Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 318.

natedawg469

newbie

Activity: 15

Merit: 0

Quote from: SMCA.cf on August 24, 2018, 03:27:00 AM

I do not know how, but based on specific mathematical calculations, I'm likely to have a golden Decimal for the address 56 between the 55484347409204500 , 53322619588066700 numbers.

http://s9.picofile.com/file/8335334000/007.JPG

your chart does not show it being in 554 ,533 range

I do not understand.

Dico88

full member

Activity: 406

Merit: 100

At the time of stopping at 2.3.5 and others. Now that we know in the first 40 private keys it's very easy to be able to enter all 40 keys and get them polynomially to fully match the 40 keys. This function cannot be predicted for the next key. Cool

cryptoeclipse

newbie

Activity: 154

Merit: 0

This drama cannot be created with fiat money. Huge kudos to the person who provided the challenges.

Andzhig

jr. member

Activity: 184

Merit: 3

Quote from: SMCA.cf on August 24, 2018, 03:27:00 AM

I do not know how, but based on specific mathematical calculations, I'm likely to have a golden Decimal for the address 56 between the 55484347409204500 , 53322619588066700 numbers.

What is the function graph Launchpad McQuack flight? Or something from http://oeis.org/ I mean that would be more detailed about "specific mathematical calculations".

SMCA.cf

newbie

Activity: 9

Merit: 0

I do not know how, but based on specific mathematical calculations, I'm likely to have a golden Decimal for the address 56 between the 55484347409204500 , 53322619588066700 numbers.

http://s9.picofile.com/file/8335334000/007.JPG

Andzhig

jr. member

Activity: 184

Merit: 3

Quote from: natedawg469 on August 20, 2018, 04:19:50 PM

Yeah well my point is Andzhig is that the key/clue/pattern may lie within the private keys and we need to look at past addresses, follow that pattern, and narrow down the search.

It remains to find a paranoid who will do this Cheesy

Quote from: Coinblaze on August 20, 2018, 09:41:06 PM

The creator of this "puzzle" already posted in this thread, said that the pk's used increasing bits but were otherwise random, and then proved it was him by saying he would increase the balance on the addresses 10x and then did just that. So don't waste your time looking for patterns, this "puzzle" is just intended to check on the progress of brute-force pk guessing.

The question of what software he used to create this (pseudo) random (in 2015).

Coinblaze

jr. member

Activity: 49

Merit: 1

The creator of this "puzzle" already posted in this thread, said that the pk's used increasing bits but were otherwise random, and then proved it was him by saying he would increase the balance on the addresses 10x and then did just that. So don't waste your time looking for patterns, this "puzzle" is just intended to check on the progress of brute-force pk guessing.

natedawg469

newbie

Activity: 15

Merit: 0

Yeah well my point is Andzhig is that the key/clue/pattern may lie within the private keys and we need to look at past addresses, follow that pattern, and narrow down the search.

tegarp90

full member

Activity: 546

Merit: 100

Quote from: Bulista on December 28, 2015, 11:12:47 AM

Hi guys,

In continuation to this thread: https://bitcointalksearch.org/topic/brute-force-on-bitcoin-addresses-video-of-the-action-1305887

While playing around with my bot, I found out this mysterious transaction:

https://blockchain.info/tx/08389f34c98c606322740c0be6a7125d9860bb8d5cb182c02f98461e5fa6cd15

those 32.896 BTC were sent to multiple addresses, all the private keys of those addresses seem to be generated by some kind of formula.

For example:

Address 2:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb
Biginteger PVK value: 3
Hex PVK value: 3

Address 3:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU76rnZwVdz
19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA
Biginteger PVK value: 7
Hex PVK value: 7

Address 4:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU77MfhviY5
1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e
Biginteger PVK value: 8
Hex PVK value: 8

Address 5:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Dq8Au4Pv
1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k
Biginteger PVK value: 21
Hex PVK value: 15

Address 6:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7Tmu6qHxS
1PitScNLyp2HCygzadCh7FveTnfmpPbfp8
Biginteger PVK value: 49
Hex PVK value: 31

Address 7:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU7hDgvu64y
1McVt1vMtCC7yn5b9wgX1833yCcLXzueeC
Biginteger PVK value: 76
Hex PVK value: 4C

Address 8:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU8xvGK1zpm
1M92tSqNmQLYw33fuBvjmeadirh1ysMBxK
Biginteger PVK value: 224
Hex PVK value: E0

Address 9:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUB3vfDKcxZ
1CQFwcjw1dwhtkVWBttNLDtqL7ivBonGPV
Biginteger PVK value: 467
Hex PVK value: 1d3

Address 10:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUBTL67V6dE
1LeBZP5QCwwgXRtmVUvTVrraqPUokyLHqe
Biginteger PVK value: 514
Hex PVK value: 202

Address 11:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUGxXgtm63M
1PgQVLmst3Z314JrQn5TNiys8Hc38TcXJu
Biginteger PVK value: 1155
Hex PVK value: 483

Address 12:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUW5RtS2JN1
1DBaumZxUkM4qMQRt2LVWyFJq5kDtSZQot
Biginteger PVK value: 2683
Hex PVK value: a7b

Address 13:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFUspniiQZds
1Pie8JkxBT6MGPz9Nvi3fsPkr2D8q3GBc1
Biginteger PVK value: 5216
Hex PVK value: 1460

Address 14:

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFVfZyiN5iEG
1ErZWg5cFCe4Vw5BzgfzB74VNLaXEiEkhk
Biginteger PVK value: 10544
Hex PVK value: 2930

and so on...

until the addresses 50 (1MEzite4ReNuWaL5Ds17ePKt2dCxWEofwk) it was already cracked by someone.

