[seqfan] Re: A115491
Bruno Berselli
berselli.bruno at yahoo.it
Fri Jul 25 10:06:14 CEST 2014
Robert, Paolo has stated that his formula is valid for n>= 0. In my opinion we can keep both versions (this has happened many times in OEIS).
Bruno
Il Venerdì 25 Luglio 2014 9:44, "israel at math.ubc.ca" <israel at math.ubc.ca> ha scritto:
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
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