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Given L=∈1⅄(1⅄Qn2/2⅄Qn1)cn
∈1⅄2Qn1 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn2/2⅄1Qn1)=(1.^6/0.^6)=2.^6 depending of cn variable factor c1
then
Ln1=∈1⅄2Qn1 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn2/2⅄1Qn1)=(1.^6/0.^6)=2.^6
Ln2=∈1⅄2Qn2 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn3/2⅄1Qn2)=(1.4/0.6)
Ln3=∈1⅄2Qn3 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn4/2⅄1Qn3)=(1.^571428/0.^714285)
Ln4=∈1⅄2Qn4 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn5/2⅄1Qn4)=(1.^18/0.^63)
Ln5=∈1⅄2Qn5 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn6/2⅄1Qn5)=(1.^307692/0.^846153)
Ln6=∈1⅄2Qn6 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn7/2⅄1Qn6)=(1.^1176470588235294/0.^7647058823529411)
Ln7=∈1⅄2Qn7 of ∈1⅄1Qn and ∈2⅄1Qn variables is (1⅄1Qn8/2⅄1Qn7)=(1.^210526315789473684/0.^894736842105263157)
Variable Factor 1⅄2Qnc1 will differ variants of ∈1⅄2Qn given a factor 1⅄1Qnc2 factor 2⅄1Qnc2 and so on . . .
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