Any ideas what's the formula behind the generation of these addresses?

Address 2, pvk decimal value: 3
Address 3, pvk decimal value: 7
Address 4, pvk decimal value: 8
Address 5, pvk decimal value: 21
Address 6, pvk decimal value: 49
Address 7, pvk decimal value: 76
Address 8, pvk decimal value: 224
Address 9, pvk decimal value: 467
Address 10, pvk decimal value: 514
Address 11, pvk decimal value: 1155
Address 12, pvk decimal value: 2683
Address 13, pvk decimal value: 5216
Address 14, pvk decimal value: 10544
Address 15 and after, pvk decimal value: ?

The prize would be ~32 BTC

EDIT: If you find the solution feel free to leave a tip

1DPUhjHvd2K4ZkycVHEJiN6wba79j5V1u3

I wish i already involved in bitcoin since 2015 :')
At that time i already know bitcoin but don't think it's a valuable thing :"
Also i really want this forum can have much games/puzzle like this, not just right now where people just post for bounty :"

Andzhig

jr. member

Activity: 184

Merit: 3

Quote from: natedawg469 on August 16, 2018, 03:21:08 PM

Found something interesting in the private keys between addresses 53,54 and 55
Address 53 15K1YKJMiJ4fpesTVUcByoz334rHmknxmT Private key: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjh 5SmqrUoK2mhwSYFV
Address 54 1KYUv7nSvXx4642TKeuC2SNdTk326uUpFy Private key: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjh HvuTMSDchRp5hktc
Address 55 1LzhS3k3e9Ub8i2W1V8xQFdB8n2MYCHPCa Private key: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYji dyYKE5NcJQZYvknA

Look at where I split it and you will see the private keys are completely the same until it changes from jh to ji

just kind of interesting to me

with 36028797018963968 already follows jj of course if there was ji it simplified the task of searching (a bit).

Quote

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j,k,l,m.

Quote

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72057594037927931 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjmTLdaTw4Wwi7Ahk8s 1Moo8SzVoETAtt5W5a4qDVGug9CeqbEsKU
72057594037927932 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjmTLdaTw4WxD4sRkRr 14QYPAe6KUo3RwtWktkAZsj4QSdH892Thm
72057594037927933 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjmTLdaTw4WxhvtcUGR 1Ly4eQVmRZ4tj89M7oKQBSr18cCELgpK6k
72057594037927934 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjmTLdaTw4WyCp3QbBN 1NamX1yuQJjzBCBYc6KQzCXN4YV6oWThVm
72057594037927935 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjmTLdaTw4WyhggrKz2 1GBDNYeCKErxHoGwfuKxS9dwz5hL2UAGoq
72057594037927936 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjmTLdaTw4WzCYzQSdy 1Dz9aHyLqXxu18dAka4RLduappJnBuXSQp

natedawg469

newbie

Activity: 15

Merit: 0

Found something interesting in the private keys between addresses 53,54 and 55
Address 53 15K1YKJMiJ4fpesTVUcByoz334rHmknxmT Private key: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjh 5SmqrUoK2mhwSYFV
Address 54 1KYUv7nSvXx4642TKeuC2SNdTk326uUpFy Private key: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjh HvuTMSDchRp5hktc
Address 55 1LzhS3k3e9Ub8i2W1V8xQFdB8n2MYCHPCa Private key: KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYji dyYKE5NcJQZYvknA

Look at where I split it and you will see the private keys are completely the same until it changes from jh to ji

just kind of interesting to me

andreyyv

newbie

Activity: 23

Merit: 0

Quote from: Coinblaze on August 03, 2018, 01:36:34 PM

The current balances are as follows:
001: 1BgGZ9tcN4rm9KBzDn7KprQz87SZ26SAMH 0.000000
002: 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb 0.000000

What is it? Where you get it?

adaseb

legendary

Activity: 3808

Merit: 1723

Been out of the loop on this contest for a while.

Checking LBC looks like they are still finding the 55th key which was already solved.

Looks like someone beat them to it or just started scanning backwards.

hari1987

newbie

Activity: 21

Merit: 0

I don't think there is a pattern in this puzzle. because the owner sends bitcoin randomly. and we know that the bitcoin address is between 2^n. and after the owner knew the puzzles were a bone of contention, the owner promised to increase the reward to 10 times and he did it.
good luck guys,there is 36,028,797,018,963,968 addresses to find puzzle number 56

Amnuazka

newbie

Activity: 15

Merit: 0

Amnuazka

newbie

Activity: 15

Merit: 0

Quote from: ZafotheNinja on August 02, 2018, 07:39:34 AM

Quote from: Amnuazka on July 30, 2018, 07:11:21 PM

I found this :

-snip-

holy_ship , Andzhig, arulbero, or anyone.., can you analize the next code ( x ) for 56 ?

