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A219164
Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.
3
5, 527, 77132286527, 35395236908668169265765137996816180039862527
OFFSET
0,1
COMMENTS
Bisection of A003487.
The next term -- a(4) -- has 175 digits. - Harvey P. Dale, Jun 09 2017
FORMULA
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)
MATHEMATICA
NestList[#^4-4#^2+2&, 5, 5] (* Harvey P. Dale, Jun 09 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 13 2012
STATUS
approved