In the numbers from 10^(n-1) through 10^n - 1, there are 9 * 10^(n-1) numbers of n digits each, so 9(n-1)10^(n-1) non-leading digits, of which one tenth, or 9(n-1)10^(n-2), are zeroes. When we change the range to 10^(n-1) + 1 through 10^n, we remove 10^(n-1) and put in 10^n, gaining one zero, so

p(n) = p(n-1) + 9(n-1)10^(n-2) + 1 with p(1)=1.

Solving the recurrence yields the closed form

p(n) = n(10^(n-1)+1) - (10^n-1)/9.

For n=6, there are 488,895 zeroes, 600,001 ones, and 600,000 of all other digits.

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