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A280344
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Number of 2 X 2 matrices with all elements in {0,...,n} with determinant = permanent^n.
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2
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0, 12, 30, 56, 90, 132, 182, 240, 306, 380, 462, 552, 650, 756, 870, 992, 1122, 1260, 1406, 1560, 1722, 1892, 2070, 2256, 2450, 2652, 2862, 3080, 3306, 3540, 3782, 4032, 4290, 4556, 4830, 5112, 5402, 5700, 6006, 6320, 6642, 6972, 7310, 7656, 8010, 8372, 8742, 9120, 9506, 9900, 10302
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OFFSET
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0,2
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COMMENTS
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For n>0, a(n) is the perimeter of a primitive Pythagorean triangle. - Torlach Rush, Jul 11 2019
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LINKS
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FORMULA
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a(n) = (((n-2)*a(n-1))/(n-4)) - (6*(3*(n-1)+1)/(n-4)) for n>=4.
a(n) = 2 + 6*n + 4*n^2 for n>0.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.
G.f.: 2*x*(6 - 3*x + x^2) / (1 - x)^3.
(End)
a(n) = (2*n+1)*(2*n+2), n>0.
a(n) = 2*((n+1)^2 + ((n+1)*n)), n>0.
(End)
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MATHEMATICA
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Table[Boole[n != 0] 2 # (2 # - 1) &[n + 1], {n, 0, 50}] (* or *)
CoefficientList[Series[2 x (6 - 3 x + x^2)/(1 - x)^3, {x, 0, 50}], x] (* Michael De Vlieger, Jan 01 2017 *)
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PROG
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(Python)
def t(n):
s=0
for a in range(0, n+1):
for b in range(0, n+1):
for c in range(0, n+1):
for d in range(0, n+1):
if (a*d-b*c)==(a*d+b*c)**n:
s+=1
return s
for i in range(0, 41):
print str(i)+" "+str(t(i))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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