927611387805
%N A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
%C A230338 For integer N, the recurrence equation a(n) = a(n-1)*sqrt((N^2 - 4)*a(n-1)^2 + 4) for n >= 1, with starting value a(0) = 1, produces an integer sequence. The present sequence is the case N = 5. Cf. A000079 (case N = 2), A058635 (case N = 3) and A071579 (case N = 4).
%C A230338 Sequence of numerators in the Engel series representation of 1/2*(7 - sqrt(21)) = 1 + 1/5 + 1 /115 + 1/60605 + .... The corresponding Engel expansion is A003487.
%C A230338 The sequence also has a description as a Pierce expansion of the quadratic irrational 1/2*(5 - sqrt(21)) to the base b := 1/sqrt(21) (see A058635 for a definition of this term).
%C A230338 The associated Pierce series representation of 1/2*(5 - sqrt(21)) to the base b begins 1/2*(5 - sqrt(21)) = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + ....
%C A230338 More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of ( 1/2*(5 - sqrt(21)) )^(2^n) to the base b = 1/sqrt(21). Some examples are given below.
%H A230338 Seiichi Manyama, <a href="/A230338/b230338.txt">Table of n, a(n) for n = 0..10</a>
%F A230338 a(n) = 1/sqrt(21)*( alpha^(2^n) - (1/alpha)^(2^n) ), where alpha = 1/2*(5 + sqrt(21)).
%F A230338 a(n) = product {k = 0..n-1} A003487(k).
%F A230338 Defining recurrence equation:
%F A230338 a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
%F A230338 Other recurrence equations:
%F A230338 a(0) = 1, a(1) = 5 and a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
%F A230338 a(0) = 1, a(1) = 5 and a(n)/a(n-1) = 21*a(n-2)^2 + 2 for n >= 2.
%F A230338 a(n) = A004254(2^n). - _Michael Somos_, Dec 06 2016
%e A230338 Let b = 1/sqrt(21) and x = 1/2*(5 - sqrt(21)). We have the following Pierce expansions to base b:
%e A230338 x = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + b^5/(1*5*115*60605*16831644835) - ....
%e A230338 x^2 = b/5 - b^2/(5*115) + b^3/(5*115*60605) - b^4/(5*115*60605*16831644835) + ....
%e A230338 x^4 = b/115 - b^2/(115*60605) + b^3/(115*60605*16831644835) - ....
%e A230338 x^8 = b/60605 - b^2/(60605*16831644835) + ....
%t A230338 a[n_] := a[n - 1]*Sqrt[21 a[n - 1]^2 + 4]; a[0] = 1; Array[a, 8, 0] (* _Robert G. Wilson v_, Mar 19 2014 *)
%t A230338 a[ n_] := If[ n < 0, 0, ChebyshevU[2^n - 1, 5/2]]; (* _Michael Somos_, Dec 06 2016 *)
%o A230338 (PARI) a(n) = if( n<0, 0, imag( (5 + quadgen(84))^2^n) / 2^(2^n - 1)); /* _Michael Somos_, Dec 06 2016 */
%Y A230338 Cf. A003487, A004254, A058635, A071579.
%K A230338 nonn,easy
%O A230338 0,2
%A A230338 _Peter Bala_, Oct 30 2013