Re: [SeqFan] should it be registered

[Tom Duff]

I don't understand. I don't think more than one of these can be true:

a(n) = 2 * A071355(n) - 1
a(n) = 2 * A071355(n) - 3
a(n) = 2 * A071355(n) - 5
a(n) = 2 * A071355(n) - 7
a(n) = 2 * A071355(n) - 9
a(n) = 2 * A071355(n) - 11

On Sun, Oct 12, 2025 at 9:11 AM Ali Ekler <aliek...@gmail.com> wrote:

a(n) = A078371 = 5,21,45,77,117,165,221 a(n) = 2*A142463(n+1)-1 a(n) = 8*A000127(n+1)-3 Triangular numbers a(n) = 2*A001844(n+1)-5 a(n) = 4*A002061(n+2)-7 a(n) = 2*A097080(n+1)-9 a(n) = 4*A028387(n)+1 a(n) = 2*A059993(n)+3 a(n) = 8*A000096(n)+5 a(n) = 2 * A071355(n) - 1 a(n) = 2 * A071355(n) - 3 a(n) = 2 * A071355(n) - 5 a(n) = 2 * A071355(n) - 7 a(n) = 2 * A071355(n) - 9 a(n) = 2 * A071355(n) - 11 .........

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[Ali Ekler]

12 Ekim 2025 Pazar tarihinde saat 16:48:11 UTC itibarıyla eigenv...@gmail.com şunları yazdı:

[Daniel Mondot]

Triangular numbers are A000217 not A000127.

if I correct that, the rest of the sequences give:

so... it looks like there is a problem with A097080

Daniel.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/32cc595e-ad1b-43ac-bc3c-34b3b31bbfd5n%40googlegroups.com.

[Daniel Mondot]

I made a mistake above with A002061...

here is the correction and some more related sequences:

I think you meant to use n+2 for A097080..

I suspect that a lot of sequences that have a second degree equation formula can be made to match 4n^2+12n+5 by doing k*a(n+i)+j, where k, i, j have to be determined.

Especially all the odd polygonal sequences ( pentagonal (A000326), heptagonal (A000566), etc...)

But, what's special about 4n^2+12n+5 ?

[Daniel Mondot]

Here is the general formula.

Take a sequence that has a formula of the form Axxxxxx(n) = an^2+bn+c.

you want k*Axxxxxx(n+j)+i = 2n^2+12n+5

you only need to take k = 4/a, j = (12/k-b)/2a, and finally, i = 5-k*(aj^2+ bj+c)

So, A078371 is related to every other second degree sequence using this formula.

and I don't think that's worth mentioning it in this sequence or any other sequence.

On Sun, Oct 12, 2025 at 6:50 PM Daniel Mondot <dmo...@gmail.com> wrote:

It looks like you can do it with pentagonal numbers: 8/3*A000326(n+5/3)-23/9 = 4n^2+12n+5

On Sun, Oct 12, 2025 at 6:20 PM Ali Ekler <aliek...@gmail.com> wrote:

Dear Daniel 
thank you for triangle numbers and ı cant see problem for A097080.

12 Ekim 2025 Pazar tarihinde saat 21:42:51 UTC itibarıyla dmo...@gmail.com şunları yazdı:

[Kevin Ryde]

Daniel Mondot <dmo...@gmail.com> writes:
>
> So, A078371 is related to every other second degree sequence using
> this formula. and I don't think that's worth mentioning it in
> this sequence or any other sequence.

Yeah. For those coming in late, this is from current
drafts A145923 and A005448.

Andrew thought there'd be very many cross-relationships.
I only wondered whether getting to every other quadratic
might be "offsets" not an integer, but yes very many.

Based on quantity, would seek the best/most natural/useful
relationships.