marketingcoin
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January 06, 2016, 04:09:27 PM |
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Its an easy way to check whether the first bit is always 1 for a given step, which it is. Thus you can limit the search space. For step n you do not need to search for any possible solutions from step n-1
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BurtW
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January 06, 2016, 04:13:09 PM |
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Its an easy way to check whether the first bit is always 1 for a given step, which it is. Thus you can limit the search space. For step n you do not need to search for any possible solutions from step n-1
This was known from page 1 of this thread.
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BurtW
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January 06, 2016, 04:18:26 PM |
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EDIT: this formula doesn't make any sense I know, just playing around Just for fun the average of this: 1.00000000 1.50000000 1.75000000 1.00000000 1.31250000 1.53125000 1.18750000 1.75000000 1.82421875 1.00390625 1.12792969 1.31005859 1.27343750 1.28710938 1.63983154 1.57196045 1.46214294 1.51572418 1.36388779 1.64664650 1.72783279 1.43408918 1.33485842 1.72003222 1.97801048 1.62538475 1.66818412 1.69600850 1.49275696 1.92441434 1.95800192 1.44051053 1.66181426 1.64530614 1.17072322 1.23364647 1.45885221 1.06935867 1.17770458 1.82563129
Is 1.482530615 So starting at 1.482530615 into the next range and doing a butterfly search might be a tiny bit faster.
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Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!
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BurtW
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January 06, 2016, 04:22:56 PM |
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Maybe these random-like numbers came from hashing operations? For example: n-th key = truncate(SHA256(f((n-1)-th key))). It will be still hopeless if the process involves a strong passphrase though.
I think this would be a good way to do it. Then the author just needs to remember the seed and algorithm instead of remembering 256 private keys. Although keeping track of 256 private keys is not that hard in the first place. The algorithm would need to "skip over" any private keys in the generated sequence that had the undesirable result of a 0 in the most significant bit after the masking operation.
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Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!
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dalek
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January 06, 2016, 04:44:46 PM |
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So starting at 1.482530615 into the next range and doing a butterfly search might be a tiny bit faster.
butterfly search... so the numbers were generated using chaos theory
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tarrant_01
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January 06, 2016, 04:56:57 PM |
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I noticed that the results for x increase then reset every 3 steps, 0-2, 3-5, 6-8, etc...except for 15-21, this pattern continues when n > 21. Just a simple observation, hopefully helpful, probably not though. Here are the results for the values we know given those 2 mentioned formulas: y = 2^n * x and so x = y / 2^n AND y = 2^n + x and so x = y - 2^n n | Known Results (y) | x = y / 2^n | x = y - 2^n | 0 | 1 | 1.00000000 | 0
| 1 | 3 | 1.50000000 | 1
| 2 | 7 | 1.75000000 | 3
| 3 | 8 | 1.00000000 | 0
| 4 | 21 | 1.31250000 | 5
| 5 | 49 | 1.53125000 | 17
| 6 | 76 | 1.18750000 | 12
| 7 | 224 | 1.75000000 | 96
| 8 | 467 | 1.82421875 | 211
| 9 | 514 | 1.00390625 | 2
| 10 | 1155 | 1.12792969 | 131
| 11 | 2683 | 1.31005859 | 635
| 12 | 5216 | 1.27343750 | 1120
| 13 | 10544 | 1.28710938 | 2352
| 14 | 26867 | 1.63983154 | 10483
| 15 | 51510 | 1.57196045 | 18742
| 16 | 95823 | 1.46214294 | 30287
