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Pattern recognition problem (format updated)

I would like to find the pattern for the elements in the following set. I especially would like to see how to use Mathematica to arrive at it. But if you can't get Mathematica to do it -- and can do it some other way-- please post your solution.

{a, a, b, c, a, a, d, a, e, d, e, a, d, a, e, a, a, b, c, a, a, d, a, e, d, e, a, d, a, e, a, a, b, c, a, a, d, a, e, a, d, e, d, a, e, a, 
a, b, c, a, a, d, a, e, a, d, e, d, a, e, a, a, b, c, a, a, a, b, c, a, d, e, d, a, e, a, a, b, c, a, a, a, b, c, a, d, e, d, e, d, e, a, 
b, c, a, a, a, b}

Here it is again, formatted differently:

  {a, a, b, c, a, a, d, a, e, d, e, a, d, a, e, a, a, b, c, a, a, d, a, e,

   d, e, a, d, a, e, a, a, b, c, a, a, d, a, e, a, d, e, d, a, e, a, a, b, 

   c, a, a, d, a, e, a, d, e, d, a, e, a, a, b, c, a, a, a, b, c, a, d, e, 

   d, a, e, a, a, b, c, a, a, a, b, c, a, d, e, d, e, d, e, a,  b, c, a, a, a, b}

a-e Actually have exact values, if that is helpful.

Here they are with the exact values:

{1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 1, 3 + 2 Sqrt[2], 1, 

 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 1, 3 + 2 Sqrt[2], 1, 

 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 1, 

 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 1, 

 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 

 17 - 12 Sqrt[2], 1, 1, 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 

 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 1, 

 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 1, 

 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 

 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 

 17 - 12 Sqrt[2], 1, 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 

 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 1, 

 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 

 17 - 12 Sqrt[2], 1, 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 

 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 

 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 1, (1 + Sqrt[2]), 

 17 - 12 Sqrt[2], 1, 1, 1, (1 + Sqrt[2])}

Here is the Mathematica notebook where they come from.(it is also attached, so you can edit and copy it.)

A formula that produces (-s/d) which = an approximation to the MRB constant:

expM[pre_] := 
  Module[{a, k, bb, c, n, end, iprec, xvals, x, pc, cores = 4, 
    tsize = 2^7, chunksize, start = 1, ll, ctab, 
    pr = Floor[1.02 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];

   d = N[(3 + Sqrt[8])^n, pr + 10];
   d = Round[1/2 (d + 1/d)];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr, pc = Min[3 pc, pr];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(*N[Exp[Log[ll]/ll],
        pr]*)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> "EvaluationsPerKernel" -> 1]];
    ctab = Table[c = b - c;
      ll = start + l - 2;
      b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
      c, {l, chunksize}];
    s += ctab.(xvals - 1);
    start += chunksize;
    , {k, 0, end - 1}];
   N[-s/d, pr]];

A small sample of high precision ss[10a]*ss[10a+20]/ss[10a+10]^2. (a=1,2,3...) In other words,(ss[10]/ss[20])/(ss(20]/ss[30]), ect These were looked up in Wolfram Alpha to get exact values.:

nx = 11; a = 0; For[xx = 1, xx <= nx, a = a + 10;

 prec = a; MRBtest2 = expM[prec]; ss[a] = s; xx++]; Table[
 ss[10 a]*ss[10 a + 20]/ss[10 a + 10]^2, {a, nx - 2}]

.

Out[9]= {1.0000000, 1.00000000000000000, 33.9705627484771405856202647, \
0.02943725152285941437973530948362305716, \
1.000000000000000000000000000000000000000000000000, \
1.0000000000000000000000000000000000000000000000000000000000, \
5.8284271247461900976033774484193961571393437507538961463533594759815,\
 1.0000000000000000000000000000000000000000000000000000000000000000000\
00000000000, \
0.17157287525380990239662255158060384286065624924610385364664052401853\
50430757859222992249}

A larger set of ss[10a]*ss[10a+20]/ss[10a+10]^2, (aa=10,20,30...). In other words,(ss[10]/ss[20])/(ss(20]/ss[30]), ect.:

nx = 100; a = 0; For[xx = 1, xx <= nx, a = a + 10;

 prec = a; MRBtest2 = expM[prec]; ss[a] = s; xx++]; stuff = 
 Table[ss[10 a]*ss[10 a + 20]/ss[10 a + 10]^2, {a, nx - 2}];
N[stuff, 2]

.

Out[10]= {1.0, 1.0, 34., 0.029, 1.0, 1.0, 5.8, 1.0, 0.17, 5.8, 0.17, 1.0, 5.8, \
1.0, 0.17, 1.0, 1.0, 34., 0.029, 1.0, 1.0, 5.8, 1.0, 0.17, 5.8, 0.17, \
1.0, 5.8, 1.0, 0.17, 1.0, 1.0, 34., 0.029, 1.0, 1.0, 5.8, 1.0, 0.17, \
1.0, 5.8, 0.17, 5.8, 1.0, 0.17, 1.0, 1.0, 34., 0.029, 1.0, 1.0, 5.8, \
1.0, 0.17, 1.0, 5.8, 0.17, 5.8, 1.0, 0.17, 1.0, 1.0, 34., 0.029, 1.0, \
1.0, 1.0, 34., 0.029, 1.0, 5.8, 0.17, 5.8, 1.0, 0.17, 1.0, 1.0, 34., \
0.029, 1.0, 1.0, 1.0, 34., 0.029, 1.0, 5.8, 0.17, 5.8, 0.17, 5.8, \
0.17, 1.0, 34., 0.029, 1.0, 1.0, 1.0, 34.}

1.0 -> 1, 34. -> b, etc.

{a, a, b, c, a, a, d, a, e, d, e, a, d, a, e, a, a, b, c, a, a, d, a, \
e, d, e, a, d, a, e, a, a, b, c, a, a, d, a, e, a, d, e, d, a, e, a, \
a, b, c, a, a, d, a, e, a, d, e, d, a, e, a, a, b, c, a, a, a, b, c, \
a, d, e, d, a, e, a, a, b, c, a, a, a, b, c, a, d, e, d, e, d, e, a, \
b, c, a, a, a, b}

a-e Replaced by exact values: enter image description here \

Out[12]= {1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 1, 3 + 2 Sqrt[2], 1, 
 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 1, 3 + 2 Sqrt[2], 1, 
 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 1, 
 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 1, 
 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 
 17 - 12 Sqrt[2], 1, 1, 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 
 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 1, 
 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 1, 
 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 
 3 + 2 Sqrt[2], 1, 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 
 17 - 12 Sqrt[2], 1, 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 
 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 1, 
 3 - 2 Sqrt[2], 1, 1, (1 + Sqrt[2]), 
 17 - 12 Sqrt[2], 1, 1, 1, (1 + Sqrt[2]), 17 - 12 Sqrt[2], 1, 
 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 
 3 + 2 Sqrt[2], 3 - 2 Sqrt[2], 1, (1 + Sqrt[2]), 
 17 - 12 Sqrt[2], 1, 1, 1, (1 + Sqrt[2])}
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POSTED BY: Marvin Ray Burns
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