If I calculated everything correctly, then the space from 2000000000000001 to 4000000000000001 can be scanned
in about 545 years if the average key generation rate is about 134 Mkey s

Confirm or disprove my calculations.

The space is hexadecimal!
2000000000000000 hex = 2305843009213693952 decimal

134000000 Keys/ sec

199164.1626255609065782 days

545.6552400700298810362 years

On a background 545 years an error per 0,5 years to make all any +/- 0,1 percents

These horror stories about the number of atoms in the universe, are used for embellishment (although the numbers are large and relatively puzzle at this stage justifiably look for 1 needle in a haystack).

Let's look at the address numbers, on the number of repetitions of numbers.

1    2    3    4    5    6    7    8   9    0

8    6   8     6   9     9    6    7   6   12   88076670097206135214755096908866554407314546020283163510152690884233313095179
4    5   5    14   7     5   11   9   5   12   57701484440690220081460730827720988043707783650447848844644231359596471509755
9    7   13   3    10   13  5    8    0   9    56852111307365822553680728638432355135106461736008687516183006036037433215266
7    5   11   7    12   8   7    9   5     6    83144555669307175352868736014304552664501628893338700237812991649157337855485
11  11   7   5    11   6    6   7   5     9    105951512903139530827431508910247856221386514825762420570832790562065784213611
3    9    8   4     5   10   9  10   4    14   2800743280837621873504738280285669010665021097040374687382697063985620056327
10   9   6   7     8    6    7   7   9     8    15155401853110903276292971961426024213516891095729877044987768503568484329302
6    8    4   10  11   7    8   9    8     5   1934145264352640584468993215160278462974765570385286888570941520948795715972
4    8   10  10   6    7    8  10   2    12   28808842000054338759447213206875406362693076403134586314430882247307076152875

there are numbers in which numbers may be absent altogether (1,2,3,4,5,6,7,8,9 = 0)

13   6   6   10   0   7   7    11  11   6   14696118311433422644887998880270171111886071679929423836479971639009928414084
11   4   0   10   6   7   12  10  10   7    84167587877791947619455975590804189912148604940284674067142717911820851066978
0    7    9   9    9   7   8    16   6    6   85056283424685094879758489628438532598948587847835803373846697232020876056473
6    9   10   9   5   10  11  10   7    0   35778492181327261174949468738285433366625387652824764739658862179647493738129

or present within 20 (very rarely 21)

20  12   4   7   5   5   6   8   5   5   14168167265037749111145324098522827112187124998114101721323801125686219581042
20   3  10   8   5   7   3   9   7   5   74210210994663119334888205459886331564143036385918170716141191111683131845191
6    3   5    5   8   4   20  6  11  7   671715073647425809453041797737097177653288097773772751099955845997107768998
10   4  10   8   9   3   4   4  20   5   91330834709955852129495199148324191949367391795990099341407196639595153298354
6   3  11   5  12  20   6   6   3   5    63563038544616496835516983610048686335751646557506626310787196666355232735676

Now look at the distribution of 40000 addresses https://imgur.com/pWdzq9T

as can see, some repeat more 6 7 8 9...

the maximum length of the number we have 115792089237316195423570985008687907852837564279074904382605163141518161494336 = 78 signs

roughly speaking 78:10 = 7 ,ie numbers can be 1,2,3,4,5,6,7,8,9,0 on 7 for mixing 1111111222222233333334444444555555566666667777777888888899999990000000. (in theory, 20-30 percent of Satoshi’s addresses should be there)

generally within length 77-78 signs numbers constituting number can float 1,2,3,4,5,6,7,8,9,0-0,9,8,7,6,5,4,3,2,1 = 3-12  

3-12 or 4-13 a lot of us, we can retreat a little from the middle (7-7-7-7-7-7-7-7-7-7) 5-10 or 6-8 etc...

(and the same sequence if you cut off the first 78 characters of degrees 2)    

Yes ,but you, my dear friend, forget that cryptography is exact mathematics ,and it does not take into account that in nature there are collisions and accidents.As soon as hundreds of millions of people are not miners, and players in the lottery <find private key > you will be surprised how many coincidences appear on the way to 264 bit keys! The Creator of Satoshi realized it 9 years ago and quietly merged with the crowd.In order not to incur the wrath of people who believe in the steadfastness of bitcoin and resistance n bust.

Well, it depends on which side to look at.
And as in a joke.

Blondie is asked how likely she is to see a live dinosaur on the street.
She answers, 50 percent!
Ask how it is ??
Well, either I will see, or I will not see!

If we speak within this puzzle. Then, having some "random" private key generation statistics, we can try to predict in which range to search for the next key. Purely mathematical analysis. At the same time, different values, repetition in bit form, in decimal, hexahedral, etc. can take part in the analysis. Either repetition, or the number of all numbers and sequences. And first, more possible keys are checked, and then less possible.

If we talk about all keys in general, then on the basis of already known private keys it will not be possible to analyze. Since different keys were generated by different software products, with different random generators, some were generated by user actions (pull the mouse, enter a phrase), others are based on random words, etc.
That is, it cannot be said that such a private key cannot exist in the current range. They are all valid in the known range.

But! They are not all valid for the required public keys or addresses! Then here it is necessary to analyze the frequency repetitions of the received keys or addresses. To find a certain dependence and then the search is divided into two levels, the search for matches, the transition to the next range of possible matches, the search by small steps until the next match is found.

But the search for these hits also takes time and power. It also reduces the speed of the search, as it is necessary to carry out an additional analysis of the result obtained.For example. The probability that in the range of 1 trillion private addresses on the output will be two addresses with a difference of only 1,2,3 bytes, almost zero. And if we have found such an address, which matches the desired one with a difference of 1,2,3 bytes. Then with great probability we can continue the search by adding 1 trillion to the current private key.

Why steping is 5???
Maybe reality collision found?

How many BTCs left in this PUZZLE ?

Please read the thread.  Go back a few pages and read forward until you find the answer to your question.

Looks like someone has found some kind of trick or regularity in calculations...
Definetely it can't be luck with stepping 5 and and can't be hashing or brute-forcing. Just only if pentagon's admin decided to test the most powerfull PC in the world with this puzzle address