-snip-

Thank you.

Thanks, you successfully made a wall of text. Please actually find a pattern instead of asking everyone else to.

Even if the next number could be said for certain, it would not solve for the entire string, therefore the difficulty of finding the private key for the next puzzle would be only marginally reduced.

Keep trying, but know that holy_ship, Andzhig, arulbero, or anyone else on this thread has probably already seen the binary version on the previous private keys, and have yet to find any pattern other than the first 0 at the beginning of each.

Thank you

Coinblaze

jr. member

Activity: 49

Merit: 1

I didn't see this update posted here, so maybe most people are not aware. (edit: actually it's way back on page 13)

As of 2017-07-11 the rewards for keys with bits 161+ were removed, and others made to be 10x the amount.

The current balances are as follows:
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002: 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb 0.000000
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029: 19EEC52krRUK1RkUAEZmQdjTyHT7Gp1TYT 0.000000
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203: 1DVWn7PuRBmsdTBKiboBdSREHdAtSoN6BB 0.000000
204: 14CMU6qvv55Y7xJdVw64ey5yfUP4BZAjct 0.000000
205: 1DHL5NuXPjwDsNnXyzAgZTAMM4aYUc1zFW 0.000000
206: 1UwsEPMF1NZTWJAuymsLVJpqetBZ3Q9sJ 0.000000
207: 16aELF1f75o464ZhtAZUwbc4ctFQZxS8qi 0.000000
208: 15TJ3wPvdviupvKQFm8hLeXSzeMSq7LSJ2 0.000000
209: 1JgrGoQbvJS7UnX6j4myiCxo6Q3gyo5Ujk 0.000000
210: 14Xm5DjBUQTJCzeGhyrhVsxkK8p35srk1S 0.000000
211: 19yhSoza8oK3ioCSydMuAGJs4Mm3FwCTht 0.000000
212: 1JXw9i8dEZGH29mUiuKjWXK9L27r2TerLQ 0.000000
213: 1CNNth43uiVypxHmZLC8hWZsb7UiP7wSkY 0.000000
214: 16ocVeZDpqcvyMvzAH1r2LR75uEzRhkyV2 0.000000
215: 15x3tRVyn9SRaxfbFUzqETmCJiz46Vs247 0.000000
216: 197K7MdYhnN88gcJouJRxMiSAHHfWuPrXC 0.000000
217: 1MDsNYfC4LErgwUDqfQ5BgFJqv5bs4Frkn 0.000000
218: 1BDTXmiyyzq9i79RGGTW7cjmYJbxKoV27e 0.000000
219: 1EiJ59LPWDezXwfAGFTcoEKdNhvRTriBXy 0.000000
220: 1rirV4Y2NxGwKNQNojJhz61jEctni8fvb 0.000000
221: 1NxgWntAdMugSNFHEXYizYwk3UjxKPxcDF 0.000000
222: 17A75vEkPPVeY9MMMXUY9M2JUHbbNyVAWC 0.000000
223: 1NUVBXgX35Uax4fpziV7VGMJyji3mUzZbt 0.000000
224: 1F6yxcbDzjumeN77DiABMXzPcvtAdnaoPF 0.000000
225: 16YvGEYhAwjf2duwvH8jbFMfVCnY6XiQTq 0.000000
226: 1HtbZg9mPjYcMDMDNyXaFjHX5e7ZPUwWAp 0.000000
227: 15a9nXpjnQzw2o5kmvGyKzv7anZkYFHwdK 0.000000
228: 18zsQK8ezT32qCqgLJQMkhqyKNwpCP3JU3 0.000000
229: 1AaBhpTnfCin9nCrai2msXzCHoVmwkAe2N 0.000000
230: 13Jtm1mm33Uke7PbmTBYGNZGU7rXUsVs3e 0.000000
231: 18aBXRctVrWN9naDGhKrLdZViNfLJfCdbF 0.000000
232: 1MEEjd99pkEyCdiqVSCa8Jqjaun7fsEXaG 0.000000
233: 1NnefTEeKQQAts37cQHzVx8oPrUu8LWyUK 0.000000
234: 1FYbLcutmRbvu4yUeLmC4TES2Q3ChhXYY 0.000000
235: 1Pp61TfkztZ9ckpdeUVCzbyA7vMqXAVsdV 0.000000
236: 1PytbQzRaf9eTpGax3c6ofKwtbxaLLSNy1 0.000000
237: 194bRtNQVRq4xi26UJZYTHexvLXijpzp3e 0.000000
238: 1DxBvzRMdvzop21DhnDJv4xJDYFGVtu7KZ 0.000000
239: 1QLHLvM6XKwygyWqo2cCPbfbf6woZzGEKH 0.000000
240: 1C8Hw6T5jypz92cNFr9Lkx4Xmr4DtP7zGA 0.000000
241: 1L7ZNx5gFPvdVPfdgFBnEUgA64woQwopqr 0.000000
242: 1EMhTbC4Kp8DYBk5zoLsTqZ2YakhiTgQYh 0.000000
243: 19CqmBBJ3H8FjxSSeSjv2Kn1cVQFXMY6AM 0.000000
244: 1KMHDrCGho2QuK59UNvfjuiPdDZSgRbneC 0.000000
245: 1HzpmnHxpu4tCJ23ZS9TfWCoX8mXfpdHiq 0.000000
246: 1LrrNP28PYg1N1z5Uo1gR8cmoPc8h4orBZ 0.000000
247: 14D2d3WnHThUxyPhGoj2AabBTxpZcoHy3t 0.000000
248: 18oqrdP6uBKsn57gvmWzLuKMAY4ShapiAU 0.000000
249: 1Bun4VzuBJ7SUoQn97dinVfDyWAS336Ldg 0.000000
250: 1Ruu3JwvGeSmhQV9GzWAnyLCz6g3evmTY 0.000000
251: 1M3u4q5Q35qtQPmDHeubVbk7APYi3VVoBX 0.000000
252: 1CaTxB3YwmXZkDnTK4rRvq61SRqs48xmui 0.000000
253: 1JqRqUPHHcQu2yrr8JZzxSYDx2jbZxEqFj 0.000000
254: 1NKkjFvXmovmjwgUujw655n3BbEvnncyza 0.000000
255: 17PEUvQmgqPkkvRkMowoR1wRDXYzre4b9Z 0.000000
256: 1FMcotmnqqE5M2x9DDX3VfPAPuBWArGisa 0.000000
Total Unclaimed: 113.400000 BTC

