Why is the 0th term of the padovan sequence 1 but for fibonacci it is 0?

Question

padovan numbers are:

P(0)=1, P(1)=1, P(2)=1, P(3)=2, P(4)=2, P(5)=3, 4, 5, 7, 9, 12, 16, 21, etc

WHERE P(n) = P(n-2) + P(n-3)

fibonacci numbers are:

F(0)=0, F(1)=1, F(2)=1, F(3)=2, F(4)=3, F(5)=5, 8, 13, 21, etc

WHERE F(n) = F(n-1) + F(n-2)

Question

My question is, why not use P(0)=0 as follows:

P(0)=0, P(1)=1, P(2)=1, P(3)=1, P(4)=2, P(5)=2, P(6)=3, 4, 5, 7, 9, 12, 16, 21, etc

Since this satisfies P(n) = P(n-2) + P(n-3)

Also, when looking at the Fibonacci squares we see the first visible term of the sequence is F(1)

With the Padovan triangles the first visible term of the sequence is P(0)?

Which seems inconsistent to me. Can anyone give a mathematical explanation as to why P(0)=1 ... many thanks

Answer 1

The OEIS sequence A000931 is "Padovan sequence (or Padovan numbers): a(n) = a(n-2) + a(n-3) with a(0)=1, a(1)=a(2)=0." In contrast, your sequence $P(n) = A000931(n+5) = A134816(n+1).$ The OEIS entry also states

The following are basically all variants of the same sequence: A000931, A078027, A096231, A124745, A133034, A134816, A164001, A182097, A228361 and probably A020720. However, each one has its own special features and deserves its own entry.

The point is that it is a matter of convenience and choice of where to start the sequence and what index to use. The same holds for the Fibonacci numbers. Some people choose $F(0)=F(1)=1$. About your specific choice of $P(0)=P(1)=P(2)=1,$ that is exactly A134816 except offset differs by 1. You are welcome to submit your different offset sequence to the OEIS, but unlikely to succeed.

However, the English Wikipedia article Padovan sequence uses your notation, initial values, and contains the spiral of triangles. It mentions the OEIS sequence A000931 but neglects to mention the indexing difference.

Answer 2

Recurrence relations like those allow for going backwards; for instance we can define $F(-1)$ by $F(1)=F(0)+F(-1)$, so $$ F(-1)=F(1)-F(0)=1 $$ Similarly, $F(-2)=F(0)-F(-1)=-1$ and so on. If we translate indices, we get a new Fibonacci-like sequence $$ F'(0)=1,\quad F'(1)=0,\quad F'(n+2)=F'(n+1)+F'(n) $$ and another one if we start from $-2$ and translate indices by $2$: $$ F''(0)=-1,\quad F'(1)=1,\quad F''(n+2)=F''(n+1)+F''(n) $$

If we apply the same idea to the Padovan sequence, we need $P(2)=P(0)+P(-1)$, so $P(-1)=P(2)-P(0)=0$ and we could define $$ P'(0)=0,\quad P'(1)=1,\quad P'(2)=1,\quad P'(n+3)=P'(n+1)+P'(n) $$ Nothing different and no real mathematical explanation. Just history.

Answer 3

That's cute. The sequence also follows $$ P(n) = P(n-1) + P(n-5) \; , \; $$ which directly explains the diagram of triangles. For example, from the diagram, $12 = 9+3$ and $16=12+4.$

Go Figure.

$$ \left( x^{5} - x^{4} - 1 \right) $$

$$ \left( x^{3} - x - 1 \right) $$

$$ \left( x^{5} - x^{4} - 1 \right) = \left( x^{3} - x - 1 \right) \cdot \color{magenta}{ \left( x^{2} - x + 1 \right) } $$