# [seqfan] Re: A090864 == A118300

Charles Greathouse charles.greathouse at case.edu
Tue Dec 13 17:47:02 CET 2011

```> A090864 and A118300 agree for the terms given for both sequences.  I believe
> them to be the same due to the comment from Perry in the related sequence
> A006906 that A006906 has the same parity as A000009.
>
> https://oeis.org/search?q=3%2C+4%2C+6%2C+8%2C+9%2C+10%2C+11%2C+13%2C+14%2C+1
> 6&sort=&language=english
>
> Any objections to merging the two sequences into A090864 and "killing"
> A118300?

It's clear that A006906 and A000009 have the same parity.  But I don't
see how that makes A090864 and A118300 the same.

If you're sure that they are, A118300 should be given keyword:dead and
name "Duplicate of A090864." once the information has been added to
A090864: the definition of A118300 and probably a capsule proof that
they're the same.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Tue, Dec 13, 2011 at 11:32 AM, Ray Chandler
<rayjchandler at sbcglobal.net> wrote:
> A090864 and A118300 agree for the terms given for both sequences.  I believe
> them to be the same due to the comment from Perry in the related sequence
> A006906 that A006906 has the same parity as A000009.
>
> https://oeis.org/search?q=3%2C+4%2C+6%2C+8%2C+9%2C+10%2C+11%2C+13%2C+14%2C+1
> 6&sort=&language=english
>
> Any objections to merging the two sequences into A090864 and "killing"
> A118300?
>
> In a related matter, the first formula entry for sequence A010815:
> a(n) = (-1)^m if n is of the form m(3m+-1)/2; otherwise a(n)=0. These values
> of n are the pentagonal numbers, A000326.
>
> https://oeis.org/A010815
>
> It is not clear to what "these values" refers in the second sentence, but
> assuming the reference is to values of n such that a(n)=+-1, then the
> resulting sequence would be A001318, generalized pentagonal numbers, rather
> than A000326.  The values of n such that a(n)=0 results in the complement of
> the generalized pentagonal numbers, i.e. the surviving sequence in the
> discussion above.
>
> Ray Chandler
>
>
>
>
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>
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