At today's prices, that is $855,631.35

Andzhig

jr. member

Activity: 184

Merit: 3

This he saw the ladder without fractional (seemingly). Some stalactites..

36028797018963968 30045390491869460 9974455244496708‬ 6763683971478124‬ 4216495639600700‬
18014398509481984 15022695245934730‬ 4987227622248354‬ 3381841985739062‬ 2108247819800350‬
9007199254740992 1 7511347622967365‬ 1 2493613811124177‬ 1 1690920992869531‬ 1054123909900175‬
4503599627370496 3755673811483682‬ 1246806905562088 ‬1 845460496434765‬ 1 527061954950087‬
2251799813685248 1 1877836905741841‬ 623403452781044‬ 422730248217382‬ 1 263530977475043‬
1125899906842624 938918452870920‬ 311701726390522‬ 1 211365124108691‬ 131765488737521‬
562949953421312 469459226435460‬ 1 155850863195261‬ 1 105682562054345‬ 65882744368760‬
281474976710656 234729613217730‬ 77925431597630‬ 52841281027172‬ 32941372184380‬
140737488355328 1 117364806608865‬ 1 38962715798815‬ 26420640513586‬ 16470686092190‬
70368744177664 58682403304432‬ 1 19481357899407‬ 1 13210320256793‬ 1 8235343046095‬
35184372088832 29341201652216‬ 1 9740678949703‬ 6605160128396‬ 1 4117671523047‬
17592186044416 14670600826108‬ 1 4870339474851‬ 3302580064198‬ 1 2058835761523‬
8796093022208 7335300413054‬ 1 2435169737425‬ 1 1651290032099‬ 1 1029417880761‬
4398046511104 1 3667650206527‬ 1217584868712‬ 1 825645016049‬ 514708940380‬
2199023255552 1 1833825103263‬ 608792434356‬ 412822508024‬ 257354470190‬
1099511627776 1 916912551631‬ 304396217178‬ 206411254012‬ 1 128677235095 ‬
549755813888 1 458456275815‬ 1 152198108589‬ 103205627006‬ 1 64338617547‬
274877906944 1 229228137907‬ 76099054294‬ 1 51602813503‬ 1 32169308773‬
137438953472 1 114614068953‬ 1 38049527147‬ 1 25801406751‬ 16084654386‬
68719476736 57307034476‬ 1 19024763573‬ 1 12900703375‬ 1 8042327193‬
34359738368 28653517238‬ 9512381786‬ 1 6450351687‬ 4021163596‬
17179869184 1 14326758619‬ 1 4756190893‬ 1 3225175843‬ 2010581798‬
8589934592 1 7163379309‬ 2378095446‬ 1 1612587921‬ 1 1005290899‬
4294967296 3581689654‬ 1 1189047723‬ 806293960‬ 1 502645449‬
2147483648 1 1790844827‬ 1 594523861‬ 403146980‬ 251322724‬
1073741824 1 895422413‬ 297261930‬ 201573490‬ 125661362‬
536870912 447711206‬ 1 148630965‬ 1 100786745‬ 1 62830681‬
268435456 1 223855603‬ 74315482‬ 50393372‬ 31415340‬
134217728 1 111927801‬ 1 37157741‬ 25196686‬ 15707670‬
67108864 55963900‬ 18578870‬ 1 12598343‬ 1 7853835‬
33554432 27981950‬ 1 9289435‬ 1 6299171‬ 1 3926917‬
16777216 1 13990975‬ 1 4644717‬ 1 3149585‬ 1963458‬
8388608 1 6995487 2322358‬ 1 1574792‬ 1 981729‬
4194304 1 3497743 1 1161179‬ 787396‬ 490864‬
2097152 1 1748871 1 580589‬ 393698‬ 245432‬
1048576 1 874435 290294‬ 1 196849‬ 122716‬
524288 1 437217 1 145147‬ 98424‬ 61358‬
262144 218608 1 72573‬ 49212‬ 1 30679‬
131072 109304 36286‬ 24606‬ 1 15339‬
65536 54652 1 18143‬ 1 12303‬ 1 7669‬
32768 27326 1 9071‬ 1 6151‬ 3834‬
16384 1 13663 1 4535‬ 1 3075‬ 1 1917‬
8192 1 6831 1 2267‬ 1 1537‬ 958‬
4096 1 3415 1 1133‬ 768‬ 1 479‬
2048 1 1707 566‬ 384‬ 1 239‬
1024 1 853 1 283‬ 192‬ 1 119‬
512 426 1 141‬ 96‬ 1 59‬
256 1 213 70‬ 48‬ 1 29‬
128 106 1 35‬ 24 14‬
64 1 53 1 17‬ 12 1 7‬
32 26 8 ‬ 6 1 3‬
16 1 13 4 1 3 1‬
8 6 2 1
4 1 3 1
2 1
1

