# [seqfan] Re: A158415

Maximilian Hasler maximilian.hasler at gmail.com
Sun Mar 22 19:35:59 CET 2009

```Sorry for my previous post ; as Richard reminds me, of course I should
have used multiplication (as I did for the other version) and not
So sequence A000992  is correct as it stands,  and the alternative
formula (a(n+1)=b(n) defined in the previous mail) is not surprising
in view of the (additional) initial 1.
So please ignore my previous mail except maybe for the comment
"
Also: number of inequivalent expressions that can be formed with n-1
symbols among {1,f,b}, where 1 is nullary, f is unary, and b is a
function symmetric in its 2 arguments.
"

Apologies,
Maximilian

On Sun, Mar 22, 2009 at 12:49 PM, Maximilian Hasler
<maximilian.hasler at gmail.com> wrote:
> A000992  a(n)= Sum_{k=1 ... floor(n/2)} a(k)a(n-k) with a(1) = 1.
> 1, 1, 1, 2, 3, 6, 11, 24, 47, 103, 214, 481, 1030, 2337, 5131, 11813,
> 26329, ...      OFFSET  1,4
>
> According to the current definition of this sequence, one should have
> a(2) = a(1)+a(1) = 2.
> unless I overlook something. Even adjusting the offset to 0 would not
> correct the problem. I don't understand ...
>
> OTOH I found the values a(n+2) when I calculated the obvious upper
> bound b(n) of Vladimir's sequence, viz
> b(n+1) = b(n) + sum( i=1..n/2, b(i) b(n-i)) ; b(1)=1
>
> So if things are adjusted such that A000992(n) = b(n+1(?)), one could
> "
> Also: number of inequivalent expressions that can be formed with n
> symbols among {1,f,b}, where 1 is nullary, f is unary, and b is a
> function symmetric in its 2 arguments.
> "
>
> Maximilian
>
>
>
> On Sat, Mar 21, 2009 at 7:27 PM, Vladimir Reshetnikov
> <v.reshetnikov at gmail.com> wrote:
>> Dear all,
>>
>> Can anybody suggest an algorithm for calculating the terms of A158415?
>>
>> Thanks