How much weight is on each person in a human pyramid? After participating in a human pyramid and learning that it's very uncomfortable to have a lot of weight on your back, I figured I'd try to write out a recurrence relation for the total amount of weight on each person in the pyramid so that I could see just how much weight I ended up carrying.
There are many different kinds of human pyramids, but the sort of pyramid I'm referring to is one that looks like this:
               *
              / \
             *   *
            / \ / \
           *   *   *
          / \ / \ / \
         *   *   *   *
               ...

Here, each star is a person and the lines represent how each person is supported.
I'm going to make the unrealistic assumption that each person weighs the same amount, which I'll call $W$.  I'm also going to assume that each person evenly transmits their weight (plus the total weight on top of them) to the two people below them.
Given these assumptions, I came up with a recurrence $w_{i, j}$ that says how much weight the $i$th person in row $j$ of the human pyramid carries on their back.  It ended up coming out like this:


*

*$w_{1, 1} = 0$.  The top person in the pyramid has no weight on them.

*$w_{1, n+1} = \frac{w_{1, n} + W}{2}$  The very first person on each row is shouldering half the weight of the person above them.  That weight is given by the weight of the person ($W$) plus the load that person carries ($w_{1, n}$).

*$w_{n+1, n+1} = \frac{w_{n, n} + W}{2}$.  The last person on each row shoulders half of the weight of the person above them.

*$w_{k+1, n+1} = \frac{w_{k+1, n} + w_{k, n} + 2W}{2}$.  Each person on a row other than the first or the last shoulders half the weight from each of the people above them.  The two people above them have $w_{k+1, n}$ and $w_{k, n}$ weight on them, and each one independently weighs $W$.  Half of each of these weights is transmitted to the person below.