| 17 | 198669 | 1.51572418 | 67597
| 18 | 357535 | 1.36388779 | 95391
| 19 | 863317 | 1.64664650 | 339029
| 20 | 1811764 | 1.72783279 | 763188
| 21 | 3007503 | 1.43408918 | 910351
| 22 | 5598802 | 1.33485842 | 1404498
| 23 | 14428676 | 1.72003222 | 6040068
| 24 | 33185509 | 1.97801048 | 16408293
| 25 | 54538862 | 1.62538475 | 20984430
| 26 | 111949941 | 1.66818412 | 44841077
| 27 | 227634408 | 1.69600850 | 93416680
| 28 | 400708894 | 1.49275696 | 132273438
| 29 | 1033162084 | 1.92441434 | 496291172
| 30 | 2102388551 | 1.95800192 | 1028646727
| 31 | 3093472814 | 1.44051053 | 945989166
| 32 | 7137437912 | 1.66181426 | 2842470616
| 33 | 14133072157 | 1.64530614 | 5543137565
| 34 | 20112871792 | 1.17072322 | 2933002608
| 35 | 42387769980 | 1.23364647 | 8028031612
| 36 | 100251560595 | 1.45885221 | 31532083859
| 37 | 146971536592 | 1.06935867 | 9532583120
| 38 | 323724968937 | 1.17770458 | 48847061993
| 39 | 1003651412950 | 1.82563129 | 453895599062
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1P95gCUCw3Tjb7yyoYtW3ARZZQyTpFgk6H
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Bulista (OP)
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January 06, 2016, 07:55:04 PM |
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I noticed that the results for x increase then reset every 3 steps, 0-2, 3-5, 6-8, etc...except for 15-21, this pattern continues when n > 21. Just a simple observation, hopefully helpful, probably not though. Here are the results for the values we know given those 2 mentioned formulas: y = 2^n * x and so x = y / 2^n AND y = 2^n + x and so x = y - 2^n n | Known Results (y) | x = y / 2^n | x = y - 2^n | 0 | 1 | 1.00000000 | 0
| 1 | 3 | 1.50000000 | 1
| 2 | 7 | 1.75000000 | 3
| 3 | 8 | 1.00000000 | 0
| 4 | 21 | 1.31250000 | 5
| 5 | 49 | 1.53125000 | 17
| 6 | 76 | 1.18750000 | 12
| 7 | 224 | 1.75000000 | 96
| 8 | 467 | 1.82421875 | 211
| 9 | 514 | 1.00390625 | 2
| 10 | 1155 | 1.12792969 | 131
| 11 | 2683 | 1.31005859 | 635
| 12 | 5216 | 1.27343750 | 1120
| 13 | 10544 | 1.28710938 | 2352
| 14 | 26867 | 1.63983154 | 10483
| 15 | 51510 | 1.57196045 | 18742
| 16 | 95823 | 1.46214294 | 30287
| 17 | 198669 | 1.51572418 | 67597
| 18 | 357535 | 1.36388779 | 95391
| 19 | 863317 | 1.64664650 | 339029
| 20 | 1811764 | 1.72783279 | 763188
| 21 | 3007503 | 1.43408918 | 910351
| 22 | 5598802 | 1.33485842 | 1404498
| 23 | 14428676 | 1.72003222 | 6040068
| 24 | 33185509 | 1.97801048 | 16408293
| 25 | 54538862 | 1.62538475 | 20984430
| 26 | 111949941 | 1.66818412 | 44841077
| 27 | 227634408 | 1.69600850 | 93416680
| 28 | 400708894 | 1.49275696 | 132273438
| 29 | 1033162084 | 1.92441434 | 496291172
| 30 | 2102388551 | 1.95800192 | 1028646727
| 31 | 3093472814 | 1.44051053 | 945989166
| 32 | 7137437912 | 1.66181426 | 2842470616
| 33 | 14133072157 | 1.64530614 | 5543137565
| 34 | 20112871792 | 1.17072322 | 2933002608
| 35 | 42387769980 | 1.23364647 | 8028031612
| 36 | 100251560595 | 1.45885221 | 31532083859
| 37 | 146971536592 | 1.06935867 | 9532583120
| 38 | 323724968937 | 1.17770458 | 48847061993
| 39 | 1003651412950 | 1.82563129 | 453895599062
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Very interesting observation!!! Thank you for sharing
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vilain
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January 06, 2016, 08:02:28 PM |
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Nice colletive work y'all! Gonna try to use matlab to make some intergers experiments.
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tarrant_01
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January 06, 2016, 08:05:27 PM |
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31-33 doesn't follow the pattern either. I missed that.
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1P95gCUCw3Tjb7yyoYtW3ARZZQyTpFgk6H
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Bulista (OP)
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January 06, 2016, 08:36:47 PM Last edit: January 06, 2016, 08:51:04 PM by Bulista |
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31-33 doesn't follow the pattern either. I missed that.
