

Variations on a
well known generalized infinite continued fraction
with a simple pattern
Preface: all the following is an amateurish heuristic and rediscovery. A thorough discussion of this matter was already lead by David Angell in [1], which I found when checked the occuring sequences of numbers in [OEIS] .
The "generalized continued fraction" (gcf) allows nonunitvalues on the numerators of the partial fractions. Here that coefficients are denoted as b with index k, where the index begins at zero and the gcfs are assumed to have a nonterminating sequence of coefficients:
Some such continued fractions having coefficients with a simple pattern are well known, for instance[1]:
Interestingly, with various offsets of that pattern we can get rational values instead of compositions of the transcendend value of e:
In the following I provide a table with some heuristics (thanks to the "wolfram alpha" team; I used their wwwbased engine[2] for most of the compositions involving the exponentials).
Denote CF([a_{0}, a_{1}, a_{2}, …],[b_{0}, b_{1}, b_{2}, …]) the generalized continued fraction. Now let the a_{k} and b_{k} be defined by simple functions f(k) and g(k) of the index k.:
x_{f,g} = CF( [a_{k} = f(k) ] , [ b_{k} = g(k) ] )
For instance, to obtain 1/(e1) we write
1/(e1) = CF( [a_{k} = k ] , [ b_{k} = 1+k ] )
The following table gives the heuristics for a family of continued fractions having different offsets. The rows define the varying offsets for the computation of bcoefficients in the function b_{k }=f_{offset}(k) = offset+k , the columns the according varying offsets for the computation of the acoefficients.
Example: the entry in row "2+k" and column "3+k" means x=CF([3,4,5,...][2,3,4,...]) and heuristically we find x = – (2*e – 5)/(4*e – 11) , where e=exp(1):
Table 1:
b\a 
0+k 
1+k 
2+k 
3+k 
4+k 
5+k 
… 
0+k 
0 
1 
2 
3 
4 
5 
… 
1+k 






… 
2+k 
1 






3+k 
4 
2*1 





4+k 
3*7 
3*3 
3*1 




5+k 
4*34 
4*13 
4*2*2 
4*1 



6+k 
5*209 
5*73 
5*3*7 
5*5 
5*1 


7+k 
6*1546 
6*501 
6*4*34 
6*31 
6*6 
6*1/1 

8+k 



7*229 
7*43 
7*7/8 

9+k 



8*1061 
8*358 
8*57/73 

r+1+k 
n(r+1) 






By the construction of the simple pattern it is obvious, that the diagonal and subdiagonals form the sequences of convergents of one similar pattern; so if we look at for instance row "8+k" and column "5+k" and follow the diagonal up to the top we find:
beginning
at x_{k} = 7*7/8 = 6+1/8 = 49/8
then x_{k1} = 4+7/(7*7/8)= 4 + 8/7 = 36/7
then x_{k2} = 3+6/(36/7) = 3+7/6 = 25/6
then x_{k3} = 2+5/(25/6) = 2+6/5 = 16/5
then x_{k4} = 1+4/(16/5) = 1+5/4 = 9/4
then x_{k5} =0+3/(9/4) = 0+4/3 = 4/3
which proceeds along the subdiagonal 3 to arrive at x=4/3
It may be easier to follow the opposite direction: if we start at some entry in the first column and travel along the according diagonal, we find the simple arithmetic progressions in the numerators (nu_{row,col}) and denominators (de_{row,col}) the same way. Example: we begin at (row,col)= (4,0).
nu_{4,0}
= 3*7 de_{4,0}
= 13 = 7 + 6
nu_{5,1} = 4*13 =
4*de_{4,0} de_{5,1}
= 21 = de_{4,0}+8
nu_{6,2} = 5*21 =
5*de_{5,1} de_{6,2}
= 31 = de_{5,1}+10
nu_{7,3} = 6*31 =
6*de_{6,2} de_{7,3}
= 43 = de_{6,2}+12
nu_{k,k4} = (k1)*de_{k1,k5} de_{k,k4} = de_{k1,k5}+2*(k1)
=(k1)+(k1)^{2}(k2) = 1 +
k(k1)
=k^{3}–4k^{2}+6k–3 = k^{2} – k + 1
For the first column a recurrenceformula for the r+1'th numerator (n_{r+1}) and denominator (d_{r+1}) seems to be
n_{r+1}
= r*d_{r} + r*n_{r}
d_{r+1} = r*d_{r}
+ 1*n_{r}
or in a matrixformula:
so for the entry n_{r+1}/d_{r+1} at row "5+k" and col "0+k" has r=4 and
4*34/73 = (4* 13 + 4*21) / (4*13 + 1*21)
where 13 and 21 are taken from the denominator and numerator of the previous entry.
The function in Pari/GP:
\\ for funcA and funcB give the formula of a functionsdefinition
\\ f(k)=<formula>
\\ as strings. for instance CF("2","1") evaluates CF([2,2,2,2,…],[1,1,1,1…])
\\ as strings. for instance CF("1+k","2+k") evaluates CF([1,2,3,4,…],[2,3,4,5…])
{CF(funcA,funcB="1",maxcoeffs=100) = local(convgts,a_k,b_k);
eval(Str("f(k)=",funcA));
eval(Str("g(k)=",funcB));
convgts = matid(2);
for(k=0,maxcoeffs1, a_k =f(k);b_k=g(k); convgts= convgts*[0,1; b_k,a_k]);
return( convgts[2,2]/ convgts[1,2]*1.0) };
\p 200
\\ Examplecomputation
result = CF("2+k","1+k") \\ = 1/(2*e  5)
The sequences of numerators and denominators are known in the OEIS, for instance we find for the first column in the table
numerator: 1,4,21,136,… http://www.research.att.com/~njas/sequences/A052852
denominator 1,3,13,73,… http://www.research.att.com/~njas/sequences/A000262
A000262 Number of "sets of lists": number of partitions of {1,..,n} into any number of lists, where a list means an ordered subset.
A052852 E.g.f.: (x/(1x))*exp(x/(1x))
A simple grammar. Number of {121,212}avoiding nary words of length n.
 R. Stephan, Apr 20 2004
Contribution from David Angell (angell(AT)maths.unsw.edu.au), Dec 18 2008:
(Start)
If n is a positive integer then the infinite continued fraction (1+n)/(1+(2+n)/(2+(3+n)/(3+...)))
converges to the rational number A052852(n)/A000262(n).
(End)
The following columns can be retrieved as well. The common aspect is, that the sequences are formed by the generating function for exp(x/(1–x)) divided by some powers of (1x).
Gottfried Helms, 28.03.2010, textversion 1.2