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Number of arrangements of n+3 numbers in 0..4 with each number being the sum mod 5 of three others
1

%I #5 Mar 31 2012 12:35:55

%S 1,241,8681,63645,363749,1905097,9681073,48678653,243871365,

%T 1220204441,6102563585,30515709437,152584140357,762931805705,

%U 3814681507825,19073453656861,95367363682085,476837016657945,2384185496162945

%N Number of arrangements of n+3 numbers in 0..4 with each number being the sum mod 5 of three others

%C Column 4 of A183892

%H R. H. Hardin, <a href="/A183887/b183887.txt">Table of n, a(n) for n = 1..62</a>

%e Some solutions for n=2

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

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

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

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

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

%K nonn

%O 1,2

%A _R. H. Hardin_ Jan 07 2011