M(M(127))

[ET_]

Originally Posted by clowns789

It's in MM127 Checkout Page.

Also, after I finish in Operation Billion Digits, do you think I should do Factor3_2 from 174 to 175? How long would this take?

It wouldn't take very long, but I think someone else has already checked that range with MFAC (that is faster).

Luigi

[Uncwilly]

Originally Posted by clowns789

It's in MM127 Checkout Page.

I could not find any mention by me of the size of (M)MM127 (other than that they are huge. I did find biwema mentioning in this post that

Originally Posted by biwema

You can print this number, if you like, but if you pack 50000 digits on a A4 page with a very small font, you need 10^33 pages

[jinydu]

Does anyone know why Terrence Law keeps posting repetitive messages about proving that MM127 is prime?

[clowns789]

Originally Posted by jinydu

Does anyone know why Terrence Law keeps posting repetitive messages about proving that MM127 is prime?

It must have something to do with him wanting us to quit trying to factor it and believe the mystery over whether or not it is really factored.

By the way, how much faster is MFAC for this? It was odd because it took 10 mins on a PII.

[antimath]

Originally Posted by Terrence Law

I calculated the first digits of MM127 using Calculator Applet. It starts with 54543129001957378843931..., because the decimal part starts with .73674004870103894404352319788994...

I calculated the last digits of MM127 using Jim howell's program, modulus a power of 10, like (2^(2^b-1)-1)%(10^n), where b equals 127 and n equals to a random number I select.

For example, the last 56 digits are 58106222489358628350506357958045615739087358437493833727.

hey man can you tell me how to calculate ALL of MM127? if possible... thx

[antimath]

Originally Posted by jinydu

M127 = 2^127 - 1 which has 39 digits.

MM127 = 2^(M127) - 1 = 2^(2^(127) - 1) - 1 which has over 5.12*10^37 digits. If each digit was stored by a single electron, all the electrons put together would weigh over 4.66*10^7 kg.

MMM127 = 2^(MM127) - 1 = 2^(2^(M127) - 1) - 1 = 2^(2^(2^(127) - 1) - 1) - 1). The number of digits is approximately (log 2) * (MM127). That's over 10^(10^37) digits. MMM127 is far larger than a googolplex. Proving that MM127 is prime (assuming that it is) would require a new primality test, far more powerful than the LL test.

From the notation at http://www.utm.edu/research/primes/mersenne/

C(4) = M127
C(5) = MM127
C(6) = MMM127

Larger numbers in the sequence are so large that the system of exponential notation becomes unwieldly, and we have to switch to "tetration".

(10 tetra 1) = 10
(10 tetra 2) = 10^10
(10 tetra 3) = 10^(10^10)
(10 tetra 4) = 10^(10^(10^10))
etc.

can u tell me the method used to calculate the number of digits are in a number...the way you found the number of digits in MM127?

[LaurV]

Originally Posted by antimath

can u tell me the method used to calculate the number of digits are in a number...the way you found the number of digits in MM127?

Why do you wake up a thread 8 years old? There are much newer discussions about this subject, generally all the same, one new guy coming and pretending he reinvented the wheel, and another 20 guys explaining that his wheel is squared.

And to answer your question, M127 has 39 digits, you can compute it with every integer calculator that has higher precision, there are hundreds on the web. I used pari/gp to get: 2^127-1=170141183460469231731687303715884105727.
Then you have MM127 is a number whose binary representation has this amount of bits. You use logarithms to express this number of bits in number of digits. The logarithm in a base B of a number N is the number of figures used in that base to represent the number. For example decimal logarithm of 5678 is about 3.x, that is, you need 4 digits to represent this number. Binary logarithm of 83 is 6.x, you need 7 bits to represent 83 in binary.
See the definition of logarithm on the web. Then see the properties of logarithms, saying that log(a^n)=n*log(a), and log(x,b)=log(x,c)/log(b,c) and apply them to:

2^M127=10^x

Left side is your number, right side is the approximative number of digits (powers of 10). Apply logarithms for both sides and solve for x giving:
log(2^M127)=log(10^x)
or
M127*log(2)=x*log(10)
or
x=M127*log(2)/log(10)
or
x=M127*0.3010299956639811952137388947
or
x=5.121*10^37 digits.

Even if you store a number with 20 digits in a single atom, you still need about billions of tones of matter to store this number.

[retina]

Originally Posted by LaurV

... 5.121*10^37 digits.

Even if you store a number with 20 digits in a single atom, you still need about billions of tones of matter to store this number.

Perhaps we will be lucky enough to have xilman explain about how to store the entire number in a single atom?

[xilman]

Originally Posted by retina

Perhaps we will be lucky enough to have xilman explain about how to store the entire number in a single atom?

Nah, technology has moved on since then.

[Batalov]

...since 1909?

Originally Posted by V. I. Lenin, 1909, Poln. sobr. soch., 5th ed., vol. 18, p. 277

...The ’essence’ of things, or ’substance,’ is also relative; it expresses only the degree of profundity of man’s knowledge of objects; and while yesterday the profundity of this knowledge did not go beyond the atom, and today does not go beyond the electron and ether, dialectical materialism insists on the temporary, relative, approximate character of all these milestones in the knowledge of nature gained by the progressing science of man. The electron is as inexhaustible as the atom, nature is infinite.

The only F I had in my diploma was in Dialectical materialism (later re-taken for a B, to get cum laude... oh, vanity of vanities, all is vanity), and a C in CPSU History. The rest were A's.

The Dialectical materialism A students, it seems, are still in power over there. :-]

[ewmayer]

Ah, yes, the esteemed Mr. Lenin ... the man did so love to hear himself write.