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%I #4 Apr 15 2015 10:47:44
%S 3,20,113,243,1024,1636,3866,6599,12387,16807,32768,41744,66291,90598,
%T 133205,161051,248832,292932,401910,501113,661703,759375,1048576,
%U 1185856,1507979,1788296,2222649,2476099,3200000,3532100,4287258,4926235,5889323
%N Number of length 5 1..(n+1) arrays with every leading partial sum divisible by 2 or 3
%C Row 5 of A257062
%H R. H. Hardin, <a href="/A257067/b257067.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = a(n-2) +a(n-3) -a(n-5) +4*a(n-6) -4*a(n-8) -4*a(n-9) +4*a(n-11) -6*a(n-12) +6*a(n-14) +6*a(n-15) -6*a(n-17) +4*a(n-18) -4*a(n-20) -4*a(n-21) +4*a(n-23) -a(n-24) +a(n-26) +a(n-27) -a(n-29)
%F Empirical for n mod 6 = 0: a(n) = (32/243)*n^5 + (32/81)*n^4 + (4/9)*n^3 + (1/9)*n^2
%F Empirical for n mod 6 = 1: a(n) = (32/243)*n^5 + (128/243)*n^4 + (487/486)*n^3 + (853/972)*n^2 + (34/243)*n + (313/972)
%F Empirical for n mod 6 = 2: a(n) = (32/243)*n^5 + (112/243)*n^4 + (355/486)*n^3 + (145/486)*n^2 + (125/486)*n + (209/243)
%F Empirical for n mod 6 = 3: a(n) = (32/243)*n^5 + (16/27)*n^4 + (8/9)*n^3 + (8/9)*n^2 + (1/3)*n
%F Empirical for n mod 6 = 4: a(n) = (32/243)*n^5 + (80/243)*n^4 + (80/243)*n^3 + (40/243)*n^2 + (10/243)*n + (1/243)
%F Empirical for n mod 6 = 5: a(n) = (32/243)*n^5 + (160/243)*n^4 + (320/243)*n^3 + (320/243)*n^2 + (160/243)*n + (32/243)
%e Some solutions for n=4
%e ..3....2....2....3....3....2....4....4....2....3....2....4....4....4....3....2
%e ..5....4....1....3....3....1....5....5....4....3....1....4....5....4....1....1
%e ..4....3....1....3....4....5....3....3....3....2....1....2....3....2....2....1
%e ..2....5....2....1....5....2....4....3....1....1....4....4....2....5....2....4
%e ..1....2....2....4....3....5....4....3....5....5....1....2....1....5....2....2
%Y Cf. A257062
%K nonn
%O 1,1
%A _R. H. Hardin_, Apr 15 2015
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