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1 decimal integer ring cycle of many

Quantum Field Fractal Polarization Math Constants

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ᐱ Y φ Θ P Q Ψ

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Given R=∈(φ/P)cn

R represents ∈(φ/P)cn

3⅄φncn/pn that path 3⅄ of φncn/pn  differ from paths 2⅄φncn/pn and 1⅄φncn/pn

Path 3⅄R=φn1cn/pn1

3⅄Rn1=1φn1/pn1=(φn1/pn1)=(0/2)=0 

3⅄Rn2=1φn2/pn2=∈(φn2/pn2)=(1/3)=0.^3

3⅄Rn3=1φn3/pn3=(φn3/pn3)=(2/5)=0.4

3⅄Rn4=1φn4/pn4=∈(φn4/pn4)=(1.5/7)=0.2^142857

3⅄Rn5=1φn5/pn5=∈(φn5/pn5)=(1.^6/11)=1.^45→. . .(φn5c2/pn5)=(1.^66/11)=0.150^9→. . . and so on

3⅄Rn6=1φn6/pn6=∈(φn6/pn6)=(1.6/13)=0.1^230769

3⅄Rn7=1φn7/pn7=∈(φn7/pn7)=(1.625/17)=0.095^5882352941176470 

3⅄Rn8=1φn8/pn8=∈(φn8/pn8)=(1.^615384/19)=0.079757^052631578421→. . .(φn8c2/pn8)=0.079757079757^052631578421→. . . and so on

3⅄Rn9=1φn9/pn9=∈(φn9/pn9)=(1.^619047/23)=0.070393^3478260865217391304

Path 1⅄R=φn2/pn1

1⅄Rn1=(φn2/pn1)=(1/2)

1⅄Rn2=(φn3/pn2)=(2/3)

1⅄Rn3=(φn4/pn3)=(1.5/5)

1⅄Rn4=(φn5/pn4)=(1.^6/7)

1⅄Rn5=(φn6/pn5)=(1.6/11)

1⅄Rn6=(φn7/pn6)=(1.625/13)

1⅄Rn7=(φn8/pn7)=(1.^615384/17)

1⅄Rn8=(φn9/pn8)=(1.^619047/19)

1⅄Rn9=(φn10/pn9)=(1.6^1762941/23)

Path 2⅄R=φn1/pn2

2⅄Rn1=(φn1/pn2)=(0/3)

2⅄Rn2=(φn2/pn3)=(1/5)

2⅄Rn3=(φn3/pn4)=(2/7)

2⅄Rn4=(φn4/pn5)=(1.5/11)

2⅄Rn5=(φn5/pn6)=(1.^6/13)

2⅄Rn6=(φn6/pn7)=(1.6/17)

2⅄Rn7=(φn7/pn8)=(1.625/19)

2⅄Rn8=(φn8/pn9)=(1.^615384/23)

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