[seqfan] Re: Help needed

[Richard Mathar]

Following http://list.seqfan.eu/pipermail/seqfan/2010-March/003990.html ,
as a first aid, one might ponder the idea to submit the following, which I
hereby publish under the global license "for discussion:"

%I A000001
%S A000001 0,1,2,12,36,125,384,1183,3528,10404,30250,87131,248832,705757,
%T A000001 1989806,5581500,15586704,43356953,120187008,332134459,915304500,
%U A000001 2516113236,6900949462,18888143927,51599794176,140718765625,383142771674
%N A000001 n times the square of Fibonacci(n).
%H A000001 D. J. Kleitman, B. Golden, <a href="http://www.jstor.org/stable/2319131">Counting trees in a certain class of graphs</a>, Amer. Math. Monthly 82 (1975), 40-44.
%H A000001 G. Baron, H. Prodinger, R. F. Tichy, F. T. Boesch, J. F. Wang, The number of spanning trees in the square of a cycle, Fibonacci Quart. 23 (1985), no. 3, 258-264 [<a href="http://www.ams.org/mathscinet-getitem?mr=806296">MR0806296</a>]
%H A000001 R. Guy, <a href="http://list.seqfan.eu/pipermail/seqfan/2010-March/003990.html">Q on papers by Kleitman, Baron et al.</a>, SeqFan list, Mar 2010
%H A000001 <a href="Sindx_Rea.html#recLCC">Index to sequences with homogeneous linear recurrences with constant coefficients</a>, signature (4,0,-10,0,4,-1)
%F A000001 a(n)= A045925(n)*A000045(n) = n*A007598(n) = n *(A000045(n))^2 .
%F A000001 a(n)= 4*a(n-1) -10*a(n-3) +4*a(n-5) -a(n-6). G.f.: x*(1-2*x+4*x^2-2*x^3+x^4)/ ((1+x)^2 * (x^2-3*x+1)^2).
%p A000001 A000001 := proc(n) n*(combinat[fibonacci](n))^2 ; end proc:
%K A000001 nonn,easy
%O A000001 0,3
%A A000001 R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 13 2010


The Fib Quart page at http://www.fq.math.ca/list-of-issues.html says that the
article "will be available...by the end of 2009" - easy to falsify unless
one lives in some sort of outer space time bubble.

RJM