I was able to write a computer program that evaluated this recurrence and I was able to get values from it, but I have no idea how to find a closed-form expression for this recurrence.  It's somewhat similar to the recurrence for combinations - each term is expressed as a sum of the two terms above it - but there's some extra junk thrown in as well.
Is there a standard approach for simplifying recurrences like this?
Thanks!
 A: Setup and change of indices
Put $W=1$. Also change the indices of $w_{i,j}$ so that $(i,j)$ points relatively to the top person at (0,0) by moving $i$ persons down-right and then $j$ persons down-left. So (i,j) points to the $i$'th person in the $(i+j)$'th row counting from zero.
The weight distribution of each person
Let us first consider how the top persons weight is distributed down through the pyramid. Then we have the recurrence
$$
w_{i,j}=\frac{w_{i-1,j}+w_{i,j-1}}{2}
$$
where $w_{0,0}=1$ and $w_{i,j}=0$ outside of the pyramid (that is $i<0$ or $j<0$). This can easily be solved to get $w_{i,j}=\frac{1}{2^{i+j}}\binom{i+j}{i}$. So this is just a variation over Pascal's Triangle where the $n$'th row is divided by $\frac{1}{2^n}$. Let VPT denote this Variation over Pascal's Triangle.
Now move on to consider how the weight of the person at position $(i,j)$ is distributed. The pattern is basically the same! Isolating each person like this, we can see how the distibution of a single persons weight forms a copy of VPT having its top corner at $(i,j)$.
Analysis of the general case
Consider the person at $(i,j)$. This person is included in all VPT's with top corners in $P=\{0,...,i\}\times\{0,...,j\}$ and carries the weight distributed according to all these VPT's. Only we have added 1 too much since the person is included as top corner in its own VPT but does not carry its own weight in the scope of this question.
Now consider $(s,t)\in P$ as one such top corner. Then the position of $(i,j)$ relative to this will be $(x,y)=(i-s,j-t)\in P$. Note here how the map $(s,t)\mapsto (i-s,j-t)$ maps $P$ to itself bijectively. Given this observation one may deduce that the distributed weight contributions form a sum that can be expressed as:
$$
w_{i,j}+1=\sum_{x=0}^i\sum_{y=0}^j\frac{1}{2^{x+y}}\binom{x+y}{x}
$$
Though not closed form, I believe this to be a very nice and symmetric expression for $w_{i,j}$. Still, I consider the answer given by robjohn just brilliant. BTW my formula yields the same values as robjohn got, and the next example of his should read:
Example (continued from robjohn)
$P_n(x)=\tfrac{31}{32}+\tfrac{87}{32}x+\tfrac{61}{16}x^2+\tfrac{61}{16}x^3+\tfrac{87}{32}x^4+\tfrac{31}{32}x^5$
Also note that the sum of coefficeints of each row in the pyramid should be a triangular number since the persons in the $n$'th row carries the weight of $T_n$ people (the $n$'th triangular number).
A: \begin{align}
\Psi_{n}\left(x\right)
&\equiv
\sum_{k = 1}^{\infty}\phi_{k,n}\,x^{k}
=
\sum_{k = 1}^{\infty}x^{k}
\left(2\phi_{k\ +\ 1,n\ +\ 1} - \phi_{k\ +\ 1,n} - 2\right)
=
2\sum_{k = 2}^{\infty}x^{k\ -\ 1}\,\phi_{k,n\ +\ 1}
\\[3mm]&-
\sum_{k = 2}^{\infty}x^{k\ -\ 1}\phi_{k,n} - 2\,{x \over 1 - x}
=
2\left\lbrack{1 \over x}\,\Psi_{n\ +\ 1}\left(x\right)
-
\phi_{1,n\ +\ 1}\right\rbrack
-
\left\lbrack%
{1 \over x}\,\Psi_{n}\left(x\right) - \phi_{1,n}
\right\rbrack
\\[3mm]&-
2\,{x \over 1 - x}
\end{align}
\begin{align}
-----&------------------------------------
\end{align}
$$
2\Psi_{n\ +\ 1}\left(x\right)
-
\left(1 + x\right)\Psi_{n}\left(x\right)
=
x\left(2\phi_{1,n\ +\ 1} - \phi_{1,n}\right) + {2x^{2} \over 1 - x}
=
x + {2x^{2} \over 1 - x}
=
{x + x^{2} \over 1 - x}
$$
\begin{align}
-----&------------------------------------
\end{align}
$$
\Psi_{n\ +\ 1}\left(x\right)
-
{x + x^{2} \over \left(1 - x\right)^{2}}
=
{1 \over 2}\left(1 + x\right)
\left\lbrack%
\Psi_{n}\left(x\right)
-
{x + x^{2} \over \left(1 - x\right)^{2}}
\right\rbrack
$$
\begin{align}
-----&------------------------------------
\end{align}
$$
\Psi_{n\ +\ 1}\left(x\right)
=
{x + x^{2} \over \left(1 - x\right)^{2}}
+
{1 \over 2^{n}}\left(1 + x\right)^{n}\,\Psi_{1}\left(x\right)
$$
\begin{align}
-----&------------------------------------
\end{align}
Multiply by $y^{n}$ and perform the sum for $n \geq 1$. With
$\Psi\left(x,y\right)
 \equiv
 \sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}\phi_{k,n}\,x^{k}y^{n}$:
$$
\Psi\left(x,y\right)
=
{x + x^{2} \over \left(1 - x\right)^{2}}\,{y^{2} \over 1 - y}
+
{y \over 1 - \left(x + 1\right)y/2}\,\Psi_{1}\left(x\right)
$$
That determines $\phi_{k,n}$ in terms of $\phi_{k,1}$. It yields an equation for $\phi_{k,1}$. It's done.
A: Let's have $W=1$.
If $n$ is number of row (starting with $0$) and $k$ is number of human (starting with $0$ from the left side), then
$$w_{k,n}=n+1-\frac{\binom{n}{k}+2S_{k,n}}{2^n}$$
Where $$S_{k,n}=\sum_{0\leq i\leq n}\binom{n}{i}\left|k-i\right|$$
I couldn't get closed form for $S_{k,n}$, but I have recurrence $S_{k,0}=\left|k\right|$ and $S_{k,n}=S_{k-1,n-1}+S_{k,n-1}$.

