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MRB constant algorithmic pattern

Posted 10 months ago
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As many of you know by now the MRB constant can be calculated by

m = 
 NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity}, WorkingPrecision -> 100,
   Method -> "AlternatingSigns"]

 (*
0.18785964246206712024851793405427323005590309490013878617200468408947\
723156466021370329665443217278.*)

Today on Dec 23, 2017 I was researching the general algorithm Sum[(-1)^x (x^(1/x) - 1), {x, 1, b}], letting

h[a_, b_] := m - (1/a - 
NSum[(-1)^x (x^(1/x) - 1), {x, 1, b}, WorkingPrecision -> 100, 
 Method -> "AlternatingSigns"])

. That is where I found this miraculous pattern from b=16 and a=

  A003592 = Union@Flatten@NestList[{2 #, 4 #, 5 #} &, 1, 5]

(* {1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, \
125, 128, 160, 200, 250, 256, 320, 400, 500, 512, 625, 640, 800, \
1000, 1024, 1250, 1280, 1600, 2000, 2500, 3125}.*)

So I entered

Table[l = A003592[[a]]; {l, h[l, 16]}, {a, 1, 36}] // TableForm

and got

enter image description here

.

That last row is 3125, 0.46794671997305155000040717923910461977317844066234420613479140573499810467672704660739099737541089.

The pattern goes on for many larger b.

The differences of the rows is found by

     t= N[
     Table[h[A003592[[a]], 16] - h[A003592[[a - 1]], 16], {a, 2, 36}]]

    (* {0.5, 0.25, 0.05, 0.075, 0.025, 0.0375, 0.0125, 0.01, \
    0.00875, 0.00625, 0.005, 0.004375, 0.003125, 0.0025, 0.002, \
    0.0001875, 0.0015625, 0.00125, 0.001, 0.00009375, 0.00078125, \
    0.000625, 0.0005, 0.000046875, 0.000353125, 0.0000375, 0.0003125, \
    0.00025, 0.0000234375, 0.000176563, 0.00001875, 0.00015625, 0.000125, \
    0.0001, 0.00008}.*)

It looks like there could be a pattern for the t's when you enter

   1/t

   (* {2., 4., 20., 13.3333, 40., 26.6667, 80., 100., 114.286, \
   160., 200., 228.571, 320., 400., 500., 5333.33, 640., 800., 1000., \
   10666.7, 1280., 1600., 2000., 21333.3, 2831.86, 26666.7, 3200., \
   4000., 42666.7, 5663.72, 53333.3, 6400., 8000., 10000., 12500.}.*)

There is a more provable pattern for all differences of the h(a,16)'s:

     tPlain = N[Table[h[a, 16] - h[a - 1, 16], {a, 2, 36}]]

   (* {0.5, 0.166667, 0.0833333, 0.05, 0.0333333, 0.0238095, \
   0.0178571, 0.0138889, 0.0111111, 0.00909091, 0.00757576, 0.00641026, \
   0.00549451, 0.0047619, 0.00416667, 0.00367647, 0.00326797, \
   0.00292398, 0.00263158, 0.00238095, 0.0021645, 0.00197628, \
   0.00181159, 0.00166667, 0.00153846, 0.0014245, 0.00132275, \
   0.00123153, 0.00114943, 0.00107527, 0.00100806, 0.00094697, \
   0.000891266, 0.000840336, 0.000793651}.*)

      1/tPlain

      (* {2., 6., 12., 20., 30., 42., 56., 72., 90., 110., 132., \
     156., 182., 210., 240., 272., 306., 342., 380., 420., 462., 506., \
     552., 600., 650., 702., 756., 812., 870., 930., 992., 1056., 1122., \
     1190., 1260.}.*)

.

My notebook is attached. It makes a little more sense to the pattern of differences, showing they are the same whether you use h[x,16] or h[x,15]. A quick mental proof shows its true for all h[x,{1,2,3...}].

Attachments:

Not so miraculous!

On second thought this pattern is not so special! When b=16, the b's were held constant, so the a's were the only free variable, and the pattern behaved accordingly! Sorry for bothering you all.

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