Yes, I see, but it was well observed. Here is a chart of those values with the ups and downs for x = y / 2^n: EDIT: and btw, this is the sequence: and this is x = y - 2^n:
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Bulista (OP)
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January 06, 2016, 09:00:38 PM |
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-snip-
I think that what you have done is to pretty much prove it is random and there is no predictive formula. I'm not sure. I got some infos that are getting me to believe that there is a possible formula behind it. I give you an example. I'm playing around by creating random formulas, and I get pretty much similar results. Yet they are predictable with a formula. All I use for inputs are 2 arrays, one with the current position and another with the sequential list of prime numbers. For example: Consider n = count, p = prime numbers, and y = sequence based on the formula 2^n + (n mod p) * Log(n+1, 2) <--- Random formula I invented. y / 2^p + 1 and y-2^p *-1 are similar formulas to what was shown before for the var x in the real sequence, their results also appear to be random... n | p | y = 2^n + (n mod p) * Log(n+1, 2) | y / 2^p + 1 | y-2^p *-1
| 0 | 2 | 1 | 1.25000000 | 3
| 1 | 3 | 3 | 1.37500000 | 5
| 2 | 5 | 7 | 1.22406016 | 25
| 3 | 7 | 14 | 1.10937500 | 114
| 4 | 11 | 25 | 1.01234752 | 2023
| 5 | 13 | 45 | 1.00548399 | 8147
| 6 | 17 | 81 | 1.00061679 | 130991
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Yet this sequence is breakable with a simple formula. EDIT: this formula doesn't make any sense I know, just playing around Am I the only one noticing that this guy makes a random formula using prime numbers and his first 3 results are exactly the ones as the sequence for this puzzle? It seems pretty obvious that there is some formula using prime numbers behind this, no? And possibly the formula he posted is not so far from the real one. Keep up the good work all! Yea, I was actually surprised by those 3 first results since I was genuinely trying random stuff. Maybe with some tweaks we can get to the right formula, if there is indeed one One thing I'm sure if we stop looking we won't know.
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shorena
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No I dont escrow anymore.
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January 06, 2016, 09:23:55 PM |
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-snip- EDIT: and btw, this is the sequence: and this is x = y - 2^n: Given the numbers we are dealing with you should probably use a log scale.
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Im not really here, its just your imagination.
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tarrant_01
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January 06, 2016, 11:20:22 PM |
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Is there any way to tell what time an address was generated? I'm asking because the results for x = y / 2^n could be converted to a time of day.
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1P95gCUCw3Tjb7yyoYtW3ARZZQyTpFgk6H
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Gleb Gamow
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January 07, 2016, 12:42:32 AM |
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Is it possible that the entity who created this is seeking employees capable of cracking it for its endeavor? I recall the NSA (or some other) putting out a contest, with the reward for solvers to become employees. Just a thought.
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Bitcoinpro
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January 07, 2016, 01:20:37 AM |
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The transactions seem to originate from address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT which had a Total received amount of 56,457.80848111 BTC and a Final Balance of 312.04932734 BTC
Definitely a very big player of some kind.
a player of one special word its called bs their is no sequence i dont think spam thread why would u try to solve a sequence of random numbers found in a transaction does the question even makes sense to anyone it may be some kind of idiot test
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CRYPTOCURRENCY CENTRAL BANK
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amaclin
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January 07, 2016, 06:10:03 AM |
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why would u try to solve a sequence of random numbers Because everyone reads only the subject of this topic it may be some kind of idiot test This board? The whole bct.o? Yes.
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Bulista (OP)
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January 07, 2016, 08:02:30 AM |
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The transactions seem to originate from address 173ujrhEVGqaZvPHXLqwXiSmPVMo225cqT which had a Total received amount of 56,457.80848111 BTC and a Final Balance of 312.04932734 BTC
Definitely a very big player of some kind.
... their is no sequence i dont think ... if you dont know if there is sequence or not, why are you talking shit?
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amaclin
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January 07, 2016, 08:05:31 AM |
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if you dont know if there is sequence or not, why are you talking shit? Because this is public resource and anyone can say whatever he wants
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Bulista (OP)
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January 07, 2016, 09:23:49 AM |
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if you dont know if there is sequence or not, why are you talking shit? Because this is public resource and anyone can say whatever he wants Good point Is there any way to tell what time an address was generated? I'm asking because the results for x = y / 2^n could be converted to a time of day.
That's not possible
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Bulista (OP)
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January 07, 2016, 12:50:37 PM |
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-snip- EDIT: and btw, this is the sequence: and this is x = y - 2^n: Given the numbers we are dealing with you should probably use a log scale. Log 2: http://imgur.com/8aW8tj6http://imgur.com/dGguJcK
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