E.G.F. for Squared Terms of Special Sequences

[Alec Mihailovs]

----- Original Message ----- 
From: Paul D. Hanna
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> Conjecture.
> The term-by-term square of the sequence generated by:
>      E.g.f.: exp(x + m/2*x^2)
> yields the sequence given by:
>      E.g.f.: exp( x/(1-m*x) )/sqrt(1 - m^2*x^2)
> and is an integer sequence whenever m is an integer.
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Proof.

Let f = exp(x + m x^2/2) = \sum a_n  x^n/n!

Then f' = (1+m x) f = \sum a_{n+1} x^n/n!

That gives the recurrence a_{n+1} = a_n + m n a_{n-1} with initial 
conditions a_0 = 1, a_1 = 1.

Squaring it, we get a_{n+1}^2 =  a_n^2 + 2mn a_n a_{n-1} + m^2 n^2 
a_{n-1}^2.

Squaring a_n - a_{n-1} = m(n-1) a_{n-2}, we get

2 a_n a_{n-1} = a_n^2 + a_{n-1}^2 - m^2 (n-1)^2 a_{n-2}^2.

Substituting that in the identity above, we get (*)

a_{n+1}^2 = (1+ mn) a_n^2 + (mn + m^2 n^2) a_{n-1}^2 - m^3 n (n-1)^2 
a_{n-2}^2.

Now, let g = exp( x/(1-mx) )/sqrt(1 - m^2 x^2) = \sum b_n x^n/n!

Then g'/g = (1 + (m+m^2) x -m^3 x^2) / (1 - m x - m^2 x^2 + m^3 x^3).

Rewriting that as (1 - m x - m^2 x^2 + m^3 x^3) g' = (1 + (m+m^2) x -m^3 
x^2) g,

we get the identity

b_{n+1} - mn b_n - m^2 n(n-1) b_{n-1} + m^3 n(n-1)(n-2) b_{n-2} =

b_n + (m+m^2) n b_{n-1} - m^3 n(n-1) b_{n-2}.

Simplifying it, we get the recurrence (**)

b_{n+1} = (1+mn) b_n + (mn + m^2 n^2) b_{n-1} - m^3 n(n-1)^2 b_{n-2}.

The conjecture follows from (*) and (**) and checking the initial 
conditions, Q.E.D.

Alec Mihailovs
http://math.tntech.edu/alec/