Life-span change: a more general form

Answer

Yes!
The proof relies on the composition of age-groups within the population of each month. Let R(M) represent the sequence called R, where the rabbits live for M months. Then, let
An represent the # of newborns in the nth month, and
Bn represent the # of 1-month olds in the nth month, and
Cn represent the # of 2-month olds in the nth month, and
..., and
Zn represent the # of M-month olds in the nth month.

Then, because the number of mature, adult rabbits in the previous month determines the number of newborns in the current month, we say that
Also, because the rabbits age as time goes on, we say that

Due to the length of the proof, it will not appear in this post. Although, I will explain how the proof works out. So, the above equations allow me to say this

because the population in the nth month is only comprised of the M age-groups.
Then, I can expand the An term and substitute for Bn through Zn to give us


We do this expansion again, to give us two of the A terms. This is very important, because in substitution, the A term is the only term which gives us all terms B through Z. Then, once we have the two A terms, we can group A through Z (with the same subscript, say n-2) as R(M)n-2. Continue to substitute and regroup until all the terms (A through Z) have been regrouped into populations (R(M)). The final result will be
This is exactly what I wanted to confirm.

Summer vacation is here

Today was the last day of my finals, for this semester. Therefore, I will be diving back into my research tomorrow.

Life-span change: a more general form

Question
Is there a general form for all the recurrence relations that describe the life-span changing sequences? In other words, ...

Let R(M) represent the sequence called R, where the rabbits live for M months.
We know that R(3)=Rn-2 + Rn-3
and that R(4)=Rn-2 + Rn-3 + Rn-4
and that R(5)=Rn-2 + Rn-3 + Rn-4 + Rn-5
See a pattern?

It appears that R(M)=Rn-2 + Rn-3 + ... +Rn-M , but is this so?

Celebration of Student Research

Today is the Celebration of Student Research, here at NMU, and I'll be presenting my research for the last time this semester. It's been a mind-opening journey, and I'm excited to continue my explorations through the summer in other areas of the Fibonacci population model.

Out of all the presentations, parts of mine made the local news. To see the story, click here.

I'll have Abstract with my coffee, please.

Yesterday, while I was tutoring for an introductory mathematics class (MA 104-College Algebra), the topic we are covering in MA 412 (Abstract Algebra II) came up. The look on his face when I said, "Hey! This is what I'm learning about now," well, in a word, fear. I'm not sure if it was my excitement that frightened him, or maybe he was wondering why on Earth was I learning the same material and any confidence he had in me was currently being tossed out of the window. Either way, I explained which class it pertained to, and why I was so excited. Signs of relief were evident on his face when I had restored his thoughts that I'm completely and happily obsessed with mathematics.

Almost there...

With it being the last month in this semester I'm not posting as much as I would like to. So, you better believe it when I say that once I complete my final exams, I'll be posting often.

As much as it pains me to put this research on hold, it would hurt worse if I dropped my GPA.

In other words, thanks for checking in and if you don't see new posts, don't worry. I didn't fall off the planet, and I'm not abandoning my blog.

-Twelve

Proof of the second formula

The proof of the second formula is carried out similarly to the proof of the first formula.

Sn = Sn-2 + Sn-3 + Sn-4 (second formula)

Let An represent the number of newborns in the nth month
Let Bn represent the number of 1-month olds in the nth month
Let Cn represent the number of 2-month olds in the nth month, and
Let Dn represent the number of 3-month olds in the nth month

Because the population in each month only consists of these age groups, we can say that Sn = An + Bn + Cn + Dn. (1)

Also, because the number of newborns depends on the number of adults (mature rabbits) in the previous month, we can say that
An = Bn-1 + Cn-1 + Dn-1. (2)

Also, because the rabbits age, we have
Bn = An-1 (3)

Cn = Bn-1 (4)

Dn = Cn-1 (5)

Now we can use formulas 1-5 to prove (or disprove) the second formula.

First, substitute (1) into each term in the second forumla:

Sn = An + Bn + Cn + Dn
Sn-2 = An-2 + Bn-2 + Cn-2 + Dn-2
Sn-3 = An-3 + Bn-3 + Cn-3 + Dn-3
Sn-4 = An-4 + Bn-4 + Cn-4 + Dn-4

Set up the LHS and RHS:

LHS: An + Bn + Cn + Dn =
RHS: An-2 + Bn-2 + Cn-2 + Dn-2 + An-3 + Bn-3 + Cn-3 + Dn-3 + An-4 + Bn-4 + Cn-4 + Dn-4

Use (2), (3), (4), and (5) to substitute into the RHS until it equals the LHS.

LHS: An + Bn + Cn + Dn =
RHS: An + Bn + Cn + Dn

Therefore, we can generate the 4MLS sequence as follows:
when n is less than or equal to 4: Fn = Fn-1 + Fn-2
when n is greater than 4: Sn = Sn-2 + Sn-3 + Sn-4