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Number of distinct solutions of sum{i=1..6}(x(2i-1)*x(2i)) = 1 (mod n), with x() in 0..n-1
1

%I #4 Mar 31 2012 12:35:46

%S 0,12,153,1200,7752,37264,158239,550860,1766580,4912624,13019959,

%T 30966523,71311499,150570542,312748103,603515488,1154632937,

%U 2072921406,3718739870,6302870939,10715702976,17307564420,28173570885,43706816100

%N Number of distinct solutions of sum{i=1..6}(x(2i-1)*x(2i)) = 1 (mod n), with x() in 0..n-1

%C Column 6 of A180813

%H R. H. Hardin, <a href="/A180808/b180808.txt">Table of n, a(n) for n=1..183</a>

%e Solutions for sum of products of 6 0..1 pairs = 1 (mod 2) are

%e (0*0 + 0*0 + 0*0 + 0*0 + 0*0 + 1*1) (0*0 + 0*0 + 0*0 + 0*0 + 0*1 + 1*1)

%e (0*0 + 0*0 + 0*0 + 0*1 + 0*1 + 1*1) (0*0 + 0*0 + 0*0 + 1*1 + 1*1 + 1*1)

%e (0*0 + 0*0 + 0*1 + 0*1 + 0*1 + 1*1) (0*0 + 0*0 + 0*1 + 1*1 + 1*1 + 1*1)

%e (0*0 + 0*1 + 0*1 + 0*1 + 0*1 + 1*1) (0*0 + 0*1 + 0*1 + 1*1 + 1*1 + 1*1)

%e (0*0 + 1*1 + 1*1 + 1*1 + 1*1 + 1*1) (0*1 + 0*1 + 0*1 + 0*1 + 0*1 + 1*1)

%e (0*1 + 0*1 + 0*1 + 1*1 + 1*1 + 1*1) (0*1 + 1*1 + 1*1 + 1*1 + 1*1 + 1*1)

%K nonn

%O 1,2

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