A060488: a(n)=n(n+1)(n+11)/6

[Jonathan Post]

Zak's version reminds us more easily of n(n+1)(n+2)/6 as:
A000292<http://www.research.att.com/~njas/sequences/A000292>
*Tetrahedral* (or pyramidal) numbers: C(n+2,3) = n(n+1)(n+2)/6.
although that can be kept in comments and xrefs.

Best,

Jonathan Vos Post

On 6/15/06, Graeme McRae <g_m at mcraefamily.com> wrote:
>
> It's a good formula if the sequence started at n=1, however that formula
> is
> no good for this sequence because its offset is 3.  You should use
> a(n)=(n-2)(n-1)(n+9)/6
>
> ----- Original Message -----
> From: "zak seidov" <zakseidov at yahoo.com>
> To: <njas at research.att.com>; <seqfan at ext.jussieu.fr>
> Sent: Thursday, June 15, 2006 3:29 AM
> Subject: A060488: a(n)=n(n+1)(n+11)/6
>
>
> > Neil, seqfans,
> > I added (simple) formula,
> > Thanks, Zak
> >
> > %I A060488
> > %S A060488 4,13,28,50,80,119,168,228,300,385,484,598
> > %F A060488 a(n)=n(n+1)(n+11)/6
> > %A A060488 Zak Seidov, Jun 15 2006
> >
> > __________________________________________________
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>
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