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Choose a base (b), and a candidate (n). Assume (n) is a number whose probability of satisfying the given condition for a given base is indistinguishable from o

Choose a base (b), and a candidate (n). Assume (n) is a number whose probability of satisfying the given condition for a given base is indistinguishable from other numbers of the same length in the given base. The probability that (n) will satisfy the condition for all bases from 2 to (b) is: product ((2^(log(k,n)-1))/((k-1)(k^(log(k,n)-1)))), k=2 to b https://www.wolframalpha.com/input/?i=product+%28%282%5E%28log%28k%2Cn%29-1%29%29%2F%28%28k-1%29*%28k%5E%28log%28k%2Cn%29-1%29%29%29%29%2C+k%3D2+to+b https://www.wolframalpha.com/input/?i=product+%28%282%5E%28l... Edit: The log operations should be floored, but this still gives one a sense of the rapidly diminishing probability.

3pt14159 11y ago
Your link does not actually lead to a graph. Can you summarize the findings of your post?
- deckar01 11y ago
  There are an infinite number of solutions for the constraint for a single base. When you consider the probability of a solution occurring amongst the set of numbers of equal length, it becomes apparent that the size of each set grows faster than the number of solutions it contains. Requiring the number to satisfy the constraint for multiple bases causes the probability to decrease rapidly. This naive heuristic analysis suggests that the probability of finding a solution is not only inversely proportional to the chosen base, but also inversely proportional to the value of the chosen candidate.
- throwaway_yy2Di 11y ago
  The original article, by Thomas Oléron Evans [0], does something similar. I made a graph of it here. It's a heuristic estimate of the expected number of integers that satisfy the property for bases (2..b), as a function of b. https://i.imgur.com/3h8GSIp.png https://i.imgur.com/3h8GSIp.png https://gist.github.com/anonymous/9f71926313e6ee5c595c https://gist.github.com/anonymous/9f71926313e6ee5c595c As the author writes, for b <= 4, the expected number diverges to +infinity. For b = 5, the expected number is less than one, and quickly decreases. So, the conjecture is that there is no satisfying solution for b >= 6. [0] http://www.mathistopheles.co.uk/maths/covering-all-the-bases/solution-covering-all-the-bases/ http://www.mathistopheles.co.uk/maths/covering-all-the-bases...

3pt14159 11y ago

Your link does not actually lead to a graph. Can you summarize the findings of your post?

deckar01 11y ago
There are an infinite number of solutions for the constraint for a single base. When you consider the probability of a solution occurring amongst the set of numbers of equal length, it becomes apparent that the size of each set grows faster than the number of solutions it contains. Requiring the number to satisfy the constraint for multiple bases causes the probability to decrease rapidly. This naive heuristic analysis suggests that the probability of finding a solution is not only inversely proportional to the chosen base, but also inversely proportional to the value of the chosen candidate.
throwaway_yy2Di 11y ago
The original article, by Thomas Oléron Evans [0], does something similar. I made a graph of it here. It's a heuristic estimate of the expected number of integers that satisfy the property for bases (2..b), as a function of b. https://i.imgur.com/3h8GSIp.png https://i.imgur.com/3h8GSIp.png https://gist.github.com/anonymous/9f71926313e6ee5c595c https://gist.github.com/anonymous/9f71926313e6ee5c595c As the author writes, for b <= 4, the expected number diverges to +infinity. For b = 5, the expected number is less than one, and quickly decreases. So, the conjecture is that there is no satisfying solution for b >= 6. [0] http://www.mathistopheles.co.uk/maths/covering-all-the-bases/solution-covering-all-the-bases/ http://www.mathistopheles.co.uk/maths/covering-all-the-bases...