Swedish Nobel Prize model

Swedish Nobel Prize model by Thomas McClure I

Introduction

This paper is written to show the Swedish Nobel Prize model for partitioning.

In the Swedish Nobel Prize model: u(c, 1 - l) = (1 - φ) log c + φ log(1 - l) Then anti-log [u(c, 1 - l)] = [c ^ (1 - φ) ]*[ (1 - l) ^ φ] ; If φ = 1, then [u(c, 1 - l)] = log (1 - l) ; let log [base (1 - l)] [(1 - l)] = 1 ; alog [base (1 -1)] [(u(c, 1 - l)] = (1 - l) to 1] = alog [base (1 - l) [1] = 1 . Hence, u(c,1 - l) = 1 If φ = 0, then [u(c, 1 - l)] = log c ; let log [base c [c]] = 1 ; alog [base (c) [(u(c, 1 - l)] = (c) to 1] = alog [base(c) [1] = 1 . Hence, u(c,1 - l) = 1 Let 1 = (1 - φ) log c + φ log(1 - l) then 1 = log c – φ* log c + φ* log(1- l)

φ*(log(1-l) – log c) = (1 – log c) φ = (1 – log c)/ (log(1-l) – log c) hence φ and (1 – φ) are the partitioning terms. Alog 1 = c^(1 – φ) * (1 – l)^( φ) e^alog 1 = e^ [c^(1 – φ) * (1 – l)^( φ)] but e^z = 2^y, where e^z is partitioned into a yth power of 2. Since z/y is the ratio of root[2] to 2 e^ [c^(1 – φ) * (1 – l)^( φ)] = 2^{ [2/root[2]]*[c^(1 – φ) * (1 – l)^( φ)]} Hence, any largest known prime as a power of two may be reduced to a lesser power of e, 2.78…. And it may be partitioned in that exponent by φ, which is a product of two factors, like c and (1 – l). And taking the natural log of that power of e, gives the exponent itself. Hence, a very large exponent of 2 minus one, may be reduced by .7 and then partitioned by a φ, which is a function of two factors, chosen like c and (1 – l).

Partitioning the Largest Known Prime

232582657-1 largest known prime becomes

e.7*32582657 –1 = e^22807860 –1 Take the half of 22807860 or 11403930. Then e^{c^(1 – φ) * (1 – l) ^(φ)} – 1, where c^(1 – φ) = 11403930 = (1 – l) ^(φ) (1 – φ) *log (base c) c = logc 11403930

(φ) * log (base (1 – l)) (1 – l) = log(1-l) 11403930 Let c = e; (1 – l) = 2; l = -1