%I #33 Apr 18 2020 22:11:52
%S 1,2,7,12,25,38,63,88,129,170,231,292,377,462,575,688,833,978,1159,
%T 1340,1561,1782,2047,2312,2625,2938,3303,3668,4089,4510,4991,5472,
%U 6017,6562,7175,7788,8473,9158,9919,10680,11521,12362,13287,14212
%N Number of 3 X 3 matrices, with elements from {0,...,n}, having the property that the middle element of each of the eight 3-element horizontal, vertical and diagonal lines equals the average of the two end elements.
%C The bisections of the first differences of {a(n)} give A001844 (the centered triangular numbers n^2+(n-1)^2).
%C Quasipolynomial of order 2. - _Charles R Greathouse IV_, Dec 07 2011
%C Also, the number of 3 X 3 magic squares with elements from {0,...,n} and with duplicate elements allowed. If [[a,b,c], [d,e,f], [g,h,i]] satisfies the property in the description of this sequence, then [[h,a,f], [c,e,g], [d,i,b]] is a magic square, and conversely. - _David Radcliffe_, Apr 13 2020
%H G. C. Greubel, <a href="/A059329/b059329.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-4,1,2,-1).
%F From _Frank Ellermann_: (Start)
%F even: a(2*n) = (4*n^3 + 6*n^2 + 8*n + 3)/3.
%F odd: a(2*n-1) = (4*n^3 + 2*n)/3. (End)
%F a(n) = Sum_{k=0..n} A109613(k)*A109613(n-k). - _Reinhard Zumkeller_, Dec 05 2009
%F From _Colin Barker_, Mar 29 2013: (Start)
%F a(n) = ((1+n)*(9+3*(-1)^n+4*n+2*n^2))/12.
%F G.f.: (x^2+1)^2 / ((x-1)^4*(x+1)^2). (End)
%F E.g.f.: (1/12)*(3*(1 - x)*exp(-x) + (9 + 21*x + 12*x^2 + 2*x^3)*exp(x)). - _G. C. Greubel_, Jan 07 2017
%t LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 2, 7, 12, 25, 38}, 50] (* _G. C. Greubel_, Jan 07 2017 *)
%o (PARI) a(n)=if(n%2,4*n^3+2*n,4*n^3+6*n^2+8*n+3)/3 \\ _Charles R Greathouse IV_, Dec 07 2011
%K nonn,easy
%O 0,2
%A _John W. Layman_, Jan 26 2001