[seqfan] Re: Highest exponent of 2 dividing A155200(n)

[franktaw at netscape.net]

Not a complete answer, but perhaps some insight.

The power of 2 dividing n! is n - A000120(n).

Thus, the conjecture is equivalent to A155200(n) * n! is divisible by 
exactly 2^n.

A155200(n) * n! is the exponential transform of 2^(n^2) * (n-1)!.

Franklin T. Adams-Watters

-----Original Message-----
From: Paul D Hanna <pauldhanna at juno.com>

Seqfans,
       Regarding sequence A155200:
http://www.research.att.com/~njas/sequences/A155200
G.f.: exp(Sum_{n>=1} 2^(n^2)*x^n/n) = 1 + 2*x + 10*x^2 + 188*x^3 + 
16774*x^4
+...+ A155200(n)*x^n +...

I propose the following
CONJECTURE:
Highest exponent of 2 dividing A155200(n) = A000120(n) = number of 1's 
in binary
expansion of n, so that A155200(n)/2^A000120(n) is odd for n>=0.

I have verified that this conjecture holds for n=0..1024.
But I do not know why it should be true.

A related question:
is there an analogous formula for the highest exponent of 3 dividing 
A155203(n):

http://www.research.att.com/~njas/sequences/A155203

Wonder if anyone has an insight into these topics.
     Paul

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