The little half puzzle proposed a “dumb’ solution in that players play a minimax strategy. There are 34 starting values less than 100 guaranteeing a sure win to dumb players. If instead the players maximise their choice at each step, the R code looks like this:

solveO=function(n){
if (n<3){ solve=(n==2)}else{
  solve=(!(solveO(n-1)))||(!solveO(ceiling(n/2)))}
solve}

and there are now 66 (=100-34, indeed!) starting values for which the starting player can win.

Incidentally, I typed

> solveO(1113)
Error: evaluation nested too deeply: infinite recursion / options(expressions=)?

which shows R cannot handle heavy recursion without further programming. Testing for the upper limit, I found that the largest acceptable value is 555 (which takes forever to return a value, predicted at more than one hour by a linear regression on the run times till 300…).