User blog comment:B1mb0w/Alpha Function/@comment-25337554-20150705084257

 This is where alpha function pass Graham’s number.

 When 0 r=4.14+t,  d=4,  i=0

 While #1.

                    d=2, Entering if and  i=1

 While #2.

                    d=1, Entering if and  i=2

 While #3.

                    d=0.5, Not entering if and  i keeps to be 2

 e=2

 For (j=1). e=4

 For (j=2). e=16

 Alpha: D(

 If true

                    D=(4.14+t-4)*4=0.56+4t

                    j=0 (t<0.001)

<p class="MsoNormal">                    Alpha: D(1

<p class="MsoNormal">                    d=0.56+4t

<p class="MsoNormal">                    For (j=1).

<p class="MsoNormal">                                      d=17(0.56+4t)=9.52+68t

<p class="MsoNormal">                                      Alpha: D(1,9                //t<0.001 therefore t<0.068 so 9.52+68t<9.588<10

<p class="MsoNormal">                                      d=0.52+68t

<p class="MsoNormal">                    For (j=2);

<p class="MsoNormal">                                      d=17(0.52+68t)=8.84+1156t

<p class="MsoNormal">                                      Alpha: D(1,9,9             //by suppose

<p class="MsoNormal">                                      d=-0.84+1156t

<p class="MsoNormal"> Alpha: D(1,9,9)

<p class="MsoNormal">

<p class="MsoNormal"> So, when the alpha function changes D(1,9,8) to D(1,9,9), 8.84+1156t=9 so t=1/7225.

<p class="MsoNormal"> Therefore, alpha function pass Graham’s number when r=2393/578.

<p class="MsoNormal">

<p class="MsoNormal"> (In decimal, it begins with 4.140138408...)

<p class="MsoNormal">

<p class="MsoNormal"> P.S. When r is exactly this value, the alpha function returns D(1,9,9) because Int(exactly 9) is 9.