$m_{k,n}$ will be "total" weight of the human (his mass and all mass he gets from higher levels of pyramid).
So $m_{0,0}=1$. And, to make anything easier I will introduce imaginary humans outside of the pyramid: $m_{k,0}=2-2\left|k\right|-
\left[k=0\right]$. 
($[k=0]$ is $1$ if $k=0$ else it's $0$).
Then $$m_{k,n}=\frac{m_{k-1,n-1}+m_{k,n-1}}{2}+1$$ and for $0\leq k\leq n$ (for real humans) everything is okay.
If $t_{k,n}=m_{k,n}-n$ then $m_{k,n}=t_{k,n}+n$ and $t_{k,n}=\frac{t_{k-1,n-1}+t_{k,n-1}}{2}$ (with $t_{k,0}=m_{k,0}$).
Let's also have $u_{k,n}=2^nt_{k,n}$. Then $t_{k,n}=\frac{u_{k,n}}{2^n}$ and $u_{k,n}=u_{k-1,n-1}+u_{k,n-1}$ (with $u_{k,n}=t_{k,0}=m_{k,0}$).
If we get formula for $u_{k,n}$, then $m_{k,n}=t_{k,n}+n=\frac{u_{k,n}}{2^n}+n$
Now I use formula $$u_{k,n}=\sum_i \binom{n}{i} u_{k-i,0}$$
$$u_{k,n}=\sum_i \binom{n}{i} \left(2-2\left|k-i\right|-
\left[k-i=0\right]\right)=\\
=2\sum_i \binom{n}{i}-2\sum_i \binom{n}{i} \left|k-i\right|- \sum_i \binom{n}{i}\left[k-i=0\right]=\\
=2^{n+1}-2\sum_i \binom{n}{i} \left|k-i\right|-\binom{n}{k}$$
$$m_{k,n}=\frac{2^{n+1}-2\sum_i \binom{n}{i} \left|k-i\right|-\binom{n}{k}}{2^n}+n=\\
=n+2-\frac{\binom{n}{k}+2\sum_i \binom{n}{i} \left|k-i\right|}{2^n}$$
And $w_{k,n}=m_{k,n}-1$.
A: For a generating function approach, Let the coefficient of $x^k$ in $P_n(x)$ be the weight on the shoulders of person $k$ (starting at $k=0$ on the left) in row $n$ (starting at $n=0$ at the top). To compute $P_n(x)$ from $P_{n-1}(x)$, we first add the weight of the people in row $n-1$ to the weights on their shoulders:
$$
P_{n-1}(x)+\frac{1-x^n}{1-x}\tag{1}
$$
Those people distribute half their weight to each person below them, so we get
$$
P_n(x)=\frac{1+x}2\left(P_{n-1}(x)+\frac{1-x^n}{1-x}\right)\tag{2}
$$
Solving this recurrence yields
$$
P_n(x)=\frac{1+x}{(1-x)^2}\left(1-2\left(\frac{1+x}{2}\right)^{n+1}+x^{n+1}\right)\tag{3}
$$
$(3)$ can be checked by noting that $P_0(x)=0$ and then by checking that $(3)$ satisfies $(2)$.
Using the binomial theorem to multiply the terms in $(3)$, we get that $w_{m+1,n+1}$, the coefficient of $x^m$ in $P_n(x)$, is given by
$$
2m+1-\frac1{2^n}\sum_{k=0}^m(k+1)\binom{n+2}{m-k}\tag{4}
$$
There might be some special function of which I am unaware (a hypergeometric function perhaps), but in terms of common functions, $(4)$ is as close to a closed form as I think there is.

