OFFSET
0,5
COMMENTS
The digits produced by the continued fraction are that of A372869.
Let c(n) be the n-th convergent resulting from A084580.
Further continued fraction terms are taken to be unknown and the worst case (maximum absolute difference) is a single additional 1 and so at worst c(n+1).
a(n) is therefore the length of agreement between c(n) and c(n+1).
Limit_{n->oo} n/a(n) seems to approach a value between 1/(2*log_10(Pi)) and Loch's constant (A086702). - Corrected by Jwalin Bhatt, Jun 29 2026
LINKS
Jwalin Bhatt, Table of n, a(n) for n = 0..5000
Wikipedia, Lochs's theorem.
EXAMPLE
For n=6, the continued fraction of 6 terms is c(n) = [0; 1,1,2,1,1,3] = 25/43 and the worst case continuation is c(n+1) = 32/55,
25/43 = 0.581 39...
32/55 = 0.581 81...
^ ^^^ a(7) = 4 places agree
PROG
(Python) # Using sample_gauss_kuzmin_distribution function from A084580.
from sympy import floor, continued_fraction_convergents
from collections import deque
from os.path import commonprefix
def reliable_digits_from_cf(coeffs, prec):
bound1, bound2 = deque(continued_fraction_convergents(coeffs+[1]), maxlen=2)
order = 10**prec
trunc_bound1 = floor(bound1*order) / order
trunc_bound2 = floor(bound2*order) / order
rel_digits = commonprefix([f'{trunc_bound1:.{prec}f}', f'{trunc_bound2:.{prec}f}'])
return len(rel_digits) - 1 if rel_digits else 0
coeffs = sample_gauss_kuzmin_distribution(100)
A390737 = [reliable_digits_from_cf([0]+coeffs[:i], 1+int(i*1.031)) for i in range(len(coeffs)+1)]
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Jwalin Bhatt, Nov 16 2025
EXTENSIONS
Edited and corrected by Jwalin Bhatt, Feb 04 2026
STATUS
approved
