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Old 2022-05-11, 23:50   #1
Graph2022
 
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Minus Pseudo Fact - Last Digits of Mersenne Primes

Hi,

See attached.
4423 and 43112609(marked in blue) are only exponents so far, where the last 39 digits lacked one of the digits.

4423 had no 4s
43112509 had no 3s
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Old 2022-05-18, 02:31   #2
Dr Sardonicus
 
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Quote:
Originally Posted by Graph2022 View Post
Hi,

See attached.
4423 and 43112609(marked in blue) are only exponents so far, where the last 39 digits lacked one of the digits.

4423 had no 4s
43112509 had no 3s
It occurred to me to wonder whether there is a k for which all k-digit endings of 2^n (with k > 9 and 2^n > 10^(k-1) of course) contain all 10 decimal digits.

I suspect there is, but don't have a clue how to approach the question.

I did a bit of checking. For k = 100, lift(Mod(2,10^100)^1088) lacks the decimal digit 2, so k > 100. Lift(Mod(2,10^100)^n) lacks at least one decimal digit for 30 values of n, 334 < n < 100000. This does not fill me with optimism regarding a brute-force assault on the question.

I checked for k = 200 up to n = 10^7 without finding any 200-digit endings lacking any of the decimal digits, but 10^7 is a tiny, tiny fraction of 4*5^199, the multiplicative order of 2 (mod 5^200).

The largest power of 2 I know of whose complete decimal expansion lacks any of the decimal digits is

2^168 = 374144419156711147060143317175368453031918731001856

which lacks the decimal digit 2.

Is it in fact the largest? I don't have a clue how to approach that question, either.
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Old 2022-05-19, 16:00   #3
Graph2022
 
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Extra pseudo-fact: I tried to analyze those digits under many amateur criteria, and there is one sequence that doesn't show up in any of those 40 Mersenne last 39-digit numbers. It only cuts work by 1/8th but still (Only need to check exponents that hold the sequence after last 39 digits are generated).

With this being said, can we treat last n-digits of Mersenne primes as random digits to see if there is any oddity even at small sample size of 40 #s? P.S. Obviously I'm not including 1st digit (last in fact) as it will be odd :)

Last fiddled with by Graph2022 on 2022-05-19 at 16:03
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Old 2022-05-19, 21:33   #4
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Quote:
Originally Posted by Graph2022 View Post
It only cuts work by 1/8th but still ...
It only cuts what work by 1/8th?
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Old 2022-05-20, 18:06   #5
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That only 1/8th of all exponents needs to be checked. This is mostly for people that like to hit that 100 Million or 1 Billion digit mark or even possibly 10 Billion (who knows when it hits LOL).
I assume that eventually it needs to happen. IF not, i'm curious why it doesn't happen (that sequence) that haven't happened for 40 times in a row(assuming that last 40 digits of Mersenne #s are ,,random").

Last fiddled with by Graph2022 on 2022-05-20 at 18:09
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Old 2022-05-20, 23:46   #6
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Extra pseudo fact. This time on bellwether.

M9689 and M25,964,951
have a 17 streak ( the longest so far) where digits (whether odd or even) correspond to each other (aka Bellwether).
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Old 2022-05-20, 23:50   #7
chalsall
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Quote:
Originally Posted by Graph2022 View Post
Extra pseudo fact. This time on bellwether.
Are you familiar with the concept of autogratification?
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Old 2022-05-21, 00:00   #8
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Quote:
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Are you familiar with the concept of autogratification?
Lol, no i do not. Don't worry though, I already requested for thread to be removed so that you are happier :)
For some reason OP of the thread can't remove it after certain time, or i somehow can't see the option.
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Old 2022-05-21, 13:58   #9
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Quote:
Originally Posted by Graph2022 View Post
<snip>
With this being said, can we treat last n-digits of Mersenne primes as random digits to see if there is any oddity even at small sample size of 40 #s?
No. The number of k-digit blocks of decimal digits is 10^k. The number of k-digit endings of 2^n, n ≥ k, is 4*5^(k-1). (4*5^(k-1))/10^k is a tiny, tiny fraction if k is at all large. For k = 40, it's about 7.276 * 10-13.

As mentioned in this recent post, a Mersenne prime can terminate in at most 3 1's.
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