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Number of length n 0..5 arrays with each partial sum starting from the beginning no more than one standard deviation from its mean.
1

%I #5 Nov 16 2016 12:37:14

%S 4,18,88,454,2424,13270,73988,384360,2126012,11983288,68380920,

%T 371177576,2103489988,12072527960,69865022356,389690375308,

%U 2242267086352,13018490319724,73199231868324,423255665348852,2467041278509640

%N Number of length n 0..5 arrays with each partial sum starting from the beginning no more than one standard deviation from its mean.

%C Column 5 of A244788

%H R. H. Hardin, <a href="/A244785/b244785.txt">Table of n, a(n) for n = 1..210</a>

%e Some solutions for n=10

%e ..3....3....3....3....3....3....3....3....3....3....3....3....3....3....3....3

%e ..0....0....0....3....3....0....0....0....3....0....0....3....3....0....3....0

%e ..3....3....3....0....3....3....3....3....0....3....3....0....0....3....0....3

%e ..2....2....2....1....2....2....2....1....2....2....1....2....2....2....2....2

%e ..5....3....5....5....2....3....3....3....3....2....3....3....5....3....4....5

%e ..3....0....3....2....0....1....5....1....4....4....3....5....2....1....0....3

%e ..1....3....2....2....3....2....3....3....0....3....2....1....0....5....4....5

%e ..3....4....5....1....4....2....5....2....3....2....5....0....2....5....3....2

%e ..2....2....3....3....1....3....2....5....4....2....0....3....2....0....0....2

%e ..1....1....1....4....1....5....3....5....4....0....0....0....5....0....5....5

%K nonn

%O 1,1

%A _R. H. Hardin_, Jul 06 2014