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%I #42 Sep 28 2022 04:11:22
%S 0,1,6,16,40,63,126,176,270,365,550,624,936,1099,1350,1664,2176,2349,
%T 3078,3280,3948,4631,5566,5712,7000,7813,8748,9520,11368,11475,13950,
%U 14592,16236,17969,19390,20304,23976,25327,27222,28400,32800,32949,37926,38896
%N The sum of all the entries in an n X n Cayley table for multiplication in Z_n.
%C Thanks to David Miller.
%F a(p) = (p-1)*(p^2-p)/2, for p prime.
%F a(n) = (n/2)*Sum_{i=1..n-1} gcd(n,i)*(n/gcd(n,i)-1). [Edited by _Richard L. Ollerton_, May 06 2021]
%F a(n) = (n^2/2)*Sum_{d|n} phi(d)*(d-1)/d, where phi = A000010. - _Richard L. Ollerton_, May 06 2021
%F From _Ridouane Oudra_, Aug 24 2022: (Start)
%F a(n) = Sum_{i=1..n} Sum_{j=1..n} (i*j mod n);
%F a(n) = n^3/2 - (n/2)*Sum_{i=1..n} gcd(n,i);
%F a(n) = n^3/2 - (n/2)*Sum_{d|n} d*tau(d)*moebius(n/d);
%F a(n) = (A000578(n) - n*A018804(n))/2. (End)
%e For n=4:
%e | 0 1 2 3
%e -+--------
%e 0| 0 0 0 0
%e 1| 0 1 2 3
%e 2| 0 2 0 2
%e 3| 0 3 2 1
%e Sum becomes 6+4+6 = 16.
%o (PARI) a(n) = (n/2)*sum(i=1, n-1, gcd(n, i)*(n/gcd(n, i)-1)); \\ _Michel Marcus_, Jun 16 2013 [edited by _Richard L. Ollerton_, May 06 2021]
%Y Cf. A000010, A000578, A018804.
%K nonn
%O 1,3
%A David Byrne (david.roggeveen.byrne(AT)gmail.com), May 06 2009
%E More terms from _Carl Najafi_, Sep 29 2011
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