## 495 Reputation

14 years, 338 days

I've been using Maple since 1997 or so.

## New record computation...

I would like to announce a new unofficial record computation of the MRB constant that was finished on Sun 21 Sep 2014 18:35:06.

I really would like to see someone beat it with Maple!

It took 1 month 27 days 2 hours 45 minutes 15 seconds. I computed 3,014,991 digits of the MRB constant, (confirming my previous 2,00,000 or more digit computation was actually accurate to 2,009,993 digits), with Mathematica 10.0. I Used my version of Richard Crandall's code:

____________________________________________________________________________

(*Fastest (at MRB's end) as of 25 Jul 2014.*)

DateString[]

prec = 3000000;(*Number of required decimals.*)ClearSystemCache[];

T0 = SessionTime[];

expM[pre_] :=

Module[{a, d, s, k, bb, c, n, end, iprec, xvals, x, pc, cores = 12,

tsize = 2^7, chunksize, start = 1, ll, ctab,

pr = Floor[1.005 pre]}, chunksize = cores*tsize;

n = Floor[1.32 pr];

end = Ceiling[n/chunksize];

Print["Iterations required: ", n];

Print["end ", end];

Print[end*chunksize]; d = ChebyshevT[n, 3];

{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" -> 4]];

ctab = ParallelTable[Table[c = b - c;

ll = start + l - 2;

b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));

c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 2];

s += ctab.(xvals - 1);

start += chunksize;

Print["done iter ", k*chunksize, " ", SessionTime[] - T0];, {k, 0,

end - 1}];

N[-s/d, pr]];

t2 = Timing[MRBtest2 = expM[prec];]; DateString[]

Print[MRBtest2]

MRBtest2 - MRBtest2M

_________________________________________________________________________.

I used a six core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz with 64 GB of RAM of which only 16 GB was used.

t2 From the computation was {1.961004112059*10^6, Null}.

## Special bounded partial sums...

For all real a, the partial sums sn= sum((-1)^k (k^(1/k) -a), k=1..n) are bounded so that their limit points form an interval [-1.+  the MRB constant +a, MRB constant] of length 1-a, where the MRB constant is limit(,N=infinity).

For all complex z, the upper limit point of  sn= sum((-1)^k (k^(1/k) -z), k=1..n) is the  the MRB constant.

We see that maple knows the basics of this because when we enter sum((-1)^k*(k^(1/k)-z), k = 1 .. n)

maple gives

marvinrayburns.com

## Solitaire Numbers...

This is a little more than a new game it potentially uncovers a new class of numbers -- though determining membership might become a hard problem.

A number that possesses the solitaire property can be written in as ...,0,...1,...2,...etc, or ...,0,...1,...10...11,...etc,(where the "0" is the first zero in the number), with a radix point anywhere. We are free to pick the base and say it is solitaire with respect to that base. After the initial 0, the subsequent ordinals (the 1,2, etc or the 1,10,11, etc) used to write the solitaire number don't have to be the first ones. For example:

pi=3.1415926535897932384626433832795
0 2884
1 971693993751058
2 0974944592
3 078163860
4 ...
etc.,

or

pi=3.1415926535897932384626433832795

0 2884197

1 6939937510582 097494459

2 3 07816

3 860
4 ...
etc.,are both acceptable. (If the number can be written as  ...,0,...1,...2,...etc, or ...,0,...1,...10...11,...etc. it is solitaire.)

The Champernowne constant with respect to base 10 has only one representation:

0.

1

2

3

4

5

6

7

8

9

10

11...

etc. .

I know Base 10 Champernowne constant is base 10 solitaire. I can not say the same with certainty for Pi.

I also propose we can measure the solitude of a number by the average amount of numbers between the 0,1,2,3..., and give a perfect solitude score to Base 10 Champernowne constant. Other constants can be given additional credit, of some kind, if the amounts of numbers between the 1,2,3... follow a specific preset pattern.

marvinrayburns.com

## A nested infinite series...

It seems that

Limit(N+(sum((-1)^n*Sum(1/n^x, x = 1 .. N), n = 1 .. infinity)), N = infinity)=log(2)

evalf(300+sum((-1)^n*(Sum(1/n^x, x = 1 .. 300)), n = 1 .. infinity), 30)

gives

0.693147180559945309417232121.

sum(1/n^x, x = 1 .. infinity)

gives

1/(n-1).

## sin(10^-k)...

Maple

There seems to be patterns for sin(10^-k) for rational k;

Here we have the "floats."

n sin(10^(-n-1/2))

1 0.03161750640

2 0.003162272390

3 0.0003162277607

4 0.00003162277660

5 0.000003162277660

6 0.0000003162277660

7 0.00000003162277660

More later on using the mantissa. You're welcome to join me.

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