[seqfan] R: Re: A074506

vincenzo.librandi at tin.it vincenzo.librandi at tin.it
Sun Jul 18 20:17:49 CEST 2010

Non cambia nulla, quindi anche se 
a(n)=7*a(n-1)-12*a(n-2)+6 with a(0)=3, a(1)=8
1^n + 3^n + 4^n:a^n+b^n+1^n=(a+b)*a(n-1)-(a*b)*a(n-2)+(a-1)*(b-1)
Best RegardsVincenzo

----Messaggio originale----
Da: mathar at strw.leidenuniv.nl
Data: 18-lug-2010 7.08 PM
A: <vincenzo.librandi at tin.it>
Ogg: Re:  A074506

vl> Return-Path: <seqfan-bounces at list.seqfan.eu>
vl> X-Original-To: mathar at strw.leidenuniv.nl
vl> Delivered-To: mathar at strw.leidenuniv.nl
vl> Date: Sun, 18 Jul 2010 08:00:13 +0200 (CEST)
vl> From: "vincenzo.librandi at tin.it" <vincenzo.librandi at tin.it>
vl> To: seqfan at seqfan.eu
vl> For the controll:
vl>  %F A074506 a(n)=7*a(n-1)-12*a(n-2)+6 with a(1)=3, a(2)=8 [From Vincenzo
vl> Librandi (vincenzo.librandi(AT)tin.it), Jul 16 2010]

What I mean is:
This is incorrect because a(0)=3, a(1)=8.

The formula a(n)=7*a(n-1)-12*a(n-2)+6 is correct (and the current
formula with the 7 should be changed).


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