server down [was: Re: G.f. for a(n) = ...]
Joerg Arndt
arndt at jjj.de
Sun Aug 19 09:44:41 CEST 2007
David Wilson wrote:
> For A006906, could a(n+1)/a(n) be approaching a limit?
> Like maybe 3^(1/3)?
The limit does not exist. The generating function for a(n) is
n 1
A(z) = SUM a(n) z = PRODUCT -------- (0)
n
n>=0 n>=1 1 - n z
for small z.
The product on the right is defined for all z with |z|<1 except for
singularities at r n^(-1/n), where n>1 is an integer and r is an
n-th root of 1. These are all poles of order 1, except for poles of
order 2 at 2^(-1/2) = 4^(-1/4) and its negative. The singularities with
the smallest absolute value occur for n=3; hence equation (0) is true
for |z| < 3^(-1/3).
So
1 A(z)
a(n) = ------ INTEGRAL ---- dz (1)
2 pi i C n+1
z
where the integral is around a circle centered at 0, with radius less
than 3^(-1/3). To estimate this, we can expand the circle, in order
to avoid the singularity at z=0. Increasing the radius of the circle
beyond 3^(-1/3) changes the value of the integral by the sum of the
residues of the integrand at the 3 singularities of absolute value
3^(-1/3):
1 A(z)
a(n) = ------ INTEGRAL ---- dz - (2)
2 pi i D n+1
z
(R(3^(-1/3)) + R(w 3^(-1/3)) + R(w^2 3^(-1/3))
where D is a circle of radius r, with 3^(-1/3) < r < 2^(-1/2),
w = (-1 + sqrt(3) i)/2 and R(z0) is the residue of A(z)/z^(n+1) at z0.
On D, the integrand is o(3^(n/3)). Since the length of D is independent
of n, the integral is also o(3^(n/3)). This will turn out to be smaller
than the sum of the residues, so we can ignore it if we just want an
asymptotic value for a(n). (For a more accurate estimate, we could
expand the circle further, taking into account the residues at the
other singularities.)
For each of the 3 relevant singularities z0 (i.e. the cube roots of 1/3),
we have
3 3
1 - 3 z = -3 (z - 1/3) = -3 (z-z0) (z-w z0) (z-w^2 z0)
so
1 1
R(z0) = lim (z - z0) ---- PRODUCT --------
z -> z0 n+1 n>=1 n
z 1 - n z
1 1 1
= ----- ---------------------------- PRODUCT ---------
n+1 -3 (z0 - w z0) (z0 - w^2 z0) n>=1 n
z0 n!=3 1 - n z0
-1 1
= -------- PRODUCT ---------
n+3 n>=1 n
9 z0 n!=3 1 - n z0
I don't know if the products on the right can be evaluated in closed
form, but they can be computed numerically:
For z0 = 3^(-1/3), the product is
p0 ~ 293769.1124891011564831268878467064420586
For z0 = w 3^(-1/3) it's
p1 ~ 0.3452412325801026189128598712870317087930 +
0.0293627889575200300383689071994562320564 i
For z0 = w^2 3^(-1/3) it's
p2 ~ 0.3452412325801026189128598712870317087930 -
0.0293627889575200300383689071994562320564 i
So
-1/3 n/3 - 1
R(3 ) = -3 p0
-1/3 n/3 - 1 2n
R(w 3 ) = -3 w p1
2 -1/3 n/3 - 1 n
R(w 3 ) = -3 w p2
Substituting these values in (2) gives an asymptotic formula for a(n):
n/3 - 1 2n n
a(n) ~ 3 (p0 + w p1 + w p2)
(3)
n/3
= 3 c(n mod 3)
where
c(0) = (p0 + p1 + p2)/3 ~ 97923.26765718877222945490452214967204067
c(1) = (p0 + w^2 p1 + w p2)/3 ~ 97922.93936857030090919621139317470903737
c(2) = (p0 + w p1 + w^2 p2)/3 ~ 97922.90546334208334447577193138206098055
Because these 3 values are different, the sequence of ratios a(n+1)/a(n) has
3 different limit points, which differ slightly from 3^(1/3).
For anyone who wants to check this computationally, here's some Mathematica
code to compute a(n); a[n,k] is the sum of the products of all partitions
of n into parts <= k.
a[0,0]=1;
a[n_,0]:=0;
a[n_,k_]:=If[k>n, a[n,n], a[n,k] = a[n,k-1] + k a[n-k,k] ];
a[n_]:=a[n,n];
Dean Hickerson
dean at math.ucdavis.edu
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