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%I #26 Nov 07 2017 03:28:59
%S 1,2,7,14,29,49,83,127,192,273,384,519,694,902,1162,1466,1835,2260,
%T 2765,3340,4011,4767,5637,6609,7714,8939,10318,11837,13532,15388,
%U 17444,19684,22149,24822,27747,30906,34345,38045,42055,46355,50996,55957,61292
%N Number of essentially different semi-magic squares of order 3 with semimagic sum n.
%D Christoph Gerber, "Zum Abzahlen semimagischer Quadrate" [Apparently unpublished. - _R. J. Mathar_, Nov 13 2011]
%D P. A. MacMahon, Combinatory Analysis, Vol II; Chelsea, New York, 1960.
%H Harvey P. Dale, <a href="/A122751/b122751.txt">Table of n, a(n) for n = 3..1000</a>
%H Christoph Gerber, <a href="http://www.phbern.ch/index.php?id=4550">More information</a> [?Broken link]
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-2,4,-4,2,2,-3,1).
%F a(n) = 1/64*n^4-1/32*n^3+1/32*n^2+d*n+e with: d:=-1/8 if n=0 or n=2 (mod 4) d:=3/32 if n=1 or n=3 (mod 4) e:=0 if n=0 (mod 4) e:=-7/64 if n=1 (mod 4) e:=1/8 if n=2 (mod 4) e:=1/64 if n=3 (mod 4).
%F G.f.: -x^3*(1-x+3*x^2-x^3+x^4) / ( (1+x^2)*(1+x)^2*(x-1)^5 ). - _R. J. Mathar_, Nov 13 2011
%F a(n) = (2*n*(n-1)*(n^2-n+1)-7*(2*n-1)*(-1)^n-8*(-1)^((2*n-1+(-1)^n)/4)+1)/128. - _Luce ETIENNE_, Oct 29 2017
%e a(4)=2 because there are 2 essentially different semi-magic squares of order 3 with semi-magic sum 4: [1,1,2; 1,2,1; 2,1,1] and [1,1,2; 2,1,1; 1,2,1].
%p A131292:=proc(n) local d,e: if (n mod 4) in {0,2} then d:=-1/8 fi: if (n mod 4) in {1,3} then d:=3/32 fi: if (n mod 4) in {0} then e:=0 fi: if (n mod 4) in {1} then e:=-7/64 fi: if (n mod 4) in {2} then e:=1/8 fi: if (n mod 4) in {3} then e:=1/64 fi: return 1/64*n^4-1/32*n^3+1/32*n^2+d*n+e: end proc:
%t LinearRecurrence[{3,-2,-2,4,-4,2,2,-3,1},{1,2,7,14,29,49,83,127,192},50] (* _Harvey P. Dale_, Jan 26 2017 *)
%K nonn
%O 3,2
%A Christoph Gerber (christoph.gerber(AT)phbern.ch), Jun 25 2007
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