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<DIV>Max (and Seqfans),</DIV>
<DIV> Interleaving my replies below ... </DIV>
<DIV>On Wed, 7 Jun 2006 20:23:29 -0700 Max <<A
href="mailto:maxale@gmail.com">maxale@gmail.com</A>> writes:<BR>><BR>>
If you bring C_k as C(2k,k)/(k+1) it is natural to split the <BR>>
formula<BR>> into three parts:<BR>> n^2/(n-1))^(n+1) / n^4 *
(2*n-1)<BR>> (n^2/(n-1))^(n+1) / n^4 * (2-n)/n * Sum_{k=0..n-1}
((n-1)/n^2)^k C(2k,k)<BR>> - (n^2/(n-1))^(n+1) / n^4 * 2 *
Sum_{k=0..n-1} ((n-1)/n^2)^k C(2k,k)/(k+1)</DIV>
<DIV> </DIV>
<DIV>This is helpful. Yet some identity is needed to simplify.
</DIV>
<DIV> </DIV>
<DIV>> These sums can give also an approximation for a(n) as n tends to
infinity.</DIV>
<DIV> </DIV>
<DIV>Good insight. </DIV>
<DIV> <BR>> > Although this is not any simpler, the presence of
the factorials <BR>> suggest<BR>> > that simplification may require
some identity of LambertW.<BR>> <BR>> Could you please explain this point
(maybe with a smaller example)?<BR>> What is the relationship between
factorials and Lambert W-function?</DIV>
<DIV> </DIV>
<DIV>Yes, my wording was vague at best. </DIV>
<DIV> </DIV>
<DIV>What I was refering to here was the following lovely </DIV>
<DIV>LambertW convolution identity: </DIV>
<DIV> </DIV>
<DIV>Sum_{k=0..n} p*(n-k+p)^(n-k-1)* q*(k+q)^(k-1)* C(n,k) </DIV>
<DIV> </DIV>
<DIV> = (p+q)*(n+p+q)^(n-1) </DIV>
<DIV> </DIV>
<DIV>which is the coefficient of x^n/n! in: </DIV>
<DIV> </DIV>
<DIV>W(x)^(p+q) = W(x)^p * W(x)^q </DIV>
<DIV> </DIV>
<DIV>where W(x) = -LambertW(-x)/x. <BR> </DIV>
<DIV>My objective was to restate your formula into a form </DIV>
<DIV>where this identity may be applied. </DIV>
<DIV> </DIV>
<DIV>
<DIV>We may restate your 3 parts into a convolution form: </DIV>
<DIV>a(n) = (n^2/(n-1))^(n+1)/n^4*(2*n-1) <BR>+(n^2/(n-1))^2/n^4*(2-n)/n *
Sum_{k=0..n-1} (n^2/(n-1))^(n-1-k) * C(2*k,k)) <BR>-(n^2/(n-1))^2/n^4*2 *
Sum_{k=0..n-1} (n^2/(n-1))^(n-1-k) * C(2*k,k)/(k+1)) <BR> </DIV>
<DIV>and then the sums include factors like (n-1)^(n-1-k) ... </DIV>
<DIV>which reminded me of LambertW. </DIV>
<DIV> </DIV></DIV>
<DIV>But LambertW may not be involved at all - </DIV>
<DIV>I will have to check. </DIV>
<DIV> <BR>Thanks,</DIV>
<DIV> Paul</DIV></BODY></HTML>