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1 decimal integer ring cycle of many
Quantum Field Fractal Polarization Math Constants
nemeth braille printable arx calc
pronounced why phi prime quotients
ᐱ Y φ Θ P Q Ψ
condensed matter
Y Phi Theta Prime Q Quotients Base Numerals 1dir 2dir 3dir cdir
numer nu mer numerical nomenclature & arcs
Given M=∈2⅄(φ/Q)cn
The definition of Q variable should be accurate to its defining base path where 1⅄Q and 2⅄Q, differ from prime base just as φ and Θ differ in paths 1⅄ and 2⅄ of Y base.
so
M=∈2⅄(φn1/1⅄Qn2)cn
and
M=∈2⅄(φn1/2⅄Qn2)cn
then
Mn1 of ∈2⅄(φn1/1⅄Qn2)=(0/1.^6)
Mn2 of ∈2⅄(φn2/1⅄Qn3)=(1/1.4)
Mn3 of ∈2⅄(φn3/1⅄Qn4)=(2/1.^571428)
Mn4 of ∈2⅄(φn4/1⅄Qn5)=(1.5/1.^18)
Mn5 of ∈2⅄(φn5/1⅄Qn6)=(1.^6/1.^307692)
Mn6 of ∈2⅄(φn6/1⅄Qn7)=(1.6/1.^1176470588235294)
Mn7 of ∈2⅄(φn7/1⅄Qn8)=(1.625/1.^210526315789473684)
if M=∈2⅄(φn1/1⅄Qn2)cn
and
M=∈2⅄(φn1/2⅄Qn2)cn
then
Mn1 of ∈2⅄(φn1/2⅄Qn2)=(0/0.6)
Mn2 of ∈2⅄(φn2/2⅄Qn3)=(1/0.^714285)
Mn3 of ∈2⅄(φn3/2⅄Qn4)=(2/0.^63)
Mn4 of ∈2⅄(φn4/2⅄Qn5)=(1.5/0.^846153)
Mn5 of ∈2⅄(φn5/2⅄Qn6)=(1.^6/0.^7647058823529411)
Mn6 of ∈2⅄(φn6/2⅄Qn7)=(1.6/0.^894736842105263157)
Mn7 of ∈2⅄(φn7/2⅄Qn8)=(1.625/0.^8260869565217391304347)
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