Base case N=1, this can be writen 3\times 0+1, so the proposition holds for N=1.

Suppose it true for N=k, then there is an n such that:

k=3n, or 3n+1 or 2n+3

Suppose the first of these is the case, then:

k+1=3n+1 and the proposition holds,

suppose the second is the case:

k+1=(3n+1)+1=3n+2, and the proposition holds,

suppose the third is the case:

k+1=(3n+2)+1=3(n+1), and the proposition holds.

Hence the proposition holds for k+1 whatever the case, and the induction goes through

RonL