Extending (arbitrarily far) the modified Trott constant of A114376
Jon Schoenfield (jonscho(AT)hiwaay.net)
April 18 2010


The Comment section in the OEIS entry for sequence A114376 ("A modified 
Trott constant") shows that, when the sequence's 107 listed terms are used 
as the terms of a simple continued fraction, the decimal digits of the value 
of that continued fraction agree with the terms of the sequence for the 
first 81 digits after the decimal point.  The section ends with the claim, 
"However, there is no obvious methodology with which to get more terms."

Here's an easy method; it won't achieve the fastest growth in the number of 
digits of agreement as terms are added, but it will -- given sufficient time 
and computing resources -- allow growing the number of digits of agreement 
to _any_ desired number:

--------------------------------------------------

1.  Let nTerms be the number of known terms in the sequence; initially, 
nTerms=107.  (For purposes of the following, I'll number them as a(1) = 3, 
a(2) = 2, a(3) = 0, ..., a(107) = 4.)  Also, supply the implied leading zero 
as a(0) = 0.

2.  Write the number x as the concatenation of all the terms of the sequence 
divided by 10^nTerms.  (This evaluates to approximately 0.3206..., but store 
the result as an exact fraction, rather than as a floating point number.)

3.  For n in [0..nTerms], let x := 1/(x-a(n)).

4.  To obtain the next term, simply truncate x, unless x >= 10; in that 
case, repeatedly use "9, 0," digit pairs, decreasing x by 9 for each pair, 
until x < 10, and then truncate x to get the next term; i.e.,

   while x >= 10 do
      nTerms := nTerms + 1
      a(nTerms) := 9
      nTerms := nTerms + 1
      a(nTerms) := 0
      x := x - 9
   end while

   nTerms := nTerms + 1
   a(nTerms) := Truncate(x)

5.  Go to step 2.

--------------------------------------------------

(Some simple changes would make the above approach more efficient, e.g., 
storing a numerator and a denominator for x, and updating them as new terms 
are added, rather than repeating the concatenation of all the terms in step 
2 each time.)

Using the above, and not counting the supplied initial zero term in the 
sequence (either as a term or as a digit), I get the following growth of the 
number of digits of agreement as a function of the number of terms:

nTerms  Digits of agreement
====   =============================
 107   81  (as stated in the Comments)
 111   82
 112   84
 123   86
...
 137   90
...
 171   100
...
 261   150
 262   151
 263   154
 265   158   (after this, there's a big jump in the number of terms required 
to get more digits of agreement, because x = 4002.556... requires 444 "9, 
0," digit pairs)
1154   161
1155   162
1156   163
...
1223   200
...
1359   250
...
1391   268
1392   269
1394   273  (after this, there's another big jump in nTerms, since x = 
8158.091... requires 906 "9, 0," digit pairs)
3207   278
3210   279
3211   280
...

Jon E. Schoenfield
Huntsville, AL