A393695 - OEIS

%I #7 Mar 03 2026 15:47:32

%S 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,

%T 23,23,26,26,27,28,28,30,30,32,32,32,33,34,37,37,37,39,39,41,42,42,43,

%U 44,46,46,48,48,48,51,51,52,52,54,55,55,57,57,58,60,60

%N Number of decimal places which are known after using n terms of A241773 and treating them as the coefficients of a continued fraction.

%C The digits produced by the continued fraction are that of A392388.

%C Let c(n) be the n-th convergent resulting from A241773.

%C 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).

%C a(n) is therefore the length of agreement between c(n) and c(n+1).

%C 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.

%C 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.

%H Jwalin Bhatt, <a href="/A393695/b393695.txt">Table of n, a(n) for n = 0..5000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Lochs%27s_theorem">Lochs's theorem</a>

%e 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,

%e      52/75 = 0.693 33...

%e     95/137 = 0.693 43...

%e              ^ ^^^ a(7) = 4 places agree

%o (Python) # Using sample_gauss_kuzmin_distribution function from A241773.

%o from sympy import floor, continued_fraction_convergents

%o from collections import deque

%o from os.path import commonprefix

%o def reliable_digits_from_cf(coeffs, prec):

%o     bound1, bound2 = deque(continued_fraction_convergents(coeffs+[1]), maxlen=2)

%o     order = 10**prec

%o     trunc_bound1 = floor(bound1*order) / order

%o     trunc_bound2 = floor(bound2*order) / order

%o     rel_digits = commonprefix([f'{trunc_bound1:.{prec}f}', f'{trunc_bound2:.{prec}f}'])

%o     return len(rel_digits) - 1 if rel_digits else 0

%o coeffs = sample_gauss_kuzmin_distribution(100)

%o A393695 = [reliable_digits_from_cf([0]+coeffs[:i], 1+int(i*1.031)) for i in range(len(coeffs)+1)]

%Y Cf. A062542, A086819, A241773, A392388.

%K nonn

%O 0,5

%A _Jwalin Bhatt_, Feb 25 2026