[seqfan] Re: wanted: solution to a recurrence
Andrew Weimholt
andrew.weimholt at gmail.com
Thu Oct 21 20:04:51 CEST 2010
On 10/21/10, Marc LeBrun <mlb at well.com> wrote:
>
> What if one requires that all the a(n) are unequal?
>
That's a contradiction with the definition
a(n+1) = a(a(n)+n+1) for n>= 1.
If for any n, we have n+1 =/= a(n)+n+1 then we have (at least) two
terms that are equal.
which contradicts all a(n) being unequal.
On the other hand, If n+1 = a(n)+n+1 for all n, then a(n) must be 0 for all n,
which also contradicts all a(n) being unequal.
Andrew
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