[seqfan] Re: id: A132045

[franktaw at netscape.net]

This formula is off by one.  Note that the offset of A132045 is 0, not 
1.  Thus the correct formula is
a(n) = 2^n - n + 1 = A000325(n) + 1,
which holds for n >= 1.

I have submitted a correction.

Franklin T. Adams-Watters

-----Original Message-----
From: avik roy <avik_3.1416 at yahoo.co.in>

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