1,3

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).

The T are coefficients in recurrences f(n)=sum_{m=1..2p} T(p,m)*f(n-m).

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.

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.

[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].

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.

Row sums: sum_{m=1..2p} T(p,m) = A098600(p).

The triangle T(p,m) with Lucas numbers on the diagonal starts

1, 1;

0, 3, 0,-1;

0, 0, 4, 0, 0, 1;

0, 0, 0, 7, 0, 0, 0,-1;

0, 0, 0, 0,11, 0, 0, 0, 0, 1;

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.

Sequence in context: A045763 A132748 A055945 this_sequence A127372 A051171 A166445

Adjacent sequences: A138120 A138121 A138122 this_sequence A138124 A138125 A138126

nonn

Paul Curtz (pcrtz(AT)free.fr), May 04 2008

Edited and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 10 2008