[seqfan] A115491

[israel at math.ubc.ca]

I see some changes to A115491 have just been published.  There are now two
competing formulas for a(n):

a(n) = -(2/5)*2^n+(32/5)*32^n, with n>=0. - _Paolo P. Lava_, Jun 17 2008 

a(n) = (32^(n+1)-16*2^(n+1))/160. - _Vincenzo Librandi_, Jul 25 2014 

Vincenzo's agrees with the data (note that a(1) = 6), but
Paolo's is for a(n+1), not a(n): he has a(1) = 204. 
So, what to do?  

Cheers,
Robert