A002449 as Solution to System of Equations?
Dean Hickerson
dean at math.ucdavis.edu
Thu Mar 15 18:09:25 CET 2007
Mostly to Paul D. Hanna:
> Help is needed to simplify the following problem.
>
> Consider the infinite system of simultaneous equations:
> A = 1 + xAB
> B = 1 + xBCD
> C = 1 + xCDEF
> D = 1 + xDEFGH
> E = 1 + xEFGHIJ
> F = 1 + xFGHIJKL
> ...
> What is the unique solution to the variables A,B,C,... as functions in x?
First, they're not really functions, since the power series don't converge
for x!=0. But the question makes sense for formal power series.
Let a(1) = A, a(2) = B, a(3) = C, ... Then the equations say that
a(n) = 1 + x a(n) a(n+1) ... a(2n)
for n>=1. Let p(n) be the product a(1)*...*a(n); in particular p(0) = 1.
Then
p(n) = p(n-1) + x p(2n).
Now let p(n) = sum_{i>=0} c(n,i) x^i. The c's can be computed by
the recurrence
c(n,i) = c(n-1,i) + c(2n,i-1) for n>=1, i>=1,
with the initial conditions
c(n,0) = 1 for n>=0,
c(0,i) = 0 for i>=1.
Since A = a(1) = p(1), you want the terms c(1,i).
As you said, computing c(1,i) takes about 2^i steps. But it turns out
that, for fixed i, c(n,i) is a polynomial in n, of degree i:
c(n,i) = sum_{0<=j<=i} d(i,j) n^j.
Getting a recurrence for the d's is straightforward but messy. (I.e. it
took me a few tries to get it right.) We find:
d(0,0) = 1,
d(i,0) = 0 for i>=1,
and
d(i,j) = 1/j [2^(j-1) d(i-1,j-1) -
sum_{j+1<=m<=i} (-1)^(j+m) binomial(m,j-1) d(i,m)] for 1<=j<=i.
Using this, we can compute the c's much more quickly. In Mathematica:
d[0,0]=1;
d[i_,0]:=0;
d[i_,j_]:= d[i,j] = 1/j (2^(j-1) d[i-1,j-1] -
Sum[(-1)^(j+m) Binomial[m,j-1] d[i,m], {m,j+1,i}] );
c[0,i_]:=d[i,0];
c[n_,i_]:=Sum[d[i,j] n^j,{j,0,i}];
Table[c[1,i],{i,0,20}]
produces:
{1, 1, 2, 6, 26, 166, 1626, 25510, 664666, 29559718, 2290267226,
314039061414, 77160820913242, 34317392762489766, 27859502236825957466,
41575811106337540656038, 114746581654195790543205466,
588765612737696531880325270438, 5642056933026209681424588087899226,
101386818783417634431603640258454473638,
3428771170016237895796931111337071731946586}
> Could someone calculate more coefficients of A
> to see if they continue to agree with A002449?
They agree at least up to A002499(50).
Dean Hickerson
dean at math.ucdavis.edu
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