%I #18 Dec 22 2023 10:39:44
%S 16,766,61184,3112500,105851488,2138413851,27990555776,262835331687,
%T 1909384608000,11319915386120,56916060868096,249702337698346,
%U 976762617522160,3464394870851125,11290721919375872,34177386571594701
%N Number of n X n arrays of squares of integers summing to 9.
%C All such sequences have holonomic recurrences (cf. comment in A159359). - _Georg Fischer_, Feb 17 2022
%H R. H. Hardin, <a href="/A159375/b159375.txt">Table of n, a(n) for n = 2..100</a>
%H <a href="/index/Rec#order_19">Index entries for linear recurrences with constant coefficients</a>, signature (19, -171, 969, -3876, 11628, -27132, 50388, -75582, 92378, -92378, 75582, -50388, 27132, -11628, 3876, -969, 171, -19, 1).
%F Empirical g.f.: -x-(1+x)*x*(1 - 4*x + 637*x^2 + 47760*x^3 + 2021602*x^4 + 54462984*x^5 + 548532899*x^6 + 2125377516*x^7 + 3360726010*x^8 + 2125377516*x^9 + 548532899*x^10 + 54462984*x^11 + 2021602*x^12 + 47760*x^13 + 637*x^14 - 4*x^15 + x^16)/(-1+x)^19. - _Vaclav Kotesovec_, Nov 30 2012
%F a(n) = binomial(n^2,1) + multinomial(n^2,1,2,(n^2-3)) + multinomial(n^2,1,5,n^2-6) + binomial(n^2,9) = (1/362880)*n^18 - (1/10080)*n^16 + (13/8640)*n^14 - (1/240)*n^12 - (1091/17280)*n^10 + (251/480)*n^8 - (95209/90720)*n^6 + (1213/2520)*n^4 + (10/9)*n^2 corresponding to the ways of obtaining 9 as a sum of n^2 squares: 9 + (n^2-1)*0, 2*4 + 1 + (n^2-3)*0, 4 + 5*1 + (n^2 - 6)*0, and 9*1 + (n^2 - 9)*0. - _Robert Israel_, Dec 18 2023
%K nonn
%O 2,1
%A _R. H. Hardin_, Apr 11 2009
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