at some steps added 1. In the python 3 it will look like

Quote

a1 = 1
a2 = a1 * 2 +1
a3 = a2 * 2
a4 = a3 * 2 +1
a5 = a4 * 2
a6 = a5 * 2 +1
a7 = a6 * 2
a8 = a7 * 2 +1
a9 = a8 * 2
a10 = a9 * 2 +1
a11 = a10 * 2 +1
a12 = a11 * 2 +1
a13 = a12 * 2 +1
a14 = a13 * 2 +1
a15 = a14 * 2
a16 = a15 * 2
a17 = a16 * 2
a18 = a17 * 2
a19 = a18 * 2 +1
a20 = a19 * 2 +1
a21 = a20 * 2 +1
a22 = a21 * 2 +1
a23 = a22 * 2 +1
a24 = a23 * 2 +1
a25 = a24 * 2
a26 = a25 * 2
a27 = a26 * 2 +1
a28 = a27 * 2 +1
a29 = a28 * 2
a30 = a29 * 2 +1
a31 = a30 * 2 +1
a32 = a31 * 2
a33 = a32 * 2 +1
a34 = a33 * 2 +1
a35 = a34 * 2
a36 = a35 * 2
a37 = a36 * 2 +1
a38 = a37 * 2 +1
a39 = a38 * 2 +1
a40 = a39 * 2 +1
a41 = a40 * 2 +1
a42 = a41 * 2 +1
a43 = a42 * 2
a44 = a43 * 2
a45 = a44 * 2
a46 = a45 * 2
a47 = a46 * 2 +1
a48 = a47 * 2
a49 = a48 * 2
a50 = a49 * 2
a51 = a50 * 2 +1
a52 = a51 * 2
a53 = a52 * 2 +1
a54 = a53 * 2
a55 = a54 * 2

print(a31,a55)

where a31 possible address (+ 200000000 or +500000000 not to smear) for scanning by Holy fork (1790844827 * 2 24 times = 30045390485061632 against 30045390491869460 the right.

Accordingly, placing correctly the pros can successfully jump.

Quote

from bitcoin import *

a1 = 1
a2 = a1 * 2 +1
a3 = a2 * 2
a4 = a3 * 2
a5 = a4 * 2 +1
a6 = a5 * 2 +1
a7 = a6 * 2
a8 = a7 * 2 +1
a9 = a8 * 2
a10 = a9 * 2
a11 = a10 * 2 +1
a12 = a11 * 2 +1
a13 = a12 * 2
a14 = a13 * 2
a15 = a14 * 2
a16 = a15 * 2 +1
a17 = a16 * 2 +1
a18 = a17 * 2 +1
a19 = a18 * 2 +1
a20 = a19 * 2
a21 = a20 * 2
a22 = a21 * 2
a23 = a22 * 2 +1
a24 = a23 * 2 +1
a25 = a24 * 2
a26 = a25 * 2 +1
a27 = a26 * 2 +1
a28 = a27 * 2 +1
a29 = a28 * 2 +1
a30 = a29 * 2
a31 = a30 * 2
a32 = a31 * 2 +1 # fork scan point *2>24
a33 = a32 * 2
a34 = a33 * 2
a35 = a34 * 2
a36 = a35 * 2
a37 = a36 * 2
a38 = a37 * 2
a39 = a38 * 2
a40 = a39 * 2
a41 = a40 * 2
a42 = a41 * 2
a43 = a42 * 2
a44 = a43 * 2
a45 = a44 * 2
a46 = a45 * 2
a47 = a46 * 2
a48 = a47 * 2
a49 = a48 * 2
a50 = a49 * 2
a51 = a50 * 2
a52 = a51 * 2
a53 = a52 * 2
a54 = a53 * 2
a55 = a54 * 2
a56 = a55 * 2

scan = a32*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2+200000000
ran = scan
myhex = "%064x" % ran
myhex = myhex[:64]
priv = myhex
pub = privtopub(priv)
pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
addr = pubtoaddr(pubkey1)

print(a32,scan,a56,addr,myhex)