Solving the recurrence
Let $Q_n(x)=\left(\frac2{1+x}\right)^nP_n(x)$, then multiplying $(2)$ by $\left(\frac2{1+x}\right)^n$ yields
$$
\begin{align}
Q_n(x)
&=Q_{n-1}(x)+\frac{1-x^n}{1-x}\left(\frac2{1+x}\right)^{n-1}\\
&=Q_{n-1}(x)+\frac12\frac{1+x}{1-x}\left(1-x^n\right)\left(\frac2{1+x}\right)^n\\
&=Q_{n-1}(x)+\frac12\frac{1+x}{1-x}\left(\left(\frac2{1+x}\right)^n-\left(\frac{2x}{1+x}\right)^n\right)\tag{5}
\end{align}
$$
Summing the geometric series gets a bit messy,
$$
\begin{align}
Q_n(x)
&=\frac12\frac{1+x}{1-x}\sum_{k=1}^n\left(\left(\frac2{1+x}\right)^k-\left(\frac{2x}{1+x}\right)^k\right)\\
&=\frac12\frac{1+x}{1-x}\left(\frac2{1+x}\frac{\left(\frac2{1+x}\right)^n-1}{\frac2{1+x}-1}-\frac{2x}{1+x}\frac{\left(\frac{2x}{1+x}\right)^n-1}{\frac{2x}{1+x}-1}\right)\\
&=\frac{1+x}{1-x}\left(\frac{\left(\frac2{1+x}\right)^n-1}{1-x}+x\frac{\left(\frac{2x}{1+x}\right)^n-1}{1-x}\right)\\
&=\frac{1+x}{(1-x)^2}\left(\left(\frac2{1+x}\right)^n-1+x\left(\frac{2x}{1+x}\right)^n-x\right)\\
&=\frac{1+x}{(1-x)^2}\left(\left(\frac2{1+x}\right)^n\left(1+x^{n+1}\right)-(1+x)\right)\tag{6}
\end{align}
$$
Multiplying by $\left(\frac{1+x}2\right)^n$, we arrive at $(3)$:
$$
P_n(x)=\frac{1+x}{(1-x)^2}\left(1+x^{n+1}-2\left(\frac{1+x}2\right)^{n+1}\right)\tag{7}
$$

Examples:
$$
\begin{align}
P_0(x)&=0\\
P_1(x)&=\frac12+\frac12x\\
P_2(x)&=\frac34+\frac32x+\frac34x^2\\
P_3(x)&=\frac78+\frac{17}{8}x+\frac{17}{8}x^2+\frac78x^3\\
P_4(x)&=\frac{15}{16}+\frac52x+\frac{25}{8}x^2+\frac52x^3+\frac{15}{16}x^4
\end{align}
$$
The bottom-middle person on a $5$-tier pyramid is supporting more than $3$ people's weight on their back.
A: Conjecture: For a large number of rows the human pyramids weight each person feels approaches 
$$
(2k+1)W
$$
Where $k$ is the person's position from the edge. Edge people $k=0$ approach $1W$, next case 1st person from the edge $k=1$ approaches $3W$, etc. This is a working upper bound for each person based on large number of rows, $n$.

Interestingly odd numbers of weights are what is felt. Estimation checked numerically for first hundred rows, $n=100$; note that this bound is overestimated when $2k+2$ numeric value is near $n$. The percent difference between actual and estimate was at ~$1$ % at $k=43$ and at a maximum of $7.6$ % difference at $k=49$ from the edge, with and estimate of 87 on 86.0856974970029 at $k=43$, and an estimate of 99 on 91.961487023895000 at $k=49$. The upper bound is within 1% about 9/10 values of $k$. The more $$2k+1<<1$$ is satisfied the more the bound becomes an accurate estimate.
As an aside since I am numbering from closest edge with 0 being the edge, there are two $k=0$ positions and so on, up to and including $k=49$, and no $k=50$ because row $n=100$ has only 100 people on the base.
