

A138123


Antidiagonal sums of a triangle of coefficients of recurrences of the Fibonacci sequence.


2



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, 0, 1363, 0, 2207, 1, 3571, 0, 5777, 0, 9349, 1, 15127, 0, 24475, 0, 39603, 1, 64079, 0, 103681, 0, 167761, 1, 271443, 0, 439203, 0, 710647, 1, 1149851, 0, 1860497, 0
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OFFSET

1,3


COMMENTS

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+nm,m).
The T are coefficients in recurrences f(n)=sum_{m=1..2p} T(p,m)*f(nm).
The recurrence for p=1, f(n)=f(n1)+f(n2), is satisfied by the Fibonacci sequence A000045. The recurrence for p=2, f(n)=3f(n2)f(n4), 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(np)(1)^p*F(n2p), any p>0, L=A000204.
[Proof: conjecture is equivalent to the existence of a g.f. of F with denominator 1L(p)x^p+(1)^p*x^(2p). Since 1xx^2 is known to be a denominator of such a g.f. of A000045, the conjecture is that 1L(p)*x^p+(1)^p*x^(2p) can be reduced to 1xx^2. One finds: {1L(p)*x^p+(1)^p*x^(2p)}/(1xx^2) = sum{n=0..p1}F(n+1)x^nsum{n=0..p2} (1)^(n+p)F(n+1)x^(2pn2) is a polynomial with integer coefficients, which is proved by multiplication with 1xx^2 and via F(n)+F(n+1)=F(n+2) and L(n)=F(n1)+F(n+1). R. J. Mathar, Jul 10 2008].
Conjecture: The Lucas sequence L also obeys all the recurrences: L(n)= L(p)*L(np)(1)^p*L(n2p), any p>0, L=A000204.


LINKS

Table of n, a(n) for n=1..60.


FORMULA

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


EXAMPLE

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+41=3.


CROSSREFS

Sequence in context: A132748 A022901 A055945 * A328382 A211868 A127372
Adjacent sequences: A138120 A138121 A138122 * A138124 A138125 A138126


KEYWORD

nonn


AUTHOR

Paul Curtz, May 04 2008


EXTENSIONS

Edited and extended by R. J. Mathar, Jul 10 2008


STATUS

approved