Or manually do not put up pluses

Quote

import secrets
from bitcoin import *

b1 = secrets.choice([0,1])
b2 = secrets.choice([0,1])
b3 = secrets.choice([0,1])
b4 = secrets.choice([0,1])
b5 = secrets.choice([0,1])
b6 = secrets.choice([0,1])
b7 = secrets.choice([0,1])
b8 = secrets.choice([0,1])
b9 = secrets.choice([0,1])
b10 = secrets.choice([0,1])
b11 = secrets.choice([0,1])
b12 = secrets.choice([0,1])
b13 = secrets.choice([0,1])
b14 = secrets.choice([0,1])
b15 = secrets.choice([0,1])
b16 = secrets.choice([0,1])
b17 = secrets.choice([0,1])
b18 = secrets.choice([0,1])
b19 = secrets.choice([0,1])
b20 = secrets.choice([0,1])
b21 = secrets.choice([0,1])
b22 = secrets.choice([0,1])
b23 = secrets.choice([0,1])
b24 = secrets.choice([0,1])
b25 = secrets.choice([0,1])
b26 = secrets.choice([0,1])
b27 = secrets.choice([0,1])
b28 = secrets.choice([0,1])
b29 = secrets.choice([0,1])
b30 = secrets.choice([0,1])
b31 = secrets.choice([0,1])

a1 = 1
a2 = a1 * 2 +b1
a3 = a2 * 2 +b2
a4 = a3 * 2 +b3
a5 = a4 * 2 +b4
a6 = a5 * 2 +b5
a7 = a6 * 2 +b6
a8 = a7 * 2 +b7
a9 = a8 * 2 +b8
a10 = a9 * 2 +b9
a11 = a10 * 2 +b10
a12 = a11 * 2 +b11
a13 = a12 * 2 +b12
a14 = a13 * 2 +b13
a15 = a14 * 2 +b14
a16 = a15 * 2 +b15
a17 = a16 * 2 +b16
a18 = a17 * 2 +b17
a19 = a18 * 2 +b18
a20 = a19 * 2 +b19
a21 = a20 * 2 +b20
a22 = a21 * 2 +b21
a23 = a22 * 2 +b22
a24 = a23 * 2 +b23
a25 = a24 * 2 +b24
a26 = a25 * 2 +b25
a27 = a26 * 2 +b26
a28 = a27 * 2 +b27
a29 = a28 * 2 +b28
a30 = a29 * 2 +b29
a31 = a30 * 2 +b30
a32 = a31 * 2 +b31 # fork scan point *2>24

scanpoint = a32*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2+200000000
ran = scanpoint
myhex = "%064x" % ran
myhex = myhex[:64]
priv = myhex
pub = privtopub(priv)
pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
addr = pubtoaddr(pubkey1)

print(a32,scanpoint,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10,b11,b12,b13,b14,b15,b16,b17,b18,b19,b20,b21,b22,b23,b24,b25,b26,b27,b28,b29,b30,b31,addr,priv)

can make autostart fork adding (fork folder and cmd file itself lie on c:\) to loop or duplicate the repetition of the code, to whom it is more convenient, you can run several forks slightly correcting a couple of characters

Quote

import subprocess
import time

oy = """cd "C:\forkfolder1" """
ey = "\nstart /min oclvanitygen.exe -D 1:0 -C -f addresses.txt -o found.txt "
f=open("C:/testpy1.cmd","w")
f.write (oy)
f.write (ey)
f.write (priv) # adds the start address of the scan
f.close()

subprocess.Popen([r"C:/testpy1.cmd"])

time.sleep(60.0) # seconds before the scanner is closed

subprocess.call("taskkill /IM oclvanitygen.exe")

or for luck 56

Quote

from bitcoin import *
import secrets

while True:
   b1 = secrets.choice([0,1])
   b2 = secrets.choice([0,1])
   b3 = secrets.choice([0,1])
   b4 = secrets.choice([0,1])
   b5 = secrets.choice([0,1])
   b6 = secrets.choice([0,1])
   b7 = secrets.choice([0,1])
   b8 = secrets.choice([0,1])
   b9 = secrets.choice([0,1])
   b10 = secrets.choice([0,1])
   b11 = secrets.choice([0,1])
   b12 = secrets.choice([0,1])
   b13 = secrets.choice([0,1])
   b14 = secrets.choice([0,1])
   b15 = secrets.choice([0,1])
   b16 = secrets.choice([0,1])
   b17 = secrets.choice([0,1])
   b18 = secrets.choice([0,1])
   b19 = secrets.choice([0,1])
   b20 = secrets.choice([0,1])
   b21 = secrets.choice([0,1])
   b22 = secrets.choice([0,1])
   b23 = secrets.choice([0,1])
   b24 = secrets.choice([0,1])
   b25 = secrets.choice([0,1])
   b26 = secrets.choice([0,1])
   b27 = secrets.choice([0,1])
   b28 = secrets.choice([0,1])
   b29 = secrets.choice([0,1])
   b30 = secrets.choice([0,1])
   b31 = secrets.choice([0,1])
   b32 = secrets.choice([0,1])
   b33 = secrets.choice([0,1])
   b34 = secrets.choice([0,1])
   b35 = secrets.choice([0,1])
   b36 = secrets.choice([0,1])
   b37 = secrets.choice([0,1])
   b38 = secrets.choice([0,1])
   b39 = secrets.choice([0,1])
   b40 = secrets.choice([0,1])
   b41 = secrets.choice([0,1])
   b42 = secrets.choice([0,1])
   b43 = secrets.choice([0,1])
   b44 = secrets.choice([0,1])
   b45 = secrets.choice([0,1])
   b46 = secrets.choice([0,1])
   b47 = secrets.choice([0,1])
   b48 = secrets.choice([0,1])
   b49 = secrets.choice([0,1])
   b50 = secrets.choice([0,1])
   b51 = secrets.choice([0,1])
   b52 = secrets.choice([0,1])
   b53 = secrets.choice([0,1])
   b54 = secrets.choice([0,1])
   b55 = secrets.choice([0,1])

