The answer is 18.

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We first remark that the sought expression is:

S = {f(2015)-f(2012)} + {f(2014) - f(2012)} + {f(2013) - f(2012)}

Which leads to understand what is the "slope" of f.

A first important remark is that the local slope belongs to {1,2,3}: for a>0:

3 = f(f(a+1 - a)) >= f(a+1-a) >= f(a+1) - f(a) since f(n) >= n (which is direct with f strict growth) hence f(a+1)-f(a) (>0) belongs to {1,2,3}.

We then at patterns for the first integers, it appears:

Hence f(1) = 2

Hence f(2) = 6

Hence f(6) = 18 ....

By recursion we can show that, for n>=0: f(3^n) = 2x3^n f(2x3^n) = 3^(n+1)

and then remark that:

in an interval [3^n, 2x3^n], f has a slope of 1 (f(p)-f(q) = p-q for all integers in this interval), since: f(2x3^n) - f(3^n) = 3^(n+1) - 2x3^n = 3^n = (2x3^n) - (3^n)

in an interval [2x3^n, 3^(n+1)], f has a slope of 3 (f(p)-f(q) = 3x(p-q) for all integers in this interval), since: f(3^(n+1)) - f(2x3^n) = 2x3^(n+1) - 3^(n+1) = 3^(n+1) = 3x { (3^(n+1)) - (2x3^n)} and the local slope can not exceed 3: f(p+1)-f(p) <= 3

Finally, 2012, 2013, 2014, 2015 all lie in the second type of intervall, meaning that

f(2015) - f(2012) = 3x3

f(2014) - f(2012) = 3x2

f(2013) - f(2012) = 3

S = 18