OFFSET
0,2
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.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..7
X. Gourdon and P. Sebah, Pythagoras' Constant.
FORMULA
a(n) is the reduced numerator of b(n) = b(n-1)*(3/2 - b(n-1)^2); b(0) = 1/2.
a(n+1) = a(n)*(3*2^(2*3^n-1)-a(n)^2). - Chai Wah Wu, Oct 11 2024
EXAMPLE
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).
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 *)
PROG
(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)
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Steven Finch, Oct 07 2024
STATUS
approved