OFFSET
1,3
COMMENTS
For number of correct decimal digits see A317908.
For the similar case of number of correct binary digits of Pi see A305879.
For the similar case of number of correct binary digits of log(2) see A317557.
The denominator of the k-th convergent obtained from a continued fraction satisfying the Gauss-Kuzmin distribution will tend to exp(k*A100199), A100199 being the inverse of Lévy's constant; the error between the k-th convergent and the constant itself tends to exp(-2*k*A100199), or in binary digits 2*k*A100199/log(2) bits after the binary point.
The sequence for quaternary digits is obtained by floor(a(n)/2), the sequence for octal digits is obtained by floor(a(n)/3), and the sequence for hexadecimal digits is obtained by floor(a(n)/4).
LINKS
A.H.M. Smeets, Table of n, a(n) for n = 1..19999
EXAMPLE
n convergent binary expansion a(n)
== ============= ========================== ====
1 2 / 1 10.0... 0
2 3 / 1 11.0... -1
3 8 / 3 10.101010... 5
4 43 / 16 10.1011... 3
5 51 / 19 10.1010111100... 9
oo lim = A317906 10.101011110111100111... --
PROG
(Python)
i, cf = 0, []
while i <= 20100:
....c = A002211(i)
....cf, i = cf+[c], i+1
p0, p1, q0, q1, i, base = cf[0], 1, 1, 0, 1, 2
while i <= 20100:
....p0, p1, q0, q1, i = cf[i]*p0+p1, p0, cf[i]*q0+q1, q0, i+1
a0 = p0//q0
p0 = p0-a0*q0
i, p0, dd = 0, p0*base, [a0]
while i < 70000:
....d, p0, i = p0//q0, (p0%q0)*base, i+1
....dd = dd+[d]
n, pn0, pn1, qn0, qn1 = 1, a0, 1, 1, 0
while n <= 20000:
....p, q = pn0, qn0
....if p//q != a0:
........print(n, "- manual!")
....else:
........i, p, di = 0, (p%q)*base, a0
........while di == dd[i]:
............i, di, p = i+1, p//q, (p%q)*base
........print(n, i-1)
....n, pn0, pn1, qn0, qn1 = n+1, cf[n]*pn0+pn1, pn0, cf[n]*qn0+qn1, qn0
CROSSREFS
KEYWORD
sign,base
AUTHOR
A.H.M. Smeets, Aug 10 2018
STATUS
approved