
#1
Apr2713, 09:21 AM

P: 339

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=(n^{2}/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 0^{n}=0, I would appreciate a function b of n such that c=(n^{2}/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] 



#2
Apr2713, 09:40 AM

Sci Advisor
Thanks
P: 2,963




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