A390737 - OEIS

%I #35 Jul 15 2026 01:08:23

%S 0,0,1,1,2,2,4,4,5,5,5,7,7,8,9,9,10,11,12,12,12,14,14,15,17,17,18,19,

%T 19,20,20,21,22,22,22,22,25,25,27,27,28,29,29,30,30,32,33,33,34,34,35,

%U 36,36,37,38,39,40,42,42,43,44,44,44,46,46,47,47,47,50

%N Number of decimal places which are known after using n terms of A084580 and treating them as the coefficients of a continued fraction.

%C The digits produced by the continued fraction are that of A372869.

%C Let c(n) be the n-th convergent resulting from A084580.

%C Further continued fraction terms are taken to be unknown and the worst case (maximum absolute difference) is a single additional 1.

%C a(n) is therefore the length of agreement between c(n) and c(n) appended with a 1. - Corrected by _Jwalin Bhatt_, Jul 13 2026

%C Limit_{n->oo} n/a(n) seems to approach a value between 1/(2*log_10(3.15)) and Lochs's constant (A086819). - Corrected by _Jwalin Bhatt_, Jun 29 2026

%H Jwalin Bhatt, <a href="/A390737/b390737.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 is c(n) = [0; 1,1,2,1,1,3] = 25/43 and the worst case continuation is 32/55,

%e     25/43 = 0.581 39...

%e     32/55 = 0.581 81...

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

%o (Python)  # Using sample_gauss_kuzmin_distribution function from A084580.

%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 A390737 = [reliable_digits_from_cf([0]+coeffs[:i], 1+int(i*1.031)) for i in range(len(coeffs)+1)]

%Y Cf. A062542, A084580, A086819, A372869.

%K nonn,base,changed

%O 0,5

%A _Jwalin Bhatt_, Nov 16 2025

%E Edited and corrected by _Jwalin Bhatt_, Feb 04 2026