EDIT: misunderstood question.

Before a solution, I think we can explain better the condition of the problem because it is not very clear... The idea is: We are given a finite sequence of positive integers. The sequence contains in total $n$ different integers appearing on it, also, if we substract 1 from any of its terms, we get a new sequence which has at least $n$ different integers on it. Whats the minimun sum of the elements of this sequence? The answer is $n=2k$, then $f(n)=3k^2+2k-1$, $n=2k+1$, then $f(n)=3k^2+5k+1$. Suppose we have an arbitrary sequence that does the job, and suppose an integer appears more than 2 times on it, clearly, leaving just 2 appearences of this integer, we get a new sequence which does the job, but with smaller sum. Hence in an optimal sequence we have every integer appearing at most 2 times. Now, if an integer $m$ appears 2 times in the sequence, but $m-1$ doesnt appears, then, removing one $m$ we decrease our sum, and get a sequence that also works (this is trivial ) . So, suppose we have a working sequence with all transformations we described, and let $a_j$ be the number of $j$ in the sequence. we get $S=1a_1+2a_2+3a_3...+ma_m$, now, this sequence will look like subsequences of consecutive terms appearing each two times, conected by sequences of terms with difference 2 among one and the next. (the smallest term claerly appears only once) for example, for $n=6$ one possible sequence is: $1,2,2,4,5,5,7,9$ with sum $35$. $1,3,5,7,9,11$ with sum $36$. Now, consider the sequence $(a_1,a_2...,a_m)$, it is build up of blocks that can be like: $1,2,2...,2$ and $0,1,0,1,0...,1,0$ that alternate. Call them $A$ and $B$ blocks. Let $A(x,y)$ and $B(x,y)$ be respectively the sum that $A$ and $B$ sequences give, starting in $x$ and finishing in $y-1$. We then have: $S=A(1,n_1)+B(n_1,n_2)+A(n_2,n_3)...$ But we also have that the number of terms in an $A$ sequence starting in $x$ and finishing in $y-1$ is $y-x$, and similary, in a $B$ sequence we have $\frac{y-x-1}{2}$ integers. The idea is: find expresions for $A(x,y)$ and $B(x,y)$, and finish with an inequality, I will finish it later I will finish it in a moment, gotta go to college

mmmm, new idea, same escenario.... Suppose we have an integer $m$ appearing once, and such that it is less than the middle of $n$. Then change it for $m-1,m-1$, and substract one form every term bigger than $m$, we get a new sequence and the sum decreases, so its not minimal. So in the minimal sum, all terms smaller than or equal to $\frac{n}{2}$ appear exactly 2 times (but 1 only once). Now, let $k$ be the biggest number, which is bigger than the half of $n$ and appears twice, changing it to $m+1$, and adding $1$ to every bigger term, we get a new sequence with smaller sum. So the sequence with smallest sum must be: $1,2,2,3,3...,k,k,k+2,k+4...,3k-2,3k$ for $n=2k$ and $1,2,2,3,3...,k,k,k+2,k+4...,3k,3k,3k+2$ for $n=2k+1$.

$1,2,2,4,5,5,7,9$ with sum $35$. $1,3,5,7,9,11$ with sum $36$ Make 1 become 0 and it doens't work