How Can You Calculate 'c' for Given 'n' in This Number Series?

[Big-Daddy]

I hope the title is not too confusing. I couldn't think how to summarize this problem.

If n=2, c=2 (sum of 1+1)
If n=3, c=3 (sum of 1+1+1)
If n=4, c=5 (sum of 1+2+1+1)
If n=5, c=7 (sum of 1+2+2+1+1)
If n=6, c=10 (sum of 1+2+3+2+1+1)

My issue is, what is c in terms of n?

So far I've had an idea:

I could propose c=(n²/4)+1, but now I need some way of removing the extra 0.25 that crops up for all odd values of n. What I need now is a little piece which =-0.25 if n is odd and =0 if n is even.

Since (-1)ⁿ=-1 if n is odd and 0ⁿ=0, I would appreciate a function b of n such that c=(n²/4)+(b)ⁿ*0.25. b would evaluate to -1 if n is odd and 0 if n is even.

Alternatively any solution would be welcomed!

Edit: I've solved it, don't worry. [SOLVED]

 

[Nugatory]

What I need now is a little piece which =-0.25 if n is odd and =0 if n is even.

How about $-(\frac{1+(-1)^{n+1}}{8})$