%I#9 2021年10月1日17:32:43

%S 2,3,4,4,9,8,5,16,25,16,6,25,62,75,30,7,3417242219,60,8,47200，

%电话：573950657120，9，6231511622833741947230，10，7948221876916，

%电话：13785148025737460,11,9867937921485940894683115866216835920,12115

%N 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与其平均值相差不超过两个标准偏差。

%C表启动

%C。。。2.....3......4.......5........6.........7.........8..........9.........10

%C。。。4.....9.....16......25.......34........47........62.........79.........98

%C。。。8....25.....62.....117......200.......315.......482........679........948

%C。。16....75....242.....573.....1162......2187......3792.......6015.......9262

%C。。30...219....950....2833.....6916.....14859.....29588......53105......91096

%C。。60...657...3744...13785....40894....102301....234128.....469965.....899962

%C.120.1947..14802…68311…241778…707567…1843104…4169823…8845276

%C.230.5737..58662.339830672…4909019..14502774…37082231…87596238

%抄送460.16835.232916.1677913.8475456.34131599.115222470.330409737.864410260

%C.920.50505.926120.8250991.50269530.235879457.912575878.2948726733.8580741682

%C以整数形式计算，使用6倍0..k平均值和36倍方差，平均值6（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<=4*j*var36（k），对于所有j=1..n。

%H R.H.Hardin，n表，n=1..9999的a（n）</a>

%e n=6 k=4的一些解

%e。。2....3....0....4....3....2....3....0....4....1....4....0....4....0....1....3

%e。。4....4....4....2....2....1....4....3....3....3....3....4....3....2....0....2

%e。。3....1....4....1....1....3....0....1....0....0....3....0....1....3....1....4

%e。。0....2....3....4....2....2....1....4....2....1....3....1....0....3....1....1

%e。。2....1....1....0....2....1....1....2....4....2....3....0....2....3....3....3

%e。。3....1....2....3....2....2....1....4....2....0....0....4....1....1....0....0

%Y行1是A000027（n+1）。

%K nonn，表

%O 1,1号机组

%A R.H.Hardin，2014年7月6日