A376870 -id:A376870 - cp's OEIS frontend

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

1, 5, 355, 94852805, 1709678476417571835487555, 9994796326591347130392203807311551183419838794447313956622219314498503205

Offset: 0

Steven Finch, Oct 07 2024

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.

			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).

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

Cf. A010503, A023365, A376870.

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

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 *)

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

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

A134799 a(n) = 3^((3^n - 1)/2).

Original entry on oeis.org

1, 3, 81, 1594323, 12157665459056928801, 5391030899743293631239539488528815119194426882613553319203

Offset: 0

Yasutoshi Kohmoto, Jan 09 2008

Number of partitions into "bus routes" of the graph G_{n+1} defined below.

These seem to be one-third the reduced denominators of Newton's iteration for 1/sqrt(3), starting with 1/3. - Steven Finch, Oct 08 2024

			.........|..................G_1
****
.......__|__................G_2
.........|
****
.__|_____|_____|__..........G_3
...|.....|.....|
.........|
.......__|__
.........|
****.
..._|_........._|_..........G_4
_|__|_____|_____|__|_
.|._|_....|...._|_.|
....|.....|.....|
......_|__|__|_
.......|._|_.|
..........|
****
G_1 = o---. = rooted tree with one edge and one leaf node. For n > 0, G_{n+1} is obtained from G_n by splitting each leaf node into three.

Andrew Howroyd, Table of n, a(n) for n = 0..7
Author?, Mitsumata tree
X. Gourdon and P. Sebah, Pythagoras' Constant.

Cf. A131709, A020760, A376870.

3^((3^Range[0, 6] - 1)/2) (* Paolo Xausa, Oct 17 2024 *)

a(n) is conjectured to be one-third the reduced denominator of b(n) = (3/2)*b(n-1)*(1 - b(n-1)^2); b(0) = 1/3. - Steven Finch, Oct 08 2024

Limit_{n -> oo} A376870(n)/(3*a(n)) = 1/sqrt(3) = A020760. - Steven Finch, Oct 08 2024

Edited by N. J. A. Sloane, Jan 29 2008

a(5) from Andrew Howroyd, Oct 07 2024

cp's OEIS Frontend

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

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