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%I #33 Sep 26 2020 10:38:50
%S 0,1,2,2,3,2,4,4,4,3,6,4,7,4,6,6,9,4,10,6,8,6,12,8,11,7,10,8,15,6,16,
%T 10,12,9,14,8,19,10,14,12,21,8,22,12,14,12,24,12,22,11,18,14,27,10,22,
%U 16,20,15,30,12,31,16,20,18,26,12,34,18,24,14,36,16,37,19,22,20,32,14,40,20,28
%N Number of distinct solutions of Sum_{i=1..1} (x(2i-1)*x(2i)) == 1 (mod n), with x() in {1,2,...,n-1}.
%C Except for the first term, this appears to be the number of pairs of integers i,j with 1 <= i <= n, 1 <= j <= i, such that i+j == i*j (mod n), for n=1,2,3,... - _John W. Layman_, Oct 19 2011
%C Layman's observation holds since i+j == i*j (mod n) is equivalent to (i-1)*(j-1) == 1 (mod n). - _Max Alekseyev_, Oct 22 2011
%C For i > 1, equal to the number of elements x relatively prime to n such that x mod n >= x^(-1) mod n. - _Jeffrey Shallit_, Jun 14 2018
%C Differs from A007897 for n = 1, 35, 45 etc. - _Georg Fischer_, Sep 20 2020
%H R. H. Hardin, <a href="/A180783/b180783.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = (A000010(n) + A060594(n)) / 2.
%e Solutions for product of a single 1..10 pair = 1 (mod 11) are (1*1) (2*6) (3*4) (5*9) (7*8) (10*10).
%Y Column 1 of A180793.
%Y Cf. A000010, A007897, A060594.
%K nonn
%O 1,3
%A _R. H. Hardin_, formula from _Max Alekseyev_ in the Sequence Fans Mailing List, Sep 20 2010
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