A219014 Numerators in a product expansion for sqrt(2).

6, 6726, 13765255184676885126

Offset: 0

Author

Peter Bala, Nov 09 2012

Keywords

• nonn
• easy

Comments

a(3) has 96 digits and a(4) has 479 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 a rapidly converging product expansion sqrt(1 + 4/x) = Product_{n >= 0} (1 + 2*a(n)/b(n)), where a(n) and b(n) are integer sequences when x is a positive integer.

The present case is when x = 4. The denominators b(n) are in A219015. See also A219010 (x = 1) and A219012 (x = 2).

Examples

			The first two terms of the product give 18 correct decimal places of sqrt(2): (1 + 2*6/29)*(1 + 2*6726/45232349) = 1.41421 35623 73095 048(5...).
		

Links

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

Crossrefs

Cf. A156035, A219010, A219012, A219015.

Programs

Mathematica

RecurrenceTable[{a[n+1] == a[n]^5 - 5*a[n]^3 + 5*a[n], a[0] == 6}, a, {n, 0, 3}] (* Amiram Eldar, Jul 20 2025 *)

Formula

Let tau = 3 + 2*sqrt(2). 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) = 6.