A219012 Numerators in a product expansion for sqrt(3).

4, 724, 198924689265124, 311489521560905211009923410118036749078665998068304623331823899816643124

Offset: 0

Author

Peter Bala, Nov 09 2012

Keywords

• nonn
• easy

Comments

a(4) has 358 digits.

Iterating the algebraic identity sqrt(1 + 4/x) = (1 + 2*(x + 2)/(x^2 + 3*x + 1)) * sqrt(1 + 4/(x*(x^2 + 5*x + 5)^2)) produces the rapidly converging product expansion sqrt(1 + 4/x) = Product_{n >= 0} (1 + 2*a(n)/b(n)); a(n) and b(n) are integer sequences when x is a positive integer.

The present case is when x = 2. The denominators b(n) are in A219013. See also A219010 (x = 1) and A219014 (x = 4).

Examples

			The first three terms of the product give 70 correct decimal places of sqrt(3): (1 + 2*4/11)*(1 + 2*724/523451)*(1 + 2*198924689265124/39571031999225940638473470251) = 1.73205 08075 68877 29352 74463 41505 87236 69428 05253 81038 06280 55806 97945 19330 (0...).
		

Links

• Amiram Eldar, Table of n, a(n) for n = 0..4

Crossrefs

Cf. A019973, A219010, A219014, A219013.

Programs

Mathematica

RecurrenceTable[{a[n+1] == a[n]^5 - 5*a[n]^3 + 5*a[n], a[0] == 4}, a, {n, 0, 3}] (* Amiram Eldar, Jul 20 2025 *)

Formula

Let tau = 2 + sqrt(3). Then a(n) = tau^(5^n) + 1/tau^(5^n).

Recurrence equation: a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with initial condition a(0) = 4.

a(n) = 2*T(5^n,2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Mar 29 2022

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. - Peter Bala, Dec 06 2022