A219161 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 5.

5, 110, 1330670, 2356194280407770990, 13080769480548649962914459850235688797656360638877986030

Offset: 0

Author

Peter Bala, Nov 13 2012

Keywords

• nonn
• easy

Comments

For some general remarks on this recurrence see A001999.

The next term (a(5)) has 166 digits. - Harvey P. Dale, Apr 23 2019

Links

• G. C. Greubel, Table of n, a(n) for n = 0..6
• E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
• N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977, 629-630.

Crossrefs

Cf. A001999, A112845, A219160.

Programs

Mathematica

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 5}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)
NestList[#^3-3#&,5,5] (* Harvey P. Dale, Apr 23 2019 *)

Formula

a(n) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n).

Product_{n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(7/3).

a(n) = 2*T(3^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A001999. - Peter Bala, Feb 01 2017