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%I #30 Nov 03 2022 05:44:24
%S 1,4,9,64,25,36,49,256,729,100,121,576,169,196,225,4096,289,2916,361,
%T 1600,441,484,529,2304,15625,676,6561,3136,841,900,961,16384,1089,
%U 1156,1225,46656,1369,1444,1521,6400,1681,1764,1849,7744,18225,2116,2209
%N Number of 3 X 3 matrices with elements from {0,...,n-1} such that the middle element of each of the eight lines of three (rows, columns and diagonals) is the square (mod n) of the difference of the end elements.
%C This sequence is multiplicative. - _Mitch Harris_, Apr 19 2005
%C The sequence enumerates the solutions of a system of polynomials equations modulo n, hence is multiplicative by the Chinese Remainder Theorem. The middle entry of the 3 X 3 is zero modulo n. - _Michael Somos_, Apr 30 2005
%H Amiram Eldar, <a href="/A059479/b059479.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A008833(n)*n^2, where A008833(n) is the largest square that divides n.
%F Multiplicative with a(p^e) = p^(3e - (e % 2)). - _Mitch Harris_, Jun 09 2005
%F Dirichlet g.f.: zeta(s-2)*zeta(2s-6)/zeta(2s-4). - _R. J. Mathar_, Mar 30 2011
%F Sum_{k=1..n} a(k) ~ zeta(3/2) * n^(7/2) / (7*zeta(3)). - _Vaclav Kotesovec_, Sep 16 2020
%F Sum_{n>=1} 1/a(n) = 15*zeta(6)/Pi^2 = A082020 * A013664 = 1.546176... . - _Amiram Eldar_, Nov 03 2022
%t f[p_, e_] := p^(3*e - (Mod[e, 2])); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Sep 16 2020 *)
%o (PARI) a(n)=if(n<1,0,n^3/core(n)) /* _Michael Somos_, Apr 30 2005 */
%Y Cf. A008833, A013664, A082020.
%K nonn,mult
%O 1,2
%A _John W. Layman_, Feb 15 2001
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