[seqfan] Re: A Remarkable Eigenfunction - A179497
Paul D Hanna
pauldhanna at juno.com
Mon Aug 2 17:18:40 CEST 2010
SeqFans,
I just realized that these results can be generalized.
Suppose G(x) satisfies:
G_n(x)^3 = G(x)^3 * G_n'(x) for all n
where G_n(x) = n-th iteration of G(x) and G(0)=0, G'(0)=1,
then G = G(x) satisfies:
G = x + G^3 + G^3*D(G^3)/2! + G^3*D(G^3*D(G^3))/3! + G^3*D(G^3*D(G^3*D(G^3)))/4! + ...
and
G_n(x) = x + n*G^3 + n^2*G^3*D(G^3)/2! + n^3*G^3*D(G^3*D(G^3))/3! + n^4*G^3*D(G^3*D(G^3*D(G^3)))/4! + ...
Explicitly,
G(x) = x + 6*x^3/3! + 540*x^5/5! + 156240*x^7/7! + 96480720*x^9/9! + 104661849600*x^11/11! + 177947471782080*x^13/13! + 439942718370355200*x^15/15! +...
(I will submit this sequence of coefficeints to OEIS tonight sometime.)
The obvious generalization is as follows.
if
G_n(x)^t = G(x)^t * G_n'(x)
then G=G(x) satisfies:
G_n(x) = x + n*G^t + n^2*G^t*D(G^t)/2! + n^3*G^t*D(G^t*D(G^t))/3! + n^4*G^t*D(G^t*D(G^t*D(G^t)))/4! + ...
I find this to be an interesting aside.
I would like to see some nice formula for the coefficients in these series.
Thanks,
Paul
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