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- jamil - Jul 25 2008 | paridade, inteiros, charadas, matemática, comunidade, orkut
Usando polinômio de Vandermond,MOD e modular
f(n) = [x(x+1)(x-1)(x+2)(x-2)(x+3)/3(3+1)(3-1)(3+2)(3-2)(3+3)].......++[x(x-1)(x+1)(x+2)(x-3)(x+3)/2(2-1)(2+1)(2+2)(2-3)(2+3)]...............+
+[x(x-1)(x-2)(x+2)(x-3)(x+3)/-1(-1-1)(-1-2)(-1+2)(-1-3)(-1+3)]..........+
+[x(x-1)(x+1)(x-2)(x-3)(x+3)/-2(-2-1)(-2+1)(-2-2)(-2-3)(-2+3)].
para n, z e x inteiros tal que n= 4z+x e |x|<4
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