Find sum of positive geometric progression numbers

Clear["Global`*"]

In the tournament, {CumulativeGamesPlayed, TeamsAdvancing} evolves as

FixedPointList[{#[[1]] + Floor[#[[2]]/2], Ceiling[#[[2]]/2]} &, {0, 7}]

(* {{0, 7}, {3, 4}, {5, 2}, {6, 1}, {6, 1}} *)

Then the sequence of {n, CumulativeGamesPlayed} is given by

seq = Table[{n, 
   FixedPoint[{#[[1]] + Floor[#[[2]]/2], Ceiling[#[[2]]/2]} &, {0, 
      n}][[1]]}, {n, 2, 15}]

(* {{2, 1}, {3, 2}, {4, 3}, {5, 4}, {6, 5}, {7, 6}, {8, 7}, {9, 8}, {10, 9}, {11,
   10}, {12, 11}, {13, 12}, {14, 13}, {15, 14}} *)

Generalizing from the sequence,

games[n_Integer?Positive] = FindSequenceFunction[seq, n]

(* -1 + n *)

There are one fewer games played than the number of teams

For n == 7

TreePlot[{1 -> 8, 2 -> 8, 3 -> 9, 4 -> 9, 5 -> 10, 6 -> 10, 7 -> 11, 8 -> 12, 
  9 -> 12, 10 -> 13, 11 -> 13, 12 -> 14, 13 -> 14}, Bottom, 
 VertexLabels -> Automatic]

For n == 8

TreePlot[{1 -> 9, 2 -> 9, 3 -> 10, 4 -> 10, 5 -> 11, 6 -> 11, 7 -> 12, 
  8 -> 12, 9 -> 13, 10 -> 13, 11 -> 14, 12 -> 14, 13 -> 15, 14 -> 15}, Bottom,
  VertexLabels -> Automatic]

Here's one way

numGames[n_] := Block[{rounds},
  rounds = Log[2, n];
  If[!IntegerQ@rounds, Print["bad bracket"]; Abort[];,
   Sum[n/2^i, {i, rounds}]]]

We assume the number of teams is a power of $2$ so you can keep dividing by $2$ each round. If it's not, we quit. Otherwise we do your sum, $$n/2+n/2/2+n/2/2/2+dots=n/2+n/2^2+n/2^3+dots=sum_i n/2^i$$

So if we run numGames[64] we get $63$ back. As @BobHanlon mentions, we get $n-1$ games for $n$ players. We can see this directly with

Sum[n/2^i, {i, Log[2, n]}]

which gives $n-1$

• In a final two teams play and one wins. Let us call this play a stage zero because it is very important.

• In semifinals four teams play and two win. Let us call this stage one.

• In quarterfinals eight teams play and four win. Let us call this stage two.

• In one-eights of a final sixteen teem play and eight win. Call this stage three.

There are not many soccer tournaments with more stages. Thus, highest stage is $s=3$. Now we can make a table:

Stage(s) Teams(n) Wins(w) Games(g) Name 
0        2        1       1        Final
1        4        2       2        Semifinals
2        8        4       4        Quarterfinals
3       16        8       8        One-eights of a final

Now everything is easy:

• Each game has a winner in a single-elimination tournament. Therefore the number of winners is equal to the number of games $w = g$.
• Number of winners at the stage $s$ is equal to $w=g=2^s; (sge0)$. This can be appreciated from the table, or one can use mathematical induction to prove this equation.
• Now we can use the power of Mathematica to compute the total number of games in a tournament starting from stage $s$:

---

gtotal = Sum[2^i, {i, 0, s}]
(* -1 + 2^(1 + s) *)

Finally, the total number of teams $n$ is equal twice the number of winners, i.e., $n=2 w$. This product too can be computed with Mathematica

n = 2 2^s
(* 2^(1 + s)*)

---

By making use of the two last calculations we arrive at the conclusion that

$$g_text{total} = n - 1, quad n>0.$$

This is the answer suggested by @JimB in the comment. However, it was not appreciated that it is only valid for $n>0$. For zero number of teams ($n=0$) there are $g_text{total}=0$ games. However, there could be a game for the third place. In this case we add one to the result above and obtain $$g_text{total with small final} = 1 + g_text{total} = n,quad n>0.$$