# Lucas n-Step Starting Numbers?

I am very interested in n-Step Lucas numbers. Trying to find, the "true starting" values seem to be contentious? I would assume $$(1,1), (1,1,1), (1,1,1,1)$$; like Fibonacci. However, 2-Step Lucas is $$(1,3), (1,3,7), (1,3,7,15)$$ one under the powers of 2. Tony Noe's paper.

Or Lucas n-Step are: $$(2,1), (2,1,3), (2,1,3,6), (2,1,3,6,12)$$ Mr. Piezas mentions $$L_k=N_{int}[r\cdot x^k]$$ here.

However, n-Step Fibonacci seems (by some) to be $$(1,1), (1,1,2), (1,1,2,4), (1,1,2,4,8)$$ Wikipedia Generalizations of Fibonacci numbers.

I admit, probably my own ignorance. Assuming Fibonaccis of any step started all at (1,1,...,1) forms. So. There you have it. Which is right for Fibonacci?, and which is right for Lucas (the complement sequence)?

If you view Mr. Sloanes OEIS?

The Fibonacci inital values are:

$$\{ 0,1\}$$

$$\{ 0,0,1\}$$

$$\{ 0,0,0,1\}$$

$$\{ 0,0,0,0,1\}$$

$$\{ 0,0,0,0,0,1\}$$

for the Lucas initial values are:

$$\{2,1 \}$$

$$\{3,1,3 \}$$

$$\{4,1,3,7 \}$$

$$\{5,1,3,7,15 \}$$

$$\{6,1,3,7,15,31 \}$$

Anyone else want to write MMA code to compute the pattern? Unfortunately, this shows Mr. Noe's table on page 3 needs updating. Otherwise, it is quite complete.

OEIS has the Lucas 2-S, 3-S, .. pattern as a sequence, too. ( Triangle read by rows, "n" followed by (n-1) terms of (1, 3, 7, 15,...) ) A143802