A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
1, 5, 115, 60605, 16831644835, 1298263252133919638045, 7723873922612696850892381990249713732303715, 273388347343560518533856033712658350781293745092679040607342582493129736504927611387805
Offset: 0
Examples
Let b = 1/sqrt(21) and x = 1/2*(5 - sqrt(21)). We have the following Pierce expansions to base b: 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) - .... x^2 = b/5 - b^2/(5*115) + b^3/(5*115*60605) - b^4/(5*115*60605*16831644835) + .... x^4 = b/115 - b^2/(115*60605) + b^3/(115*60605*16831644835) - .... x^8 = b/60605 - b^2/(60605*16831644835) + ....
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10
Programs
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Mathematica
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 *) a[ n_] := If[ n < 0, 0, ChebyshevU[2^n - 1, 5/2]]; (* Michael Somos, Dec 06 2016 *)
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PARI
a(n) = if( n<0, 0, imag( (5 + quadgen(84))^2^n) / 2^(2^n - 1)); /* Michael Somos, Dec 06 2016 */
Formula
a(n) = 1/sqrt(21)*( alpha^(2^n) - (1/alpha)^(2^n) ), where alpha = 1/2*(5 + sqrt(21)).
a(n) = product {k = 0..n-1} A003487(k).
Defining recurrence equation:
a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
Other recurrence equations:
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = 21*a(n-2)^2 + 2 for n >= 2.
a(n) = A004254(2^n). - Michael Somos, Dec 06 2016
Comments