Fwd: Asymptotics about A005228

Maximilian Hasler maximilian.hasler at gmail.com
Wed Jun 4 06:29:14 CEST 2008

Since others might be interested,
and after checking that my second-order approximation

A030124(n) = n + sqrt(2n) + o(n^(1/2))

A005228(n) = n²/2 + (1/3) (2n)^(3/2) + o(n^(3/2))
 = n²/2 * (1 + (4/3)sqrt(2/n) + o(1/sqrt(n)))

reproduces quite well the numerical values computed by T.D.Noe,
I finally send a copy of my reply also to the list, and if we were not
in OEIS' "summer mode", I'd have proposed to Neil to add the formulae
in the respective sequences, if he considered it worth while...

---------- Forwarded message ----------
From: Maximilian Hasler <maximilian.hasler at gmail.com>
Date: Tue, Jun 3, 2008 at 6:02 PM
Subject: Re: Asymptotics about A005228
To: "David W. Wilson" <wilson.d at anseri.com>
Cc: A.N.W.Hone at kent.ac.uk

a little complement to my previous mail:
obviously, f(x) = x²/2 is a solution to the equation
f¯¹(x) + (f')¯¹(x) ~ x
where I replaced = by ~, i.e. asymptotically equivalent ;
f'(x) = x = (f')¯¹(x)
and f¯¹(x) = sqrt(2x) is negligible w.r.t. x.

To get "higher order" terms, one can make asymptotic developments with
a "perturbation term",
f(x) = x²/2 + h(x)
with h(x) = o(x²), and we can also require h'(x) to be small.

To get rid of the sqrt(2x) which is too much in l.h.s. of (5)~
we need to make (f')¯¹ a bit smaller, i.e. f' a bit larger.
Let me try
 f'(x) = x + sqrt(2x)
to get
 (f')¯¹(y) = y - sqrt(2y+1) + 1
 = y - sqrt(2y) + o(sqrt(y))
 f(x) = x²/2 + (1/3) (2x)^(3/2)
 = x²/2 * (1+(4/3)sqrt(2/x))
 f¯¹(y) = x = sqrt( 2y / (1+(4/3)sqrt(2/x)) )
 = sqrt(2y) + o(sqrt(y))

So the sqrt(2x) term would cancel,
and the game can be continued with the next term
(which would be O(y^(1/4)) I think).


On Tue, Jun 3, 2008 at 11:56 AM, Maximilian Hasler
<maximilian.hasler at gmail.com> wrote:
>>Let f = A005228, and let g = complement of f.
>>    [4]  f' = g     [corrected]
> I admit that I did not follow earlier discussion about this sequence,
> maybe someone already made a more refined analysis
> and I may be completely wrong....
> but at a first glance, it seems obvious to me that g(x) ~ x
> (since f grows much faster, the possibilities taken away by f are negligible)
> and thus f(x) ~ x^2/2.
> Maximilian

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