[seqfan] Re: A003773 = 16*(A001109)^3
maxale at gmail.com
Wed Feb 4 07:56:34 CET 2009
On Fri, Jan 30, 2009 at 4:20 PM, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Is the fact of the title well known? R.
I'm not sure if it is known but it easy to prove from the given
recurrences of the two sequences.
Namely, it is easy to see that
16*A001109(n)^3 = ( A001109(3n) - 3*A001109(n) ) / 2.
On the other hand, the characteristic polynomial f(x) = x^2 - 6x + 1
of A001109 divides both the characteristic polynomial g(x) = x^4 -
204x^3 + 1190x^2 -204x + 1 of A003773 and the polynomial g(x^3).
Therefore, g(x) is a multiple of the characteristic polynomials of
each of the sequences A001109(n), A001109(3n), and 16*A001109(n)^3.
Hence, the characteristic polynomials of 16*A001109(n)^3 and A003773
coincide and assuming that the first four terms of these sequences
also coincide (I did not check that), we must have the equality
A003773(n) = 16*A001109(n)^3.
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