Help Me Understand The Perrin Numbers Sequence

[mathgroupy]

Quite frankly I feel like a cretin not being able to understand the Perrin numbers sequence. I looked at the Wiki and various other pages but the explanations there are always the same.

Here is what it states:

P(n) = P(n − 2) + P(n − 3) for n > 2,with initial values
P(0) = 3, P(1) = 0, P(2) = 2.The sequence of Perrin numbers starts with
3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39...

Okay - I just don't get it. Let's say n = 3 then:

P(3) = P(3 − 2) + P(3 − 3) = 1 + 0 = 1

n = 4 then:

P(4) = P(4 − 2) + P(4 − 3) = 2 + 1 = 3

n = 5 then:

P(5) = P(5 − 2) + P(5 − 3) = 3 + 2 = 5

What am I missing here?

 

[Deleted member 4993]

Quite frankly I feel like a cretin not being able to understand the Perrin numbers sequence. I looked at the Wiki and various other pages but the explanations there are always the same.

Here is what it states:

P(n) = P(n − 2) + P(n − 3) for n > 2,with initial values
P(0) = 3, P(1) = 0, P(2) = 2.The sequence of Perrin numbers starts with
3, , 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39

Okay - I just don't get it. Let's say n = 3 then:

P(3) = P(3 − 2) + P(3 − 3) = 1 + 0 = 1 ← Incorrect



P(3) = P(1) + P(0) = 0 + 3 = 3



n = 4 then:

P(4) = P(4 − 2) + P(4 − 3) = 2 + 1 = 3

n = 5 then:

P(5) = P(5 − 2) + P(5 − 3) = 3 + 2 = 5

What am I missing here?

.

 

[mathgroupy]

P(3) = P(1) + P(0) = 0 + 3 = 3

That's where my mental hangup seems to be. HOW do you get to P(1) = 0 and P(0) = 3?? HOW does n-2 and n-3 come into play? If n is 3 then (3-2) =1 and (3-3) = 0

I just don't get it, extremely frustrating.

 

[stapel]

That's where my mental hangup seems to be. HOW do you get to P(1) = 0 and P(0) = 3?

As the previous helper pointed out with the highlighting, isn't that what they gave you initially?

HOW does n-2 and n-3 come into play?

How else is one to move past the three initial values in order to find further values?

If n is 3 then (3-2) =1 and (3-3) = 0

Yes, which is why the previous helper plugged in the values for P(1) and P(0).

I just don't get it, extremely frustrating.

Try trying something, doing something with what you've been given. If you work with it for a few minutes, by, for instance, trying to find the value of, say, P(3) from what you've been given, I suspect it will starting making much more sense!

By the way:

...the Perrin numbers sequence...:

P(n) = P(n − 2) + P(n − 3) for n > 2,

with initial values P(0) = 3, P(1) = 0, P(2) = 2....

Okay - I just don't get it. Let's say n = 3 then:

P(3) = P(3 − 2) + P(3 − 3) = 1 + 0 = 1

Why did you reverse the order of the values here? You had:

. . . . .P(3) = P(3 - 2) + P(3 - 3) = P(1) + P(0)

...which, in the same order, is:

. . . . .P(1) + P(0) = 0 + 3

So, besides making a mistake somehow in copying in the value of P(0), you reversed the order of the values. This sort of thing can tick off graders; just FYI... ;-)

 

[mathgroupy]

Finally!!

Okay, I finally got the Perrin sequence but only after looking up 'recurrence relation' in the Wiki. This was based on my misunderstanding of how to interpret the formula given. What I was actually attempting to do (and you'll probably laugh now) is to deduct 2 and 3 from the current number. Which is what caused the confusion. When I saw formula defining the Fibonacci sequence it quickly made sense to me:

I'm an old dog and never ascended to calculus in my youth - I'm teaching myself and I didn't realize that this was to be understood as a number series and the n-1 and n-2 refer to the previous two positions in the evolving number series. The initializing numbers here are 0 and 1.

F_n = F_{n − 2} + P_{n − 3} for n > 2,

So with the Perrin number we have 3, 0, 2 as the initializer and here we are shifting back -2 and -3 instead of -1 and -2 in Fibonacci. Which produces:

3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, etc.

Crystal clear now.

 

[stapel]

Okay, I finally got the Perrin sequence...

Crystal clear now.

Thank you for replying and letting us know!