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A257066 Number of length 4 1..(n+1) arrays with every leading partial sum divisible by 2 or 3.

%I A257066 #6 Jun 02 2025 11:46:35
%S A257066 2,11,45,81,256,364,738,1149,1905,2401,4096,4912,7172,9297,12685,
%T A257066 14641,20736,23436,30344,36455,45633,50625,65536,71872,87438,100767,
%U A257066 120141,130321,160000,172300,201782,226521,261745,279841,331776,352944,402848
%N A257066 Number of length 4 1..(n+1) arrays with every leading partial sum divisible by 2 or 3.
%C A257066 Row 4 of A257062
%H A257066 R. H. Hardin, <a href="/A257066/b257066.txt">Table of n, a(n) for n = 1..210</a>
%F A257066 Empirical: a(n) = a(n-1) +4*a(n-6) -4*a(n-7) -6*a(n-12) +6*a(n-13) +4*a(n-18) -4*a(n-19) -a(n-24) +a(n-25)
%F A257066 Empirical for n mod 6 = 0: a(n) = (16/81)*n^4 + (4/9)*n^3 + (1/3)*n^2
%F A257066 Empirical for n mod 6 = 1: a(n) = (16/81)*n^4 + (50/81)*n^3 + (107/108)*n^2 + (44/81)*n - (113/324)
%F A257066 Empirical for n mod 6 = 2: a(n) = (16/81)*n^4 + (46/81)*n^3 + (83/108)*n^2 - (7/162)*n + (25/81)
%F A257066 Empirical for n mod 6 = 3: a(n) = (16/81)*n^4 + (20/27)*n^3 + (7/9)*n^2 + (2/3)*n
%F A257066 Empirical for n mod 6 = 4: a(n) = (16/81)*n^4 + (32/81)*n^3 + (8/27)*n^2 + (8/81)*n + (1/81)
%F A257066 Empirical for n mod 6 = 5: a(n) = (16/81)*n^4 + (64/81)*n^3 + (32/27)*n^2 + (64/81)*n + (16/81)
%e A257066 Some solutions for n=4
%e A257066 ..3....4....2....4....3....3....3....4....4....3....2....2....3....3....4....4
%e A257066 ..5....5....2....4....3....5....5....5....5....1....1....2....5....1....5....4
%e A257066 ..2....1....5....2....2....1....1....5....3....2....3....2....2....2....3....4
%e A257066 ..4....2....1....2....2....3....1....2....4....2....2....4....5....3....3....2
%Y A257066 Cf. A257062
%K A257066 nonn
%O A257066 1,1
%A A257066 _R. H. Hardin_, Apr 15 2015