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<DIV>Seqfans,</DIV>
<DIV> Final note regarding solutions for
A(x) that satisfies: </DIV>
<DIV>(1) r*A(x) = c + b*x + A(x)^n </DIV>
<DIV> </DIV>
<DIV>Max Alekseyev wrote:<BR>> <BR>> > (2) A(x) =
Sum_{i>=0} C(n*i,i)/(n*i-i+1)*(c+bx)^(n*i-i+1)/r^(n*i+1)<BR>>
[...]<BR>> > Yet I am still interested in an explicit formula for a(n)
...<BR>> <BR>> Why not simply express a(k) as the coefficient of x^k in
(2)? That is<BR>> <BR>> a(k) = Sum_{i>=0} C(n*i,i) * C(n*i-i+1,k)
* b^k * c^(n*i-i+1-k) / (n*i-i+1) / r^(n*i+1)<BR>> <BR>>
Max<BR> </DIV>
<DIV>I did not prefer that infinite series expression for a(n) because
</DIV>
<DIV>it converges quite slowly, and also I was looking for a finite
sum. </DIV>
<DIV> </DIV>
<DIV>By inspecting the triangle of coefficients of
t generated by: </DIV>
<DIV> A(x) = 1 + Series_Reversion[ (1 + (t+n)*x - (1+x)^n )/b ]
</DIV>
<DIV>it seems that a finite expression for a(n) is not far away. </DIV>
<DIV>In the above, setting r=t+n simplifies the series rather
substantially. </DIV>
<DIV> </DIV>
<DIV>Note that in the cases that I am studying, c=r-1 since A(0)=1.</DIV>
<DIV> </DIV>
<DIV>Well, enough said on this subject. </DIV>
<DIV>Thanks, </DIV>
<DIV> Paul </DIV>
<DIV> </DIV>
<DIV>P.S.: </DIV>
<DIV>To derive the series reversion formula:</DIV>
<DIV> A(x) = 1 + Series_Reversion[ (1 + r*x - (1+x)^n )/b ]
</DIV>
<DIV>from </DIV>
<DIV>(1) r*A(x) = c + b*x + A(x)^n </DIV>
<DIV>one may proceed as follows. </DIV>
<DIV> </DIV>
<DIV>Let A=A(x) and since A(0)=1 then set c=r-1 in (1): </DIV>
<DIV> r*A = r-1 + bx + A^n </DIV>
<DIV>or </DIV>
<DIV> 1 + r(A - 1) = bx + A^n .</DIV>
<DIV> </DIV>
<DIV>Now let y = A-1 so that </DIV>
<DIV> x = [ (1 + ry - (1+y)^n )/b ] .</DIV>
<DIV>Thus</DIV>
<DIV> y = Series_Reversion[ (1 + rx - (1+x)^n )/b ]</DIV>
<DIV>and so </DIV>
<DIV> A = 1 + Series_Reversion[ (1 + rx - (1+x)^n )/b ]. </DIV>
<DIV>END. </DIV></BODY></HTML>