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A244903
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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 sqrt(2) standard deviations from its mean
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15
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2, 3, 4, 4, 7, 6, 5, 14, 21, 12, 4, 19, 52, 59, 24, 5, 22, 89, 198, 163, 42, 6, 33, 118, 431, 764, 447, 84, 7, 42, 199, 646, 2009, 2740, 1341, 168, 8, 57, 316, 1299, 3602, 9257, 10484, 3905, 312, 9, 68, 469, 2336, 8695, 20358, 45207, 40696, 11271, 624, 10, 87, 624
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OFFSET
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1,1
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COMMENTS
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Table starts
...2.....3......4.......5........4.........5.........6..........7..........8
...4.....7.....14......19.......22........33........42.........57.........68
...6....21.....52......89......118.......199.......316........469........624
..12....59....198.....431......646......1299......2336.......3949.......5912
..24...163....764....2009.....3602......8695.....17392......33803......55282
..42...447...2740....9257....20358.....56609....130548.....282785.....518428
..84..1341..10484...45207...116272....382743...1019188....2438447....5021116
.168..3905..40696..215411...669516...2617609...7890336...21257775...48258938
.312.11271.159332.1061375..3880216..17614513..61161968..182555691..464345786
.624.32357.627156.5141999.22605350.117799671.475132844.1597877557.4477346412
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<=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=6 k=4
..4....1....4....4....4....1....3....1....3....3....2....3....2....3....2....4
..0....1....1....1....1....4....2....4....1....3....3....1....1....3....0....1
..2....4....0....1....1....1....3....1....1....0....2....0....3....0....4....3
..2....2....0....0....4....0....0....4....4....2....0....3....1....3....4....2
..2....4....4....3....4....4....3....2....3....1....4....0....1....0....4....3
..3....0....0....3....2....1....3....1....0....3....4....2....1....3....1....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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