Help to define a sequence 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 $i > N-2$
and
$\sum N_i = 1$
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.
 A: 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
$$f(N,R,i) = \left\{\begin{array}{ll}
\frac{R(1+R)^{i-1}}{(1+R)^N-1}&\text{if }1\leq i\leq N\\
\text{undefined}&\text{otherwise.}
\end{array}\right.$$
A: 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.
A: 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 $r/2$.