   a1 = 1
   a2 = a1 * 2 +b1
   a3 = a2 * 2 +b2
   a4 = a3 * 2 +b3
   a5 = a4 * 2 +b4
   a6 = a5 * 2 +b5
   a7 = a6 * 2 +b6
   a8 = a7 * 2 +b7
   a9 = a8 * 2 +b8
   a10 = a9 * 2 +b9
   a11 = a10 * 2 +b10
   a12 = a11 * 2 +b11
   a13 = a12 * 2 +b12
   a14 = a13 * 2 +b13
   a15 = a14 * 2 +b14
   a16 = a15 * 2 +b15
   a17 = a16 * 2 +b16
   a18 = a17 * 2 +b17
   a19 = a18 * 2 +b18
   a20 = a19 * 2 +b19
   a21 = a20 * 2 +b20
   a22 = a21 * 2 +b21
   a23 = a22 * 2 +b22
   a24 = a23 * 2 +b23
   a25 = a24 * 2 +b24
   a26 = a25 * 2 +b25
   a27 = a26 * 2 +b26
   a28 = a27 * 2 +b27
   a29 = a28 * 2 +b28
   a30 = a29 * 2 +b29
   a31 = a30 * 2 +b30
   a32 = a31 * 2 +b31 # fork scan point *2>24
   a33 = a32 * 2 +b32
   a34 = a33 * 2 +b33
   a35 = a34 * 2 +b34
   a36 = a35 * 2 +b35
   a37 = a36 * 2 +b36
   a38 = a37 * 2 +b37
   a39 = a38 * 2 +b38
   a40 = a39 * 2 +b39
   a41 = a40 * 2 +b40
   a42 = a41 * 2 +b41
   a43 = a42 * 2 +b42
   a44 = a43 * 2 +b43
   a45 = a44 * 2 +b44
   a46 = a45 * 2 +b45
   a47 = a46 * 2 +b46
   a48 = a47 * 2 +b47
   a49 = a48 * 2 +b48
   a50 = a49 * 2 +b49
   a51 = a50 * 2 +b50
   a52 = a51 * 2 +b51
   a53 = a52 * 2 +b52
   a54 = a53 * 2 +b53
   a55 = a54 * 2 +b54
   a56 = a55 * 2 +b55
   ran = a56
   myhex = "%064x" % ran
   myhex = myhex[:64]
   priv = myhex
   pub = privtopub(priv)
   pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
   addr = pubtoaddr(pubkey1)
   if addr == "17aPYR1m6pVAacXg1PTDDU7XafvK1dxvhi":
   print ("found!!!",ran,addr,myhex)
   s1 = myhex
   s2 = addr
   f=open(u"C:/a.txt","a")
   f.write(s1)
   f.write(s2)
   f.close()
   break
   else:
   print ("searching...",ran,addr,myhex)

ZafotheNinja

jr. member

Activity: 30

Merit: 1

Quote from: Amnuazka on July 30, 2018, 07:11:21 PM

I found this :

-snip-

holy_ship , Andzhig, arulbero, or anyone.., can you analize the next code ( x ) for 56 ?

-snip-

Thank you.

Thanks, you successfully made a wall of text. Please actually find a pattern instead of asking everyone else to.

Even if the next number could be said for certain, it would not solve for the entire string, therefore the difficulty of finding the private key for the next puzzle would be only marginally reduced.

Keep trying, but know that holy_ship, Andzhig, arulbero, or anyone else on this thread has probably already seen the binary version on the previous private keys, and have yet to find any pattern other than the first 0 at the beginning of each.

