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A244788
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T(n,k)=Number of length n 0..k arrays with each partial sum starting from the beginning no more than one standard deviation from its mean.
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15
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2, 1, 2, 2, 3, 4, 3, 6, 7, 8, 4, 13, 20, 17, 12, 5, 18, 51, 72, 41, 24, 4, 23, 88, 201, 268, 123, 36, 5, 26, 135, 454, 941, 876, 345, 72, 6, 33, 166, 851, 2424, 4153, 3236, 953, 144, 7, 48, 243, 1116, 5101, 13270, 18143, 12284, 2613, 252, 6, 57, 382, 1919, 7730, 30359, 73988
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OFFSET
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1,1
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COMMENTS
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Table starts
...2....1......2.......3........4........5.........4.........5..........6
...2....3......6......13.......18.......23........26........33.........48
...4....7.....20......51.......88......135.......166.......243........382
...8...17.....72.....201......454......851......1116......1919.......3162
..12...41....268.....941.....2424.....5101......7730.....14519......26882
..24..123....876....4153....13270....30359.....54842....118709.....249138
..36..345...3236...18143....73988...197949....396488....940083....2280068
..72..953..12284...87353...384360..1246529...2909656...7875465...21006940
.144.2613..42396..405423..2126012..8310879..21610932..64293499..195017988
.252.7149.159876.1865359.11983288.53948531.162100016.547533077.1823805952
Computation in integer form, using 6 times the 0..k mean and 36 times the variance, mean6(k)=3*k; var36(k)=6*k*(2*k+1)-mean6(k)^2; then (6*sum{x(i),i=1..j}-j*mean6(k))^2<=j*var36(k) for all j=1..n
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LINKS
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EXAMPLE
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Some solutions for n=10 k=4
..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....3....2....3....2....3....3....3....4....3....4....3....2....2....4....4
..4....4....4....2....3....2....3....3....1....4....0....1....4....4....3....1
..3....3....2....3....2....1....1....3....1....3....3....4....2....2....1....2
..2....0....3....1....3....4....2....4....4....0....3....2....3....1....1....3
..4....4....3....2....4....3....4....2....0....0....1....4....3....4....4....0
..0....4....4....1....0....0....1....1....1....2....1....2....0....4....1....0
..4....1....0....3....0....0....2....4....3....3....3....4....3....0....4....4
..3....0....0....3....1....4....4....0....3....2....1....2....1....4....3....3
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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