[seqfan] id: A132045

[avik roy]

I've suggested a general formula for generating terms in A132045, it is given by a(n+1)=2^n-n+1; n>=1
the proof is given here:

It is easy to observe, the rows of A132044 are simply produced by keeping the 1st and last 1's untouched in Pascal's triangle and by reducing the intermediate terms by 1.

Thus the sum of (n+1)th rows is given by, 
1 + [C(n,1) + C(n,2) + C(n,3) + ... + C(n-1)] - (n-1) + 1
= 2+2^n-2-n+1=2^n-n+1




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