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@BillFoster
Created February 23, 2012 12:14
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problems with subarrays
variables: {
j:"random(0..6)"
t6: "[6,1,[[3,2]],[[7,14],[9,18],[0]]]"
t10: "[10,1,[0],[[11,22],[0],[0]]]"
t12:"[12,1,[0],[[13,26],[0],t6]]"
t18:"[18,1,[[3,3]],[[19,38],[27,54],[0]]]"
t20:"[20,0,[[5,2]],[[0],[25,50],t10]]"
t22:"[22,1,[0],[[23,46],[0],[0]]]"
t28:"[28,1,[0],[[29,58],[0],[0]]]"
t:"[t6,t10,t12,t18,t20,t22,t28]"
si:"t[j]"
m:"t[j][0]"
m1:"m/2"
primess:"if(si[1]=1,'Since $\var{m}+1=\var{m+1}$ is a prime number we have solutions, $n=\var{m+1}$ and $n=2\times \var{m+1}=\var{2*(m+1)}$. Next we look to see if $\var{m}$ is a value in the table.',
'Since $\var{m}+1=\var{m+1}$ is not a prime number we next look to see if $\var{m}$ is a value in the table.')"
valmess:"if(si[2][0]=0,'Looking at the table we see that there is no value in the table such that $ \phi(\var{si[2][0][0]}^{\var{si[2][0][1]}})=\var{m}$','Looking at the table we see that $ \phi(\var{si[2][0][0]}^{\var{si[2][0][1]}})=\var{m}$.
Hence we have $n=\var{si[2][0][0]^si[2][0][1]}$ and $n=\var{2*si[2][0][0]^si[2][0][1]}$ are solutions.')"
factmess:"if(abs(si[3][2])=1,'The factorization $\var{m} = 2 \times \var{m1}$ does not lead to any other solutions as $\var{m1}$ is odd',1)"
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