Talk:Padovan sequence

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The use of anti-Fibonacci number for "Fibonacci number with negative index" requires a source; in fact, that "anti-Fibonacci number" exists, other than as a nonce-word, requires a source. Septentrionalis PMAnderson 23:34, 29 January 2007 (UTC)Reply

No source, so I've removed the term and replaced it with "negative index values". Gandalf61 10:57, 30 January 2007 (UTC)Reply

Should there be some link to "Generalizations of Fibonacci numbers?" —Preceding unsigned comment added by Sa zbarsky (talk • contribs)

The general formula for P(n) is not correct, I think, as you can see in mathworld for instance.

155.210.85.186 18:37, 9 October 2007 (UTC)Reply

I don't know that it is worth listing random recurrences satisfied.

Given a linear recurrence relation one can create other recurrences satisfied by every sequence satisfying the initial recurrence. Indeed you can start off as you wish and then uniquely tack on a few term at the front to make it true (as many additional terms as the degree of the recurrence). In this case take ANY combination of P(n-4),P(n-5) etc that you wish and then there is a unique way to add AP(n-1)+BP(n-2)+CP(n-3) to make it all equal to P(n): Here is everything (or at least lots of cases) where the coefficients (in absolute value) add to 5 or less (going out to P(n-20) ):

a

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

P(n) = P(n-1) + P(n-5)

b

P(n) = P(n-1) + P(n-2) - P(n-4)

P(n) = P(n-3) + P(n-4) + P(n-5)

P(n) = P(n-2) + P(n-5) + P(n-6)

P(n) = 2 P(n-2) - P(n-7)

P(n) = 2 P(n-3) + P(n-7)

P(n) = P(n-1) + P(n-7) + P(n-8)

c

P(n) = 2 P(n-1) - P(n-4) - P(n-6)

P(n) = 2 P(n-2) - P(n-4) + P(n-6)

P(n) = 2 P(n-3) + P(n-4) - P(n-6)

P(n) = P(n-4) + 2 P(n-5) + P(n-6)

P(n) = P(n-3) + P(n-5) + P(n-6) + P(n-7)

P(n) = P(n-2) + P(n-6) + P(n-7) + P(n-8)

P(n) = P(n-1) + P(n-8) + P(n-9) + P(n-10)

P(n) = 2 P(n-2) - P(n-9) - P(n-10)

P(n) = 2 P(n-3) + P(n-9) + P(n-10)

P(n) = 3 P(n-4) + P(n-13)

P(n) = -P(n-1) + P(n-2) + 2 P(n-3) + P(n-4)

P(n) = 2 P(n-1) - P(n-3) - P(n-4) + P(n-5)

P(n) = 2 P(n-2) + P(n-3) - P(n-4) - P(n-5)

P(n) = 2 P(n-1) - 2 P(n-6) - P(n-7)

P(n) = 2 P(n-4) + 2 P(n-5) - P(n-7)

P(n) = 2 P(n-5) + 2 P(n-6) + P(n-7)

P(n) = P(n-3) + P(n-6) + 2 P(n-7) + P(n-8)

P(n) = 3 P(n-3) - P(n-6) - P(n-8)

P(n) = 2 P(n-4) + P(n-7) + 2 P(n-8)

P(n) = P(n-2) + P(n-7) + 2 P(n-8) + P(n-9)

P(n) = 3 P(n-4) + P(n-8) - P(n-9)

P(n) = P(n-1) + P(n-9) + 2 P(n-10) + P(n-11)

P(n) = 2 P(n-2) - P(n-10) - P(n-11) - P(n-12)

P(n) = 2 P(n-3) + P(n-10) + P(n-11) + P(n-12)

P(n) = 3 P(n-4) + P(n-10) - P(n-12)

P(n) = 4 P(n-5) + P(n-14)

P(n) = 3 P(n-4) + P(n-15) + P(n-16)

--Gentlemath (talk) 05:23, 16 March 2009 (UTC)Reply

As there are closed functions for Fibonacci numbers like F(x)=((1+sqrt 5)/2)^x-((-1)^x/((1+sqrt 5)/2)^x)))/sqrt 5, I was wondering if there where any closed (finite) functions that generate the xth Padovan number? Robo37 (talk) 22:49, 30 July 2011 (UTC)Reply