A219165 -id:A219165 - cp's OEIS frontend

A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3.

3, 47, 4870847, 562882766124611619513723647

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A001566. Compare the following remarks with A001999.
The present sequence is the case x = 3 of the following general remarks. For other cases see A219163 (x = 4), A219164 (x = 5) and A219165 (x = 6).
Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(4^n) + (1/alpha)^(4^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^4 + 4*a(n)^2 - 2 with the initial condition a(0) = x.
We have the product expansion sqrt((x + 2)/(x - 2)) = Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)).

Cf. A000032, A001566, A001999, A002814, A145502, A219163, A219164, A219165.

a(n)={if(n==0,3,a(n-1)^4-4*a(n-1)^2+2)} \\ Edward Jiang, Sep 11 2014

Let alpha = 1/2*(3 + sqrt(5)) then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A001566(2*n) = A000032(2*4^n).
Product {n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(5).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
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. (End)

A219163 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 4.

Original entry on oeis.org

4, 194, 1416317954, 4023861667741036022825635656102100994

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A003010.

a(4) has 147 digits and a(5) has 586 digits. - Harvey P. Dale, Mar 03 2020

Cf. A001999, A003010, A003500, A219162, A219164, A219165.

NestList[#^4-4#^2+2&,4,5] (* Harvey P. Dale, Mar 03 2020 *)

a(n)={if(n==0,4,a(n-1)^4-4*a(n-1)^2+2)} \\ Edward Jiang, Sep 11 2014

Let alpha = 2 + sqrt(3). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003010(2*n) = A003500(4^n).
Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(3).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 4. 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. (End)

A219164 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.

Original entry on oeis.org

5, 527, 77132286527, 35395236908668169265765137996816180039862527

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A003487.

The next term -- a(4) -- has 175 digits. - Harvey P. Dale, Jun 09 2017

Cf. A001999, A003487, A003501, A219162, A219163, A219165.

NestList[#^4-4#^2+2&,5,5] (* Harvey P. Dale, Jun 09 2017 *)

Let alpha = 1/2*(5 + sqrt(21)). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003487(2*n) = A003501(4^n).
Product_{n>=0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(7/3).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 5. 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. (End)

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A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3.

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A219163 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 4.

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A219164 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.

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