[seqfan] Re: Filling a triangle with numbers
Ron Hardin
rhhardin at att.net
Fri Jul 6 16:18:28 CEST 2012
I get 12 for a(4)
.....5........6........6........5........5........3........3........4..
....3.7......3.7......2.7......2.7......3.6......2.6......2.7......3.7..
...1.4.8....1.4.8....1.4.8....1.4.8....1.4.8....1.5.8....1.5.8....1.5.8..
..0.2.6.9..0.2.5.9..0.3.5.9..0.3.6.9..0.2.7.9..0.4.7.9..0.4.6.9..0.2.6.9..
..
.....4........4........5........4..
....2.6......2.7......2.6......3.6..
...1.5.8....1.5.8....1.4.8....1.5.8..
..0.3.7.9..0.3.6.9..0.3.7.9..0.2.7.9..
rhhardin at mindspring.com
rhhardin at att.net (either)
----- Original Message ----
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Fri, July 6, 2012 9:51:14 AM
> Subject: [seqfan] Filling a triangle with numbers
>
> I heard this problem in Hungary recently:
>
> How many ways are there to arrange the numbers 1 through n(n+1)/2 in
> a triangle so that each term is between the two numbers
> immediately below it (and each row is increasing)?
>
> Examples:
> .2.
> 1.3
>
> ..4..
> .2.5.
> 1.3.6
>
> ..3..
> .2.5.
> 1.4.6
>
> I get a(1)=a(2)=1, a(3)=2, a(4)=6?
>
> The only related sequences I can find in the OEIS are A064049, A064059,
> which are different. However, it is quite likely this is a well-studied
> problem.
>
> Also, what if one omits the clause in parentheses in the definition?
>
> Neil
>
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