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%I #5 Nov 16 2016 12:37:14
%S 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,
%T 123,36,5,26,135,454,941,876,345,72,6,33,166,851,2424,4153,3236,953,
%U 144,7,48,243,1116,5101,13270,18143,12284,2613,252,6,57,382,1919,7730,30359,73988
%N 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.
%C Table starts
%C ...2....1......2.......3........4........5.........4.........5..........6
%C ...2....3......6......13.......18.......23........26........33.........48
%C ...4....7.....20......51.......88......135.......166.......243........382
%C ...8...17.....72.....201......454......851......1116......1919.......3162
%C ..12...41....268.....941.....2424.....5101......7730.....14519......26882
%C ..24..123....876....4153....13270....30359.....54842....118709.....249138
%C ..36..345...3236...18143....73988...197949....396488....940083....2280068
%C ..72..953..12284...87353...384360..1246529...2909656...7875465...21006940
%C .144.2613..42396..405423..2126012..8310879..21610932..64293499..195017988
%C .252.7149.159876.1865359.11983288.53948531.162100016.547533077.1823805952
%C 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
%H R. H. Hardin, <a href="/A244788/b244788.txt">Table of n, a(n) for n = 1..9999</a>
%e Some solutions for n=10 k=4
%e ..1....1....1....1....1....1....1....1....1....1....1....1....1....1....1....1
%e ..1....1....1....1....1....1....1....1....1....1....1....1....1....1....1....1
%e ..2....3....2....3....2....3....3....3....4....3....4....3....2....2....4....4
%e ..4....4....4....2....3....2....3....3....1....4....0....1....4....4....3....1
%e ..3....3....2....3....2....1....1....3....1....3....3....4....2....2....1....2
%e ..2....0....3....1....3....4....2....4....4....0....3....2....3....1....1....3
%e ..4....4....3....2....4....3....4....2....0....0....1....4....3....4....4....0
%e ..0....4....4....1....0....0....1....1....1....2....1....2....0....4....1....0
%e ..4....1....0....3....0....0....2....4....3....3....3....4....3....0....4....4
%e ..3....0....0....3....1....4....4....0....3....2....1....2....1....4....3....3
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_, Jul 06 2014
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