[seqfan] Re: A019303 "Pascal sweep" for k=2: draw a horizontal line ...
Neil Sloane
njasloane at gmail.com
Fri Oct 12 17:18:34 CEST 2018
I don't have a problem with the definition
draw a horizontal line through the 1 at C(2,0) in Pascal's triangle; rotate
this line and record the sum of the numbers on it (excluding the initial 1).
As the line starts rotating, the first time we hit some numbers is when the
line passes through C(2,0) = 1 (which we don't count), C(3,2)=3, and C(4,4)
= 1, total is 3+1 = 4
When we rotate it a bit more, the next event is when we hit C(5,5) = 1, so
we get a total of 1
and so on
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 Fri, Oct 12, 2018 at 10:23 AM Georg.Fischer <georg.fischer at t-online.de>
wrote:
> Hello Seqfans,
>
> this sequence says nothing (Formula, comment, crossref.) but:
>
> A019303 "Pascal sweep" for k=2: draw a horizontal line through the 1 at
> C(k,0) in Pascal's triangle; rotate this line and record the sum of the
> numbers on it (excluding the initial 1). 0
> 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14,
> 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 1, 20, 1, 21, 1, 22, 1, 23, ...
> OFFSET 0,2
>
> It seems to be a subsequence of:
>
> A179820 a(n) = n-th triangular number mod (n+2).
> 0, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1,
> OFFSET 0,3
>
> I wonder whether I'm the only one who does not understand
> the name of A019303. I googled "Pascal sweep" - I seems
> not to be a common term.
>
> Best regards - Georg
>
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>
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