[seqfan] Triangle T(n, k) = C(n, k)^m (m-th Powers of Binomial Coefficients)
Paul D Hanna
pauldhanna at juno.com
Sun Oct 31 02:14:49 CEST 2010
SeqFans,
In researching series related to Franel numbers (A000172), I was surprised to find that the triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, was not in the OEIS.
I have submitted the triangle as A181543, and arrived at this formula:
(1) G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1).
We can see the cofficients C(n,k)^3 in the expansion of (1):
A(x,y) = 1 +
(y + 1)*x +
(y^2 + 8*y + 1)*x^2 +
(y^3 + 27*y^2 + 27*y + 1)*x^3 +
(y^4 + 64*y^3 + 216*y^2 + 64*y + 1)*x^4 +
(y^5 + 125*y^4 + 1000*y^3 + 1000*y^2 + 125*y + 1)*x^5 +
(y^6 + 216*y^5 + 3375*y^4 + 8000*y^3 + 3375*y^2 + 216*y + 1)*x^6 +...
However, I was unable to find such a g.f. for the triangle T(n,k) = C(n,k)^m where m>3.
Can anyone find a similar g.f. for triangle T(n,k) = C(n,k)^4 ?
>From (1), we now have this g.f. for Franel numbers (A000172):
(2) Sum_{n>=0} A000172(n)*x^n = Sum_{n>=0} (3n)!/n!^3 * x^(2n)/(1-2x)^(3n+1)
where A000172(n) = Sum_{k=0..n} C(n,k)^3 forms the row sums of A181543.
Also, (2) leads me to a rather complex functional equation for the g.f. of the Franel numbers (A000172):
(3) G.f.: A(x) = G( x^2/(1-2*x)^3 )/(1-2*x) where G(x) satisfies:
G(x^3) = G( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x)
and G(x) is the g.f. of A006480.
Can anyone improve/simplify the functional equation of (3)?
Thanks,
Paul
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