User talk:Paul Barry

Paul, have you noticed that the Motzkin numbers have a Hankel Transform that's a bisection of the fibonacci numbers. I placed an entry towards the bottom A001006.

A187012

Hello; I noticed a small issue with your formula on the sequence page of A187012:

a(n) = sum{k=0..floor(n/2), 0^(k(n-2k))*(n-k+1)}

This summation gives a(n+2) not a(n). The offset of the sequence is 2 so, for example, a(6)=8 and a(8)=11. But substituting n=6 into your formula gives:

 sum{k=0..floor(n/2), 0^(k(n-2k))*(n-k+1)}
  = sum{k=0..3, 0^(k*(6-2*k))*(6-k+1)}
  = 7 + 0 + 0 + 4 = 11

I believe the formula should be amended to: a(n+2) = Sum_{k=0..floor(n/2)} 0^(k*(n-2*k))*(n-k+1)

Maybe in 2013 when the sequence was added to the OEIS, there was a last-minute change to the offset and the formula was not correctly updated at that time.

Mathew Englander (talk) 22:54, 5 February 2024 (EST)