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Number of distinct solutions of sum{i=1..2}(x(2i-1)*x(2i)) = 0 (mod n), with x() in 0..n-1
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%I #4 Mar 31 2012 12:35:46

%S 1,4,8,20,28,56,64,122,150,218,216,424,346,544,667,863,733,1305,1000,

%T 1752,1715,1944,1728,3258,2535,3142,3495,4520,3382,6254,4096,6486,

%U 6243,6812,7315,10959,6868,9400,10121,13922,9271,16388,10648,16520,17805

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

%C Column 2 of A180803

%H R. H. Hardin, <a href="/A180794/b180794.txt">Table of n, a(n) for n=1..999</a>

%e Solutions for sum of products of 2 0..4 pairs = 0 (mod 5) are

%e (0*0 + 0*0) (0*0 + 0*1) (0*0 + 0*2) (0*0 + 0*3) (0*0 + 0*4) (0*1 + 0*1)

%e (0*1 + 0*2) (0*1 + 0*3) (0*1 + 0*4) (0*2 + 0*2) (0*2 + 0*3) (0*2 + 0*4)

%e (0*3 + 0*3) (0*3 + 0*4) (0*4 + 0*4) (1*1 + 1*4) (1*1 + 2*2) (1*1 + 3*3)

%e (1*2 + 1*3) (1*2 + 2*4) (1*3 + 3*4) (1*4 + 2*3) (1*4 + 4*4) (2*2 + 2*3)

%e (2*2 + 4*4) (2*3 + 3*3) (2*4 + 3*4) (3*3 + 4*4)

%K nonn

%O 1,2

%A _R. H. Hardin_, suggested by _Max Alekseyev_ in the Sequence Fans Mailing List, Sep 20 2010