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%I #13 Apr 16 2024 03:59:09
%S 1,1,3,0,3,0,7,1,11,0,17,0,29,1,47,0,75,0,123,1,199,0,321,0,521,1,843,
%T 0,1363,0,2207,1,3571,0,5777,0,9349,1,15127,0,24475,0,39603,1,64079,0,
%U 103681,0,167761,1,271443,0,439203,0,710647,1,1149851,0,1860497,0
%N Antidiagonal sums of a triangle of coefficients of recurrences of the Fibonacci sequence.
%C Consider the irregular sparse triangle T(p,p) = A000204(p), T(p,2p)= -A033999(p)=(-1)^(p+1), T(p,m) =0 else; 1<=m<=2p, p>=1. Then a(n)=sum_{m=1..[2(n+1)/3]} T(1+n-m,m).
%C The T are coefficients in recurrences f(n)=sum_{m=1..2p} T(p,m)*f(n-m).
%C The recurrence for p=1, f(n)=f(n-1)+f(n-2), is satisfied by the Fibonacci sequence A000045. The recurrence for p=2, f(n)=3f(n-2)-f(n-4), is satisfied by A005013, A005247, A075091, A075270, A108362 and A135992.
%C Conjecture: The Fibonacci sequence F obeys all the recurrences: A000045(n)=F(n)= L(p)*F(n-p)-(-1)^p*F(n-2p), any p>0, L=A000204.
%C [Proof: conjecture is equivalent to the existence of a g.f. of F with denominator 1-L(p)x^p+(-1)^p*x^(2p). Since 1-x-x^2 is known to be a denominator of such a g.f. of A000045, the conjecture is that 1-L(p)*x^p+(-1)^p*x^(2p) can be reduced to 1-x-x^2. One finds: {1-L(p)*x^p+(-1)^p*x^(2p)}/(1-x-x^2) = sum{n=0..p-1}F(n+1)x^n-sum{n=0..p-2} (-1)^(n+p)F(n+1)x^(2p-n-2) is a polynomial with integer coefficients, which is proved by multiplication with 1-x-x^2 and via F(n)+F(n+1)=F(n+2) and L(n)=F(n-1)+F(n+1). - _R. J. Mathar_, Jul 10 2008].
%C Conjecture: The Lucas sequence L also obeys all the recurrences: L(n)= L(p)*L(n-p)-(-1)^p*L(n-2p), any p>0, L=A000204.
%F Row sums: Sum_{m=1..2p} T(p,m) = A098600(p).
%F Conjectures from _Chai Wah Wu_, Apr 15 2024: (Start)
%F a(n) = a(n-2) - a(n-3) + a(n-4) + a(n-5) + a(n-7) for n > 7.
%F G.f.: x*(-x^5 - 2*x^2 - x - 1)/((x + 1)*(x^2 - x + 1)*(x^4 + x^2 - 1)). (End)
%e The triangle T(p,m) with Lucas numbers on the diagonal starts
%e 1, 1;
%e 0, 3, 0,-1;
%e 0, 0, 4, 0, 0, 1;
%e 0, 0, 0, 7, 0, 0, 0,-1;
%e 0, 0, 0, 0,11, 0, 0, 0, 0, 1;
%e The antidiagonal sums are a(1)=1. a(2)=0+1=1. a(3)=0+3=3. a(4)=0+0+0=0. a(5)=0+0+4-1=3.
%K nonn
%O 1,3
%A _Paul Curtz_, May 04 2008
%E Edited and extended by _R. J. Mathar_, Jul 10 2008
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