[seqfan] Re: Simple empirical formulas, proof?

Ron Hardin rhhardin at att.net
Tue May 6 04:18:18 CEST 2014


The corresponding problem with two free elements instead of one has quartic polynomial rows instead of quadratic, and quadratic order column recurrences instead of linear order, so far as I've tested it.

Without nice coefficients on either.:

/tmp/eil
T(n,k)=Number of length n+k+2 0..k arrays with every value 0..k appearing at least once in every consecutive k+3 elements, and new values 0..k introduced in order

Table starts
....7....25.....65.....140.....266.....462......750.....1155.....1705.....2431
...13....61....185.....440.....896....1638.....2766.....4395.....6655.....9691
...24...145....503....1300....2801....5334.....9290....15123....23350....34551
...44...337...1316....3648....8231...16194....28897....47931....75118...112511
...81...781...3398...10012...23486...47466....86381...145443...230647...348771
..149..1829...8801...27368...66366..137166...253674...432331...692113..1054531
..274..4269..23069...75236..187671..395166...740496..1274419..2055676..3150991
..504..9957..60197..208976..533801.1141290..2161503..3749211..6083896..9369751
..927.23233.156887..577964.1530356.3312546..6326951.11042115.18002245.27827211
.1705.54225.408962.1596216.4371836.9669270.18590776.32600811.53341987.82686971

Some.solutions.for.n=5.k=4..
..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0..
..1....1....0....1....1....1....1....1....1....1....1....0....1....0....1....1..
..0....2....1....1....2....2....1....2....2....2....1....0....2....1....0....1..
..2....1....2....2....3....2....0....1....3....3....2....1....3....2....2....2..
..3....3....3....3....0....3....2....3....1....4....3....2....4....0....0....0..
..4....0....0....0....2....2....3....4....0....2....2....3....3....3....3....3..
..1....4....4....4....4....4....4....0....4....0....4....4....2....4....4....4..
..2....1....2....4....1....0....1....2....3....3....0....1....0....1....4....4..
..0....1....0....1....1....1....1....3....3....1....1....2....1....1....1....1..
..3....2....1....2....3....4....0....1....2....1....4....0....3....3....2....2..
..1....0....4....0....2....4....0....1....2....2....4....4....2....2....0....4..

Empirical for column k:
k=1: a(n)=a(n-1)+a(n-2)+a(n-3)
k=2: a(n)=a(n-1)+2*a(n-2)+2*a(n-3)+2*a(n-4)-a(n-5)-a(n-6)
k=3: [order 10]
k=4: [order 15]
Empirical for row n:
n=1: a(n) = (1/8)*n^4 + (11/12)*n^3 + (19/8)*n^2 + (31/12)*n + 1
n=2: a(n) = (5/8)*n^4 + (35/12)*n^3 + (39/8)*n^2 + (43/12)*n + 1
n=3: a(n) = (21/8)*n^4 + (89/12)*n^3 + (67/8)*n^2 + (55/12)*n + 1
n=4: a(n) = (77/8)*n^4 + (179/12)*n^3 + (103/8)*n^2 + (67/12)*n + 1
n=5: a(n) = (261/8)*n^4 + (245/12)*n^3 + (163/8)*n^2 + (79/12)*n + 1
n=6: a(n) = (845/8)*n^4 - (73/12)*n^3 + (343/8)*n^2 + (91/12)*n + 1 for n>1
n=7: a(n) = (2661/8)*n^4 - (2263/12)*n^3 + (1059/8)*n^2 + (103/12)*n + 1 for n>2

Column 1 is A000073(n+5)
Row 1 is A001296(n+1)


 
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rhhardin at att.net (either)




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