[seqfan] Re: A161527 and A254196

[Jamie Morken]

Hi Seqfans,
I tried to research this and think the two sequences may be connectedwith this formula:1/((x(n)/A161527(n)/A038110(n+1))+1)=A161527(n)/A060753(n+1),
 where x={1,1,16,64,256,36864,9437184,...}
 ie for n =3: 
1/((16/A161527(3)/A038110(4))+1)=A161527(3)/A060753(4)=1/((16/11/4)+1)=11/15.

  For n=4: 
1/((64/A161527(4)/A038110(5))+1)=A161527(4)/A060753(5)=1/((64/27/8)+1)=27/35.Sequence x is actually interesting so I think it is worth submitting sequence x to OEIS and merging the two sequences A161527 and A254196.cheers,Jamie
----- Original Message -----
From: Neil Sloane <njasloane at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Mon, 05 Aug 2019 09:38:53 -0600 (MDT)
Subject: [seqfan] Re: A161527 and A254196

Michel, good catch! No one seems to have noticed this before. But the
corresponding denominators
of the two sequences are different, A038110 vs A060753. I've added more
cross-references
between these sequences.

I think someone should do some research here and figure our exactly what is
going on before we merge them. I've added this to my list of "sequences
that need work".

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com



On Mon, Aug 5, 2019 at 5:23 AM <michel.marcus at free.fr> wrote:

> Dear All,
>
>
> A254196 appears to be a duplicate of A161527.
> Right ?
>
>
> Best.
>
>
> MM
>
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> Seqfan Mailing list - http://list.seqfan.eu/
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