Bitcoin puzzle transaction ~32 BTC prize to who solves it

[marketingcoin]

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

[1714767572]

1714767572

[1714767572]

1714767572

[1714767572]

1714767572

[1714767572]

1714767572

[BurtW]

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.

[BurtW]

EDIT: this formula doesn't make any sense I know, just playing around

Just for fun the average of this:

Code:

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.
 

[BurtW]

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.

[dalek]

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 

[tarrant_01]

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

[Bulista]

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

Very interesting observation!!! Thank you for sharing

[vilain]

Nice colletive work y'all! Gonna try to use matlab to make some intergers experiments.

[tarrant_01]

31-33 doesn't follow the pattern either.  I missed that.

[Bulista]

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:

[Bulista]

-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

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.

[shorena]

-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.

[tarrant_01]

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.

[Gleb Gamow]

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.

[Bitcoinpro]

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

[amaclin]

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.

[Bulista]

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?

[amaclin]

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

[Bulista]

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

[Bulista]

-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/8aW8tj6
http://imgur.com/dGguJcK