A note on A001450

					Peter Bala, Oct 10 2015

Clearly, we have

	binomial(5*n,n) = [x^n] (1 + x)^(5*n).

We claim

	A001450(n) = binomial(5*n,2*n) = [x^n] ( (1 + x)*C(x) )^(5*n),

where C(x) = (1 – sqrt(1 – 4*x))/(2*x) is the o.g.f for the Catalan numbers A000108.

Proof.

Define
	s(n) = [x^n] ( (1 + x)*C(x) )^(5*n).

We recall the following expansion for the powers of the generating function of the Catalan

numbers [WIL, 2.5.16, p. 54]

	C(x)^m =  Sum_{k >= 0} m/(m + k)*binomial(m + 2*k – 1,k)*x^k.

Thus

	( (1 + x)*C(x) )^(5*n) = (1 + x)^(5*n) * C(x)^(5*n)

	= ( Sum_{i >= 0} binomial(5*n,i)*x^i ) * ( Sum_{k >= 0} 5*n/(5*n + k)*binomial(5*n + 2*k - 1,k)*x^k ) ....(1)

For n >= 1, the coefficient s(n) of x^n in the rhs of equation (1) is

	s(n) = Sum_{k = 0..n} binomial(5*n,n - k) * 5*n/(5*n + k)*binomial(5*n + 2*k - 1,k)

with the initial value s(1) = 10.

The Maple command

	sumrecursion( binomial(5*n,n - k)*5*n/(5*n + k)*binomial(5*n + 2*k - 1, k), k, s(n) );

returns the recurrence

	6 n(3 n - 1)(2 n - 1)(3 n – 2)s(n) = 5(5 n - 4)(5 n - 3)(5 n - 2)(5 n - 1)s(n - 1)  

for the sequence s(n).

It easy to check that binomial(5*n,2*n) satisfies the same recurrence with the same initial value at

n = 1 as s(n). Hence s(n) = binomial(5*n,2*n).

End proof.

One consequence of the above result is that 

	exp( Sum_{n >= 1} 1/5 * A001450(n)*x^n/n ) = 1 + 2*x + 23*x^3 + 377*x^4 + ...

is the o.g.f. for the sequence of Duchon numbers A060941. This result was observed by Knuth (posted

on Oct 05 2014). [BAL] has further information on representing a sequence in the form [x^n]G(x)^n

for some power series G(x).

REFERENCES

[BAL] Peter Bala, Representing a sequence as [x^n]G(x)^n.

		  uploaded to A066398

[WIL] Herbert S. Wilf, Generatingfunctionology, Academic Press, Inc.

		       available online at https://www.math.upenn.edu/~wilf/gfologyLinked2.pdf

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Further thoughts.

In an exactly similar manner to the above we can show

	A001448(n) = binomial(4*n,2*n) = [x^n] ( (1 + x)^2/(1 – x) )^(2*n).

Results of this type suggest that it may be possible to express the binomial coefficient binomial(a*n,b*n)

for a >= b in the form

	binomial(a*n,b*n) = [x^n] ( G(x) )^(p*n)	n = 0,1,2,3,...		(2)

for some power series G(x) = G(a,b,x) with integer coefficients, and some value of p, which depends

on a and b. Empirically, it appears that the value of p in (2) can be chosen as	p = a/GCD(a,b).
	
Here are some calculations that support this idea:	
	
	binomial(6*n,2*n) = [x^n](1 + 5*x + 20*x^2 + 96*x^3 + 528*x^4 + 3136*x^5 + 19584*x^6 + 126720*x^7 + ...)^(3*n) 

	binomial(6*n,3*n) = [x^n](1 + 10*x + 81*x^2 + 720*x^3 + 7056*x^4 + 73872*x^5 + 809424*x^6 + 9167184*x^7 + ...)^(2*n)

	binomial(7*n,2*n) = [x^n](1 + 3*x + 13*x^2 + 94*x^3 + 810*x^4 + 7667*x^5 + 76998*x^6 + 805560*x^7 + ...)^(7*n) 

	binomial(7*n,3*n) = [x^n](1 + 5*x + 52*x^2 + 880*x^3 + 17856*x^4 + 399296*x^5 + 9491008*x^6 + 235274240*x^7 + ...)^(7*n) 

	binomial(8*n,2*n) = [x^n](1 + 7*x + 56*x^2 + 616*x^3 + 7960*x^4 + 112328*x^5 + 1676792*x^6 + 26030376*x^7 + ...)^(4*n) 

	binomial(8*n,3*n) = [x^n](1 + 7*x + 133*x^2 + 4140*x^3 + 154938*x^4 + 6398717*x^5 + 281086555*x^6 + 12882897819*x^7 + ...)^(8*n) 

	binomial(8*n,4*n) = [x^n](1 + 35*x + 1380*x^2 + 66276*x^3 + 3601060*x^4 + 211234212*x^5 + 13042493412*x^6 + 835295682660*x^7 + ...)^(2*n) 

	binomial(9*n,2*n) = [x^n](1 + 4*x + 34*x^2 + 494*x^3 + 8615*x^4 + 165550*x^5 + 3380923*x^6 + 71999763*x^7 + ...)^(9*n) 

We searched the database for a match to the series occurring in the above calculations and only found

a match for the first one: the sequence of coefficents [1, 5, 20, 96, 528, ...] appears to be A182959,

which is defined as the expansion of the o.g.f. 2*(1 + x)^2/(1 - 2*x + sqrt(1 - 8*x)).

This o.g.f. can be rewritten in terms of the Catalan numbers generating function C(x) as (1 + x)*(2*C(2*x) - 1).

Thus, conjecturally, we have the identity
	
	binomial(6*n,2*n) = [x^n] ( (1 + x)*(2*C(2*x) - 1) )^(3*n)
	
to add to our two previous identities
	
	binomial(4*n,2*n) = [x^n] ( (1 + x)*(1 + x)/(1 – x) )^(2*n)

and
	
	binomial(5*n,2*n) = [x^n] ( (1 + x)*C(x) )^(5*n).
	
These three results point towards the possibility that the functions G(x) occurring in (2) may be algebraic

for each choice of a and b.

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