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Imagine a sequence of $N$terms where each term is $R$ percent larger then pair of following elements, and the sum of all terms is constant, say $1$?

For example:

$R = \dfrac{N_i}{N_{i+i} + N_{i+2}}$

$N_i = R \times (N_{i+1}+N_{i+2})$ if $i > N-2$


$\sum N_i = 1$

I wanted to create a function $f$ such as provided with number of elements in group $N$ , ratio between element and the sum of to following elements $R$ and index of an element $i$ would provide $i^{th}$ element:

$f(N, R, i) = N_i$

EDIT: Example:

For example a sequence (that doesn't add to 1), with $N = 5$, $R = 0.8$:

$7.68, 5.6, 4, 3, 2$

Note that $7.68 = 0.8 \times (5.6 + 4)$, $5.6 = 0.8 \times (4 + 3)$ and $4 = 0.8 \times (3+2)$

EDIT #2: Fixed first sentence to reflect what I wanted to know.

I'm a novice in math, and this question is totally over my head, so any help would be welcome.

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Are you sure you want to post this in Group Theory? Seems to me more Calculus-like. – Nicky Hekster Jul 11 '12 at 12:08
@NickyHekster - no, I'm not sure about where this question is suited best. Will add calculus for good measure. – zzandy Jul 11 '12 at 12:10
Percentages work with numbers...but a group is a more abstract concept. – fretty Jul 11 '12 at 12:10
What you ask for is not quite clear. Maybe you could use a percentage, like $20\%$, or anything you like, and supply a specific four-term numerical sequence $a_1,a_2,a_3,a_4$ that qualifies. You needn't bother to make their sum equal to $1$. – André Nicolas Jul 11 '12 at 12:18
I removed the group theory and finite groups tags, as this is not related to any sort of group theory. – Tobias Kildetoft Jul 11 '12 at 12:19
up vote 0 down vote accepted

I am a little unsure what you are asking.

In your first sentence you say "each term is R percent larger then previous element", which leads to the recurrence $a_n = (1+R)a_{n-1}$ as DonAntonio and Arturo have analyzed.

But in your example each term is actually a constant factor larger than the sum of the previous two elements, which is $a_n = (1+R)(a_{n-1} + a_{n-2})$. This is a more difficult problem.

For convenience write $a_1 = \alpha$, $a_2 = \beta$, and $(1+R) = r$. Then the two-term recurrence gives \begin{align*} a_1 &= \alpha \\ a_2 &= \beta \\ a_3 &= r(a_1 + a_2) = r\alpha + r\beta \\ a_4 &= r(a_2 + a_3) = r^2\alpha + (r + r^2)\beta \\ a_5 &= r(a_3 + a_4) = (r^2 + r^3)\alpha + (2r^2 + r^3)\beta \\ a_6 &= r(a_4 + a_5) = (2r^3 + r^4)\alpha + (r^2 + 3r^3 + r^4)\beta \end{align*} and so on. The corresponding sums $S_N = \sum_{i=1}^N a_i$ are \begin{align*} S_1 &=\alpha \\ S_2 &=\alpha + \beta \\ S_3 &=(1+r)\alpha + (1+r)\beta \\ S_4 &=(1+r+r^2)\alpha + (1 + 2r + r^2)\beta \\ S_5 &=(1+r+2r^2+r^3)\alpha + (1+2r+3r^2+r^3)\beta \\ S_6 &=(1+r+2r^2+3r^3+r^4)\alpha + (1+2r+4r^2+4r^3 + r^4)\beta \end{align*} It is possible that we can spot a pattern here. Perhaps the coefficients in the expression before $\alpha$ are simply $1,1, 2, 3, ..., N-3, 1$ (though I don't see a nice pattern in the $\beta$ expression). Then, if we are lucky, we could find a closed form expression for $S_N$, which we could use to choose $\alpha$ and $\beta$ to give us a sum of 1.

But I am not optimistic.

Edit My lack of optimism was misplaced. If we take $\alpha = 2$ and $\beta = r$ (and index starting at zero rather than one) we have a (non-integer) Lucas sequence $V_n(r,-r)$: \begin{align*} V_0(r, -r) &= 2, \\ V_1(r, -r) &= r,\\ V_n(r, -r) &= rV_{n-1}(r,-r) + rV_{n-2}(r,-r). \end{align*} The analysis in the Wikipedia page shows that $V_n = a^n + b^n$ where $$a = \frac{r + \sqrt{\Delta}}{2}, \quad b = \frac{r - \sqrt{\Delta}}{2}, \quad \Delta=r^2 + 4r$$ and so the sum $S_N$ is \begin{align*} S_N &= \sum_{n=0}^N a^n + b^n \\ &= \frac{a^{N+1}-1}{a-1} + \frac{b^{N+1} -1}{b-1}. \end{align*} Then, by starting our sequence with $2/S_N$ and $r/S_N$ rather than with $2$ and $r$ we will get a sequence with a sum of 1.

Note, though, that this method doesn't allow us to choose $a_1/a_0$: the ratio must be $r/2$.

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Thanks for the answer, the sequence I meant have this relationship: $N_i = R(N_{i+1} + N_{i+2})$, sorry for confusion. – zzandy Jul 12 '12 at 8:57

Let $\,a_1,a_2,...,a_N\, $ be the sequence, such that every term is $\,R\%\,$ larger than the precedent one, i.e. such that $$a_{n+1}=\left(1+\frac{R}{100}\right)a_n$$ The above means, of course, that we have a geometric sequence with quotient $\,q:=1+R/100\,$ and we want its sum to equal $\,1$:

$$1=S_N=\sum_{k=0}^N a_1q^k=a_1\frac{q^N-1}{q-1}\Longrightarrow a_1=\frac{q-1}{q^n-1}$$ and since you know $\,q\,$ we're done.

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I'm sorry for confusion, here is wat I actually meant: $N_i = R(N_{i+1} + N_{i+2})$ – zzandy Jul 12 '12 at 9:00
Yes, too bad...but I bet now you have new ideas how to tackle the question. – DonAntonio Jul 12 '12 at 11:23

Note: this answer addresses the question as it existed before the edit that corrected the statement. Before, rather than $a_{n+2}=(1+R)(a_{n}+a_{n+1})$, the statement only indicated each term depended only on the previous one.

If the first terms is $a_1$, the second will be $(1+R)a_1$, the third will be $(1+R)^2a_1$, and so on, until the $N$th term is $(1+R)^{N-1}a_1$; I am assuming that $R$ is given as a fraction, $0\leq R$ (so if you want the increase to be $22$%, you would have $R=\frac{22}{100}$).

The sum of the terms is $$\begin{align*} a_1+\cdots+a_N &= a_1 + (1+R)a_1+\cdots+(1+R)^{N-1}a_1\\ &= a_1\left( 1 + (1+R) + \cdots + (1+R)^{N-1}\right)\\ &= a_1\left(\frac{1 - (1+R)^N}{1-(1+R)}\right)\\ &= \frac{a_1\left((1+R)^N-1\right)}{R}. \end{align*}$$ For this to add up to $1$, you need $$a_1 = \frac{R}{(1+R)^N-1}.$$

So your function will be $$f(N,R,i) = \left\{\begin{array}{ll} \frac{R(1+R)^{i-1}}{(1+R)^N-1}&\text{if }1\leq i\leq N\\ \text{undefined}&\text{otherwise.} \end{array}\right.$$

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Beg pardon, but one helpul soul actually confused the question. The sequence have to have this relationship: $N_i = R(N_{i+1} + N_{i+2})$ – zzandy Jul 12 '12 at 8:59

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