Andzhig

jr. member

Activity: 184

Merit: 3

Amnuazka we so transmutation to reptilian humanoids from Zeta Reticules.

try correct for binary input

Quote

from bitcoin import *
import secrets

while True:
   c1 = secrets.choice("456")
   c2 = secrets.choice("0987654321")
   c3 = secrets.choice("0987654321")
   c4 = secrets.choice("0987654321")
   c5 = secrets.choice("0987654321")
   c6 = secrets.choice("0987654321")
   c7 = secrets.choice("0987654321")
   c8 = secrets.choice("0987654321")
   c9 = secrets.choice("0987654321")
   c10 = secrets.choice("0987654321")
   c11 = secrets.choice("0987654321")
   c12 = secrets.choice("0987654321")
   c13 = secrets.choice("0987654321")
   c14 = secrets.choice("0987654321")
   c15 = secrets.choice("0987654321")
   c16 = secrets.choice("0987654321")
   c17 = secrets.choice("0987654321")
   ran = int(c1+c2+c3+c4+c5+c6+c7+c8+c9+c10+c11+c12+c13+c14+c15+c16+c17)
   myhex = "%064x" % ran
   myhex = myhex[:64]
   priv = myhex
   pub = privtopub(priv)
   pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
   addr = pubtoaddr(pubkey1)
   if addr == "1LzhS3k3e9Ub8i2W1V8xQFdB8n2MYCHPCa":
   print ("found!!!",ran,addr,myhex)
   s1 = myhex
   s2 = addr
   f=open(u"C:/a.txt","a")
   f.write(s1)
   f.write(s2)
   f.close()
   break
   else:
   print ("searching...",ran,addr,myhex)

something like

Quote

c1 = secrets.choice("01")
c2 = secrets.choice("01")
c3 = secrets.choice("01")
c4 = secrets.choice("01")
c5 = secrets.choice("01")
c6 = secrets.choice("01")
c7 = secrets.choice("01")
c8 = secrets.choice("01")
c9 = secrets.choice("01")
c10 = secrets.choice("01")
c11 = secrets.choice("01")
c12 = secrets.choice("01")
c13 = secrets.choice("01")
c14 = secrets.choice("01")
c15 = secrets.choice("01")
c16 = secrets.choice("01")
c17 = secrets.choice("01")
c18 = secrets.choice("01")
c19 = secrets.choice("01")
c20 = secrets.choice("01")
c21 = secrets.choice("01")
c22 = secrets.choice("01")
c23 = secrets.choice("01")
c24 = secrets.choice("01")
c25 = secrets.choice("01")
c26 = secrets.choice("01")
c27 = secrets.choice("01")
c28 = secrets.choice("01")
c29 = secrets.choice("01")
c30 = secrets.choice("01")
c31 = secrets.choice("01")
c32 = secrets.choice("01")
c33 = secrets.choice("01")
c34 = secrets.choice("01")
c35 = secrets.choice("01")
c36 = secrets.choice("01")
c37 = secrets.choice("01")
c38 = secrets.choice("01")
c39 = secrets.choice("01")
c40 = secrets.choice("01")
c41 = secrets.choice("01")
c42 = secrets.choice("01")
c43 = secrets.choice("01")
c44 = secrets.choice("01")
c45 = secrets.choice("01")
c46 = secrets.choice("01")
c47 = secrets.choice("01")
c48 = secrets.choice("01")
c49 = secrets.choice("01")
c50 = secrets.choice("01")
c51 = secrets.choice("01")
c52 = secrets.choice("01")
c53 = secrets.choice("01")
c54 = secrets.choice("01")
c55 = secrets.choice("01")
c56 = secrets.choice("01")

or from random.randint(0,1)...

if you have fractional use Decimal

Quote

from bitcoin import *
import secrets
from decimal import *

while True:
   a = str("127.")
   c1 = str (secrets.choice("1234"))
   c2 = str (secrets.choice("0123456789"))
   c3 = str (secrets.choice("0123456789"))
   c4 = str (secrets.choice("0123456789"))
   c5 = str (secrets.choice("0123456789"))
   c6 = str (secrets.choice("0123456789"))
   c7 = str (secrets.choice("0123456789"))
   c8 = str (secrets.choice("0123456789"))
   c9 = str (secrets.choice("0123456789"))
   c10 = str (secrets.choice("0123456789"))
   c11 = str (secrets.choice("0123456789"))
   c12 = str (secrets.choice("0123456789"))
   c13 = str (secrets.choice("0123456789"))
   c14 = str (secrets.choice("0123456789"))
   c15 = str (secrets.choice("0123456789"))
   c16 = str (secrets.choice("0123456789"))
   c17 = str (secrets.choice("0123456789"))
   b = Decimal(a+c1+c2+c3+c4+c5+c6+c7+c8+c9+c10+c11+c12+c13+c14+c15+c16+c17)
   rrrr = int(72057594037927936 - (((((((((b * 128) * 128) * 128)* 128) * 128) * 128)* 128) - 71494644084506624) * 128))
   ran = rrrr
   myhex = "%064x" % ran
   myhex = myhex[:64]
   priv = myhex
   pub = privtopub(priv)
   pubkey1 = encode_pubkey(privtopub(priv), "bin_compressed")
   addr = pubtoaddr(pubkey1)
   if addr == "17aPYR1m6pVAacXg1PTDDU7XafvK1dxvhi":
   print ("found!!!",ran,addr,myhex)
   s1 = myhex
   s2 = addr
   f=open(u"C:/a.txt","a")
   f.write(s1)
   f.write(s2)
   f.close()
   break
   else:
   print ("searching...",b,ran,addr,myhex)

55 to bin 01101010101111100001111110011011011001111110000100010100‬ and in 56.56 digits you have 48...

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 318. (Read 228764 times)