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A180775 Number of distinct solutions of sum{i=1..4}(x(2i-1)*x(2i)) = 0 (mod n), with x() only in 1..n-1 1

%I #3 Mar 31 2012 12:35:46

%S 0,1,3,34,145,522,1518,4041,9150,19970,38555,74370,128040,224434,

%T 358988,587014,876114,1372578,1941624,2912816,4001868,5742391,7599933,

%U 10831065,13788935,18946564,24080514,32270596,39619720,53256875,63655605,83580675

%N Number of distinct solutions of sum{i=1..4}(x(2i-1)*x(2i)) = 0 (mod n), with x() only in 1..n-1

%C Column 4 of A180782

%H R. H. Hardin, <a href="/A180775/b180775.txt">Table of n, a(n) for n=1..380</a>

%e Solutions for sum of products of 4 1..3 pairs = 0 (mod 4) are

%e (1*1 + 1*1 + 1*1 + 1*1) (1*1 + 1*1 + 1*1 + 3*3) (1*1 + 1*1 + 1*2 + 2*2)

%e (1*1 + 1*1 + 1*3 + 1*3) (1*1 + 1*1 + 2*2 + 2*3) (1*1 + 1*1 + 3*3 + 3*3)

%e (1*1 + 1*2 + 1*2 + 1*3) (1*1 + 1*2 + 1*3 + 2*3) (1*1 + 1*2 + 2*2 + 3*3)

%e (1*1 + 1*3 + 1*3 + 3*3) (1*1 + 1*3 + 2*2 + 2*2) (1*1 + 1*3 + 2*3 + 2*3)

%e (1*1 + 2*2 + 2*3 + 3*3) (1*1 + 3*3 + 3*3 + 3*3) (1*2 + 1*2 + 1*2 + 1*2)

%e (1*2 + 1*2 + 1*2 + 2*3) (1*2 + 1*2 + 1*3 + 3*3) (1*2 + 1*2 + 2*2 + 2*2)

%e (1*2 + 1*2 + 2*3 + 2*3) (1*2 + 1*3 + 1*3 + 2*2) (1*2 + 1*3 + 2*3 + 3*3)

%e (1*2 + 2*2 + 2*2 + 2*3) (1*2 + 2*2 + 3*3 + 3*3) (1*2 + 2*3 + 2*3 + 2*3)

%e (1*3 + 1*3 + 1*3 + 1*3) (1*3 + 1*3 + 2*2 + 2*3) (1*3 + 1*3 + 3*3 + 3*3)

%e (1*3 + 2*2 + 2*2 + 3*3) (1*3 + 2*3 + 2*3 + 3*3) (2*2 + 2*2 + 2*2 + 2*2)

%e (2*2 + 2*2 + 2*3 + 2*3) (2*2 + 2*3 + 3*3 + 3*3) (2*3 + 2*3 + 2*3 + 2*3)

%e (3*3 + 3*3 + 3*3 + 3*3)

%K nonn

%O 1,3

%A _R. H. Hardin_ Sep 20 2010

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