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1 decimal integer ring cycle of many
Quantum Field Fractal Polarization Math Constants
nemeth braille printable arx calc
ᐱ Y φ Θ P Q Ψ
condensed matter
Complex sets of Y Phi Theta Prime Q Psi Quotient Based Numerals
nu mer numer i call numerical nomenclature & arc ratio constants
Given H=∈2⅄(Q/Θ)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
H=∈2⅄(1⅄Qn1/Θn2)cn
and
H=∈2⅄(2⅄Qn1/Θn2)cn
then
Hn1 of 2⅄(1⅄Qn1/Θn2)=(1.5/1)=1.5
Hn2 of 2⅄(1⅄Qn2/Θn3)=(1.^6/0.5)=3.2 and Hn2 of 2⅄(1⅄Qn2c2/Θn3)=(1.^66/0.5)=3.32 and Hn2 of 2⅄(1⅄Qn2c3/Θn3)=(1.^666/0.5)=3.332
Hn3 of 2⅄(1⅄Qn3/Θn4)=(1.4/0.^6)=2.^3 or 2.333 and Hn3 of 2⅄(1⅄Qn3/Θn4)=(1.4/0.^66)=^2.1 or 21.2121 and so on... for Hn3 of 2⅄(1⅄Qn3/Θn4cn)
Hn4 of 2⅄(1⅄Qn4/Θn5)=(1.^571428/0.6)=2.61904^6 or2.619046 and Hn4 of 2⅄(1⅄Qn4c2/Θn5)=(1.^571428571428/0.6)=2.61904761904^6 or 2.61904761904666 and so on... for Hn4 of 2⅄(1⅄Qn4cn/Θn5)
Hn5 of 2⅄(1⅄Qn5/Θn6)=(1.^18/0.625)=1.888 and Hn5 of 2⅄(1⅄Qn5c2/Θn6)=(1.^1818/0.625)=1.89088 and Hn5 of 2⅄(1⅄Qn5c3/Θn6)=(1.^181818/0.625)=1.8909088 and Hn5 of 2⅄(1⅄Qn5c4/Θn6)=(1.^18181818/0.625)=1.890909088 and so on for Hn5 of 2⅄(1⅄Qn5cn/Θn6)
Hn6 of 2⅄(1⅄Qn6/Θn7)=(1.^307692/0.^615384)=2.1^250016 or 2.1250016 and Hn6 of 2⅄(1⅄Qn6c2/Θn7)=(1.^307692307692/0.^615384)=^2.12500 or 2.12500212500 and Hn6 of 2⅄(1⅄Qn6/Θn7c2)=(1.^307692/0.^615384615384)=^2.12499950000 or 2.12499950000212499950000 and Hn6 of 2⅄(1⅄Qn6c2/Θn7c2)=(1.^307692307692/0.^615384615384)=2.1^250000000016 or 2.125000000001621250000000016 and so on for Hn6 of 2⅄(1⅄Qn6cn/Θn7cn)
Hn7 of 2⅄(1⅄Qn7/Θn8)=(1.^1176470588235294/0.^619047)=1.8054316696^850633312171773710235248696787158325619864081402542941004479466017927556389094 or 1.8054316696850633312171773710235248696787158325619864081402542941004479466017927556389094 and
Hn7 of 2⅄(1⅄Qn7/Θn8c2)=(1.^1176470588235294/0.^619047619047)=1.8054^298642551990760181013529221719475067683257936606144796398144606334859683067873321221529411782759990950244298452488705836914027167375375565628913837104090452298642551990760181013529221719475067683257936606144796398144606334859683067873321221529411782759990950244298452488705836914027167375375565628913837104090452 or 1.8054298642551990760181013529221719475067683257936606144796398144606334859683067873321221529411782759990950244298452488705836914027167375375565628913837104090452298642551990760181013529221719475067683257936606144796398144606334859683067873321221529411782759990950244298452488705836914027167375375565628913837104090452298642551990760181013529221719475067683257936606144796398144606334859683067873321221529411782759990950244298452488705836914027167375375565628913837104090452298642551990760181013529221719475067683257936606144796398144606334859683067873321221529411782759990950244298452488705836914027167375375565628913837104090452 and Hn7 of 2⅄(1⅄Qn7c2/Θn8)=(1.^11764705882352941176470588235294/0.^619047)=1.80543166968506335022172126^244524 or 1.80543166968506335022172126244524244524244524 and Hn7 of 2⅄(1⅄Qn7c2/Θn8c2)=(1.^11764705882352941176470588235294/0.^619047619047)=1.805429864255199095^0226262398190 or 1.805429864255199095022626239819002262623981900226262398190 and so on for Hn7 of 2⅄(1⅄Qn7cn/Θn8cn)
Hn8 of 2⅄(1⅄Qn8/Θn9)=(1.^210526315789473684/0.6^1764705882352941)=1.9598997493734335892187611886860007311012224438647639936225403158040876008263056642021550499799209453394906189902503200175922447340485333835968897163291430007530182371308847640562425822787183734940264255582429718876945492140275387267463310876977296954657078696125610346639272465120791466588397519392737523585897674455440543578755260348877743558348362901555457785757227337777498435496840012697614577610019083897946412219102144470323082530768031819970711992670567104678224740963525060985404021800547793291630538477755599880849157555492190135759497777 with an extended shell of decimal having a potential of 61,764,705,882,352,940 total digits in the final quotient based on probability.
then
if H=∈2⅄(1⅄Qn1/Θn2)cn
and
H=∈2⅄(2⅄Qn1/Θn2)cn
then
Hn1 of 2⅄(2⅄Qn1/Θn2)=(0.^6/1)=0.6
Hn2 of 2⅄(2⅄Qn2/Θn3)=(0.6/0.5)=1.2
Hn3 of 2⅄(2⅄Qn3/Θn4)=(0.^714285/0.^6)=1.190475
Hn4 of 2⅄(2⅄Qn4/Θn5)=(0.^63/0.6)=1.05
Hn5 of 2⅄(2⅄Qn5/Θn6)=(0.^846153/0.625)=1.3538448
Hn6 of 2⅄(2⅄Qn6/Θn7)=(0.^7647058823529411/0.^615384)=1.2426483014718^307593 or 1.2426483014718307593307593307593 and so on...
Hn7 of 2⅄(2⅄Qn7/Θn8)=(0.^894736842105263157/0.^619047)=1.445345574900^230769 or 1.445345574900230769230769 and so on...
Hn8 of 2⅄(2⅄Qn8/Θn9)=(0.^8260869565217391304347/0.6^1764705882352941)=1.337474120082815738810868914522330682316768327206659092333623792019025978096067977197217080274479934849191657927085528140547594077387223258707411649677780739164033284793659254754380813696169299298230896274769426566373989356484075903925683875668788 with an extended shell of decimal having a potential of 61,764,705,882,352,940 total digits in the final quotient based on probability.
And so on for function set H=∈2⅄(Q/Θ)cn