# [seqfan] Re: sequence based on A050873 triangular array for GCD(u, v)

Bob Selcoe rselcoe at entouchonline.net
Mon Jul 31 23:35:00 CEST 2017

```Hi Jamie & Seqfans,

I gather a(n) is the number of rows it takes for a repeating pattern to
occur, so surely these two conditions are true:

i.  a(n) must be a multiple of n.
ii.  when prime(k) < n < prime(k+1), then primorial(k) < a(n) <
primorial(k+1).

Are you sure a(9) = 612??  a(n) satisfies the first two conditions, but I
get this result:

row
9      1  1  3  1  1  3  1  1  9
10    1  2  1  2  5  2  1  2  1
...
621  1  1  3  1  1  3  1  1  9
622  1  2  1  2  1  2  1  2  1

If I understand correctly, if a(n) = 612, then row 10 should be equivalent
to row 622.

Cheers,
Bob Selcoe

--------------------------------------------------
From: "Olivier Gerard" <olivier.gerard at gmail.com>
Sent: Monday, July 31, 2017 4:04 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: sequence based on A050873 triangular array for GCD(u,
v)

> I resend Jamie Morken's message at his suggestion
> hoping it will appear this time will all the necessary
> line feeds.
>
> ====================
>
>
> Hi,
>
> For the sequence A050873, I'd like to propose a sequence based on it,
> but am not sure what the formula for the sequence is, and need help to
> find some more terms.
>
> Here is the sequence so far:
>
> n a(n)
> 1 1
> 2 2
> 3 6
> 4 12
> 5 30
> 6 60
> 7 210
> 8 840
> 9 612
> 10 420
> 11 2310
>
> For prime n, a(n) is a Primorial number.
>
> Example of how to construct the sequence from A050873, for n = 3
>
> Use the first 3 columns of A050873 starting at row 3 and identify the
> count
> of consecutive rows until a repeating pattern occurs.
>
> 1 1 3
> 1 2 1
> 1 1 1
> 1 2 3
> 1 1 1
> 1 2 1
>
> The first three columns of A050873 starting at the third row is a
> repetition of those 6 rows, so the next row is 1 1 3.
>
> Example for n = 11
>
> Use the first 11 columns of A050873 starting at row 11:
>
> 1 1 1 1 1 1 1 1 1 1 11
> 1 2 3 4 1 6 1 4 3 2 1
>
> 1 1 1 1 1 1 1 1 1 1 1
> 1 2 1 2 1 2 7 2 1 2 1
> ...
>
> It takes 2310+1 rows before the patterns shows up and then the first
> eleven
> columns starting
> at the eleventh row is a repetition of those 2310 rows.
>
> The prime rows have a Pn row count for the repetition, but calculating the
> non prime column
> row counts like above seems harder, I don't know if there is a formula for
> it like for the prime
> columns row counts, but maybe someone can find it or could calculate more
> values of the
> sequence.  Thanks.
>
> cheers,
> Jamie
>
> =======
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>

```

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