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URL: https://oeis.org/A393695

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A393695
Number of decimal places which are known after using n terms of A241773 and treating them as the coefficients of a continued fraction.
1
0, 0, 1, 1, 3, 4, 4, 5, 6, 6, 8, 9, 10, 10, 11, 12, 13, 14, 14, 17, 17, 17, 19, 19, 21, 21, 22, 23, 23, 26, 26, 27, 28, 28, 30, 30, 32, 32, 32, 33, 34, 37, 37, 37, 39, 39, 41, 42, 42, 43, 44, 46, 46, 48, 48, 48, 51, 51, 52, 52, 54, 55, 55, 57, 57, 58, 60, 60
OFFSET
0,5
COMMENTS
The digits produced by the continued fraction are that of A392388.
Let c(n) be the n-th convergent resulting from A241773.
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).
It is expected n/a(n) -> A086819 due to Lochs's theorem, but proving that depends on not having runs of consecutive equal terms long enough to push the ratio to some higher constant infinitely often.
Runs of consecutive equal terms occur when c(n) and c(n+1) span a run of 0 digits or 9 digits in A392388 and only several more continued fraction terms can narrow that interval enough to be sure whether 0's or 9's.
LINKS
EXAMPLE
For n=6, the continued fraction of 6 terms from A241773 is c(n) = [0; 1,2,3,1,4,1] = 57/75 and the worst case continuation is c(n+1) = 95/137,
52/75 = 0.693 33...
95/137 = 0.693 43...
^ ^^^ a(7) = 4 places agree
PROG
(Python) # Using sample_gauss_kuzmin_distribution function from A241773.
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)
A393695 = [reliable_digits_from_cf([0]+coeffs[:i], 1+int(i*1.031)) for i in range(len(coeffs)+1)]
CROSSREFS
KEYWORD
nonn
AUTHOR
Jwalin Bhatt, Feb 25 2026
STATUS
approved