You can replace the "1/x" with "any rational number/x" if you also replace the "^x" with "^one of the right rational numbers*x."

x        (10 Mantissa[sin(10^(-100 -3.14159260000000000/x))])^(5000000*x)

1       1.*  10^4292037

2       1.*  10^4292037

3      1.*  10^14292037

etc.

Here we use the mantissa.

Noticing that 3.162277660^2~=10 we have here a more subtle and beautiful pattern for sin(10^-k), using sufficiently large integral value for k; here we use 100 but 9 is usually sufficient.

x        (10 Mantissa[sin(10^(-100 - 1/x))])^x

1       1.*  10^1

2       1.*  10^1

3       1.* 10^2

4       1.* 10^3

5       1.* 10^4

6       1.* 10^5

etc.

x        (10 Mantissa[sin(10^(-100 - 2/x))])^x

1        1.* 10^0

2        1.* 10^2

3       1.* 10^1

4       1.* 10^2

5       1.* 10^3

6       1.* 10^4

etc.

x        (10 Mantissa[sin(10^(-100 - 3/x))])^x

1       1.*  10^1

2       1.*  10^1

3      1.*  10^3

4      1.*  10^1

5      1.*  10^2

6      1.*  10^3

etc.

x        (10 Mantissa[sin(10^(-100 - 4/x))])^x

1       1.*  10^1

2       1.*  10^0

3      1.*  10^2

4      1.*  10^4

5      1.*  10^1

6      1.*  10^2

7      1.*  10^3

etc.

x        (10 Mantissa[sin(10^(-100 - 5/x))])^x

1       1.*  10^1

2       1.*  10^1

3      1.*  10^1

4      1.*  10^3

5      1.*  10^4

6      1.*  10^1

7      1.*  10^2

8      1.*  10^3

etc.

The Mathematica code for this is 

Table[{x, (10 MantissaExponent[N[Sin[10^(-100 - 1/x)], 10]][[1]])^  x}, {x, 1, 10}] // TableForm

Change 1/x to 2/x,3/x, etc .

Can anyone figure out the pattern here?

Replacing 1/x with (3/2)/x  and ^x to ^(2x) we find

x        (10 Mantissa[sin(10^(-100 - (3/2)/x))])^(2x)

1       1.*  10^1

2      1.*   10^1

3      1.*  10^3

4      1.*  10^5

5      1.*  10^7

6      1.*  10^9

etc.

Replacing 1/x with (5/2)/x  and ^x to ^(2x) we find

x        (10 Mantissa[sin(10^(-100 - (5/2)/x))])^(2x)

1      1.*   10^1

2      1.*   10^3

3      1.*  10^1

4      1.*  10^3

5     1.*   10^5

6     1.*   10^7

etc.

Replacing 1/x with (5/3)/x  and ^x to ^(3x) we find

x        (10 Mantissa[sin(10^(-100 - (5/3)/x))])^(3x)

1      1.*   10^1

2      1.*   10^1

3      1.*  10^4

4      1.*  10^7

5      1.*  10^10

6      1.*  10^12

etc.

Another partial repeating decimal sequece:

n=22,25,28,30,31,33,34,36,37,39,40,42,43,45,46,48,49,51,52,54,55,57,58,60,61,

63,64,66,67,69,70,72,73,75,76,78,79,81,82,84,85,87,88,90,91,93,94,96,97,99,100 and some others 

give 41991753102864213975325086436197547308658.

Two more repeating decimal sequences are

864213975325086436197547308658419769530880641991753102

and

038965150076261187372298483409594520705631816742927854

There is at least one more that has the digits 5085671481 in it.

The Taylor series of sin(x) about x=0 is 

x-x^3/6+x^5/120-x^7/5040+x^9/362880-x^11/39916800+x^13/6227020800-x^
15/1307674368000+x^17/355687428096000-x^19/121645100408832000+O(x^20)

Can anyone describe the pattern here?

marvinrayburns.com

Let c be the Hafner-Sarnak-McCurley Constant = 0.3532363718549959845435165504326820112801... and g= Gamma[Pi] = 2.28803779534003241795958890906023392289...

m/g - (30780 c - 10003)/(7 (37 c + 1500)) = -7.99658771873458...*10^-21

I wish I knew how Wolfram Alpha does it! 

I tried reposting the table in my comment but it didn't work. So I just made the corrections in the table

 Here a sample Mathematica code I used:

WolframAlpha["7^6^5^4^3^2", {{"LastFewDigits", 1}, "Plaintext"}]

It returned

"...9058585601"

Good catch. The ones with only 9 digits have a missing first or last displayed digit. I replaced the missing digits.

I computed a little over 1,200,000 digits of the MRB constant in 11 days, 21 hours, 17 minutes, and 41 seconds.

Ricard Crandall computed 1,048,576 digits in a lighting fast 76.4 hours.

Here are my best timings of computing digits of the MRB constant using Crandall's method in seconds:

I use both Maple and Mma, but I'm a student so licensing is not very expensive. For example, with a few exceptions, I use Maple for convenience (right click options, typesetting and teaching other people what I discovered, and tutors and assistants), while I use Mma as the workhorse. In general I try to use Maple where it is stronger and Mma where it is stronger.

I feel like a child whose parents are always fighting! One way to see through the all the marketing is to look closely into the documentation, and try some things out for yourself. 

I checked the above formulas with greater precision and they seem to be prety accurate.

Perhaps this isn't important to the general public, but I would like to know if Maple is using the convergent Sum((-1)^n*(n^(1/n)-1)) on purpose to give a result for those divergent series? or does maple's "summing" just happen to fit the identies I described?

marvinrayburns.com

Actually some of the above calculations are off in Maple. I will find out if  it is my formulae or Maple that is in error!