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| A244940型 |
| T（n，k）=长度为n 0..k的数组的数量，每个部分和从一开始就不超过其平均值的sqrt（3）个标准偏差 |
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14
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2, 3, 4, 4, 9, 8, 5, 14, 25, 14, 6, 23, 52, 69, 28, 7, 34, 111, 198, 207, 56, 8, 43, 190, 517, 764, 603, 104, 9, 58, 295, 1076, 2529, 2976, 1741, 208, 10, 75, 444, 1939, 6370, 12497, 11668, 5223, 416, 11, 94, 631, 3358, 13139, 37364, 60773, 45960, 15445, 796, 12, 109
(列表;桌子;图表;参考;听;历史;文本;内部格式)
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表格开始
...2.....3......4.......5........6.........7.........8..........9.........10
…4……9……14……23……34……43………58……75……94
...8....25.....52.....111......190.......295.......444........631........896
..14....69....198.....517.....1076......1939......3358.......5405.......8450
..28...207....764....2529.....6370.....13139.....26256......47837......81956
..56...603...2976...12497....37364.....89937....203496.....414005.....802938
.104..1741..11668...60773...219382....619567...1606524....3633503....7812680
.208..5223..45960..293467..1290578...4214681..12610616...32062641...76978472
.416.15445.181652.1452027..7608118..29077603..98974880..284037099..754859772
.796.45423.719784.7098491.44939408.198628937.777361848.2523617923.7399299882
以整数形式计算，使用0..k平均值的6倍和方差的36倍，mean6（k）=3*k；var36（k）=6*k*（2*k+1）-平均值6（k）^2；则（6*sum{x（i），i=1..j}-j*mean6（k））^2<=3*j*var36（k），对于所有j=1..n
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示例
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n=6 k=4的一些解
..4….1….2….0….4….3….0….1….1….3….1….0….4….3….4
..1....3....1....4....0....0....1....3....3....0....4....4....4....1....3....1
..3....1....0....2....4....4....4....2....3....3....3....0....2....2....0....1
..3....3....4....3....3....2....4....1....1....2....2....4....4....1....3....4
..4....1....1....2....3....4....0....1....1....2....3....4....3....0....0....0
..3....3....0....4....3....3....0....0....1....3....0....4....2....1....3....4
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