A317908 Number of decimal places to which the n-th convergent of the continued fraction expansion of Khintchine's constant matches the correct value.

0, -1, 1, 2, 2, 3, 3, 4, 4, 6, 5, 8, 8, 9, 11, 13, 12, 14, 15, 16, 16, 16, 18, 21, 21, 23, 24, 24, 25, 25, 26, 27, 28, 29, 30, 30, 32, 32, 33, 33, 36, 35, 36, 37, 37, 38, 39, 39, 40, 41, 42, 42, 43, 44, 45, 44, 46, 47, 48, 48, 49, 50, 51, 54, 55, 56, 56, 58, 58, 60

Offset: 1

Author

A.H.M. Smeets, Aug 10 2018

Keywords

• sign
• base

Comments

Decimal expansion of Khintchine's constant in A002210.

For the similar case of the number of correct decimal digits of Pi see A084407.

For the similar case of the number of correct decimal digits of log(2) see A317558.

For the number of correct binary places see A317907.

Examples

			   n   convergent     decimal expansion    a(n)
  ==  =============  ====================  ====
   1     2 / 1       2.0                     0
   2     3 / 1       3.0                    -1
   3     8 / 3       2.66...                 1
   4    43 / 16      2.687...                2
   5    51 / 19      2.684...                2
   6    94 / 35      2.6857...               3
   7   239 / 89      2.6853...               3
   8   333 / 124     2.68548...              4
   9   572 / 213     2.68544...              4
  10  2049 / 763     2.6854521...            6
  oo  lim = A002210  2.685452001065306...   --
		

Links

• A.H.M. Smeets, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A002210, A002211, A086702, A100199, A240995, A317907.

Programs

Python

i,cf = 0,[]
while i <= 20100:
    c = A002211(i)
    cf,i = cf+[c],i+1
p0,p1,q0,q1,i,base = cf[0],1,1,0,1,10
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 < 21000:
    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

Formula

Limit_{n -> oo} (a(n)/n) = 2*log(A086702)/log(10) = 2*A100199/log(10) = 2*A240995.