[seqfan] Re: n-partitions of sum of powers of 0..n-1

Max Alekseyev maxale at gmail.com
Fri Aug 24 22:30:09 CEST 2012


Ron,
In many rows the last non-1 equals 2.
Can you check what other solution this 2 counts?
Do these other solutions show any pattern across the rows?
Max

On Fri, Aug 24, 2012 at 8:17 PM, Ron Hardin <rhhardin at att.net> wrote:
> Where the infinite row of 1's will begin, vs n?
> (1..n gives a similar problem, if I get to it)
>
> T(n,k)=Number of nondecreasing arrays of 0..n-1 integers with the sum of their k
> powers equal to sum(i^k,i=0..n-1)
>
> Table starts
> ........1......1....1...1.1.1.1.1.1.1.1.1.1.1.1.1.1.1
> ........1......1....1...1.1.1.1.1.1.1.1.1.1.1.1.1.1..
> ........2......1....1...1.1.1.1.1.1.1.1.1.1.1.1.1....
> ........5......1....1...1.1.1.1.1.1.1.1.1.1.1.1......
> .......12......2....1...1.1.1.1.1.1.1.1.1.1.1........
> .......32......4....1...1.1.1.1.1.1.1.1.1.1..........
> .......94.....14....3...2.1.1.1.1.1.1.1..............
> ......289.....37....8...3.1.1.1.1.1.1................
> ......910....105...18...6.1.1.1.1.1..................
> .....2934....309...42..12.1.1.1.1....................
> .....9686....939..100..24.1.2.1......................
> ....32540...2903..265..63.2.2........................
> ...110780...8865..775.164.7..........................
> ...381676..28163.2241.424............................
> ..1328980..90648.6709................................
> ..4669367.297615.....................................
> .16535154............................................
>
> All solutions for n=8 k=3
> ..0....0....1....0....1....2....1....0..
> ..0....1....1....1....4....3....2....0..
> ..2....1....1....2....4....4....2....0..
> ..4....4....2....3....4....4....3....1..
> ..4....5....5....4....5....4....3....2..
> ..6....5....6....5....5....5....3....6..
> ..6....5....6....6....5....6....7....6..
> ..6....7....6....7....6....6....7....7..
>
>
>  rhhardin at mindspring.com
> rhhardin at att.net (either)
>
>
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