A376867 Reduced numerators of Newton's iteration for 1/sqrt(2), starting with 1/2.

1, 5, 355, 94852805, 1709678476417571835487555, 9994796326591347130392203807311551183419838794447313956622219314498503205

Offset: 0

Author

Steven Finch, Oct 07 2024

Keywords

• nonn
• frac

Comments

An explicit formula for a(n) is not known, although it arises from a recurrence and the corresponding denominators are simply 2^(3^n) = A023365(n+1).

Next term is too large to include.

Examples

			a(1) = 5 because b(1) = (1/2)*(3/2 - 1/4) = 5/8.
1/2, 5/8, 355/512, 94852805/134217728, ... = a(n)/A023365(n+1).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..7
• X. Gourdon and P. Sebah, Pythagoras' Constant.

Crossrefs

Cf. A010503, A023365, A376870.

Programs

Maple

b:= proc(n) b(n):= `if`(n=0, 1/2, b(n-1)*(3/2-b(n-1)^2)) end:
a:= n-> numer(b(n)):
seq(a(n), n=0..5);  # Alois P. Heinz, Oct 07 2024

Mathematica

a[0]=1/2; a[n_]:=a[n-1](3/2-a[n-1]^2); Numerator[Array[a,6,0]] (* Stefano Spezia, Oct 15 2024 *)

Python

from itertools import count, islice
def A376867_gen(): # generator of terms
    p = 1
    for k in count(0):
        yield p
        p *= ((3<<((3**k<<1)-1))-p**2)
A376867_list = list(islice(A376867_gen(),6)) # Chai Wah Wu, Oct 11 2024

Formula

a(n) is the reduced numerator of b(n) = b(n-1)*(3/2 - b(n-1)^2); b(0) = 1/2.

Limit_{n -> oo} a(n)/A023365(n+1) = 1/sqrt(2) = A010503.

a(n+1) = a(n)*(3*2^(2*3^n-1)-a(n)^2). - Chai Wah Wu, Oct 11 2024