[seqfan] Re: A019575 challenge

[Robert Gerbicz]

2010/8/23 Ron Hardin <rhhardin at att.net>

> Is this "Empirical" or is there an obvious reason
>
> n indistinguishable balls into n boxes
> T(n,k) = ways to get exactly k balls in the most-filled box
> table starts
>
> 1
> 1 2
> 1 6 3
> 1 18 12 4
> 1 50 50 20 5
> 1 140 195 90 30 6
> 1 392 735 392 147 42 7
> 1 1106 2716 1652 672 224 56 8
> 1 3138 9912 6804 2970 1080 324 72 9
> 1 8952 35850 27600 12825 4950 1650 450 90 10
> 1 25652 128865 110715 54450 22022 7865 2420 605 110 11
> 1 73788 461175 440374 228294 96030 36036 12012 3432 792 132 12
> 1 212940 1645215 1740024 948090 412698 160888 56784 17745 4732 1014 156 13
> 1 616226 5855941 6838832 3907995 1754116 705341 259896 86632 25480 6370
> 1274 182
> 14
> 1 1787606 20810153 26762645 16011905 7391475 3050775 1162800 406980 128520
> 35700
> 8400 1575 210 15
>
> take the series a(n) = T(n,n-k) of the kth term from the right end of each
> line.
>
> Empirical: The (k+1)th difference of T(n,n-k) equals k+1 for n > 2k.
>
> (I'd guess the possibility of two equally filled max boxes ruins the
> counting)
>
> see http://rhhardin.home.mindspring.com/current4.txt for the full b-file
>
>
>
>  rhhardin at mindspring.com
> rhhardin at att.net (either)
>
>
>
> ----- Original Message ----
> > From: Ron Hardin <rhhardin at att.net>
> > To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> > Sent: Thu, August 19, 2010 9:12:59 PM
> > Subject: [seqfan] Re: A019575 challenge
> >
> > Curiously the number of NON-distinguishable balls problem seems not to be
> in
> > OEIS
> > (either that or I have a bug in computing it)
> >
> > so far,  triangle  tabl
> >
> > 1,1,2,1,6,3,1,18,12,4,1,50,50,20,5,1,140,195,90,30,6,1,392,735,392,147,
> > 42,7,1,1106,2716,1652,672,224,56,8,1,3138,9912,6804,2970,1080,324,72,9,
> > 1,8952,35850,27600,12825,4950,1650,450,90,10,1,25652,128865,110715
> > T(n,k)=Number  of arrangements of n indistinguishable balls in n boxes
> with the
> >
> > maximum  number of balls in any box equal to k
> > Table starts
> > 1
> > 1 2
> > 1 6 3
> > 1  18 12 4
> > 1 50 50 20 5
> > 1 140 195 90 30 6
> > Table of n, a(n) for  n=1..423
> >
> > current and irregularly updated b-file at
> > http://rhhardin.home.mindspring.com/current4.txt
> >
> >
> > rhhardin at mindspring.com
> > rhhardin at att.net (either)
> >
> >
> >
> > _______________________________________________
> >
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> >
>
>
>
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>
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>

That suggests that there can be a polynom for each k value, and in fact, my
conjecture is that: T(n,k)=n*binomial(2*n-k-2,n-2) if 2*k>n, so for your
a(n)=T(n,n-k)=n*binomial(n+k-2,k)=n^(k+1)/k!+O(n^k) (it is true, if n>2*k),
that gives for the (k+1)-th difference sequence: k+1, what we needed.