%I #42 Jul 27 2019 16:30:15
%S 0,12,30,56,90,132,182,240,306,380,462,552,650,756,870,992,1122,1260,
%T 1406,1560,1722,1892,2070,2256,2450,2652,2862,3080,3306,3540,3782,
%U 4032,4290,4556,4830,5112,5402,5700,6006,6320,6642,6972,7310,7656,8010,8372,8742,9120,9506,9900,10302
%N Number of 2 X 2 matrices with all elements in {0,...,n} with determinant = permanent^n.
%C For n>0, a(n) is the perimeter of a primitive Pythagorean triangle. - _Torlach Rush_, Jul 11 2019
%H Indranil Ghosh, <a href="/A280344/b280344.txt">Table of n, a(n) for n = 0..995</a>
%F a(0) = A002939(0) = 0; a(n) = A002939(n+1), for n>=1.
%F a(n) = (((n-2)*a(n-1))/(n-4)) - (6*(3*(n-1)+1)/(n-4)) for n>=4.
%F Conjectures from _Colin Barker_, Jan 01 2017: (Start)
%F a(n) = 2 + 6*n + 4*n^2 for n>0.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
%F G.f.: 2*x*(6 - 3*x + x^2) / (1 - x)^3.
%F (End)
%F From _Torlach Rush_, Jul 11 2019: (Start)
%F a(n) = (2*n+1)*(2*n+2), n>0.
%F a(n) = 2*((n+1)^2 + ((n+1)*n)), n>0.
%F (End)
%t Table[Boole[n != 0] 2 # (2 # - 1) &[n + 1], {n, 0, 50}] (* or *)
%t CoefficientList[Series[2 x (6 - 3 x + x^2)/(1 - x)^3, {x, 0, 50}], x] (* _Michael De Vlieger_, Jan 01 2017 *)
%o (Python)
%o def t(n):
%o s=0
%o for a in range(0,n+1):
%o for b in range(0,n+1):
%o for c in range(0,n+1):
%o for d in range(0,n+1):
%o if (a*d-b*c)==(a*d+b*c)**n:
%o s+=1
%o return s
%o for i in range(0,41):
%o print str(i)+" "+str(t(i))
%Y Same as both A002939 and A118239 without A002939(1) = 2.
%Y Cf. A016754, A280343.
%K nonn
%O 0,2
%A _Indranil Ghosh_, Jan 01 2017
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