number of surjective functions f:[k]→[n] :
N=i=0∑n(ni)(−1)i(n−i)k
in our case:
N=i=0∑n(3i)(−1)i(3−i)n=3n−3⋅2n+3
minimum number of function such that f(1) > f(2) is 0
maximum number of function such that f(1) > f(2) is 3:
f(2)=1,f(2)=2,f(1)=3
so, θ estimate for an :
N−3≤an≤N
3n−3⋅2n<an<3n−3⋅2n+3
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