The new header logo broke the tabs.

[edit] Oh, actually, no.. tabs have now been fixed by kaeladan.

(Thanks!)

This counting puzzle from Matt Parker looks like the kind you should write a program to solve:

Imagine f(n) that counts the occurrences of the digit “1″ in {1,2,3…n}.

Eg f(13)=6

For what values of n does f(n)=n?

That’s all well and good – until you find yourself in a car park, waiting twenty minutes for a taxi. Under these circumstances, it’s reasonable to consider the first few values of f(n)…

Well, f(n)=1 for n=1,2,3,…9. Then you get a lot of changes in a row – f(10)=2, f(11)=4 and so on up to f(19)=12. And, since the next changes are far apart – 21,31,41… – that’s the closest f(n) and n will be for some time.

It’s been ages since the last post. So, here’s a puzzle. And, if you want to compare solutions, look here.