Bitcoin puzzle transaction ~32 BTC prize to who solves it

[kTimesG]

Hello friends, it's time to make it public.
I hope these findings will help you find the keys.

Any series of strings in continuous increasing length starting from some center point and spiraled around in ascending length order will produce arms.
Why? Because they increase in same length, so for any element X(i) its length is length of X(i-1) * someConstant, so eventually their starting points on the circle's perimeter will get aligned, depending on how far from the center you position it.

This happens for any rational and non-rational number that exists, except for phi (1+sqrt(5))/2 and its family. That one is the single possible ratio that always wraps in full uniform distribution and is seen all over in nature.

[1714497742]

1714497742

[1714497742]

1714497742

[1714497742]

1714497742

[CY4NiDE]

Well, since we are here I'd like to ask what y'all think of this:

Code:

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import splev, splrep
from decimal import Decimal

sequence = [
1, 3, 7, 8, 21, 49, 76, 224, 467, 514, 1155, 2683, 5216, 10544, 26867, 51510, 95823, 198669, 357535, 863317, 1811764, 3007503, 5598802, 14428676, 33185509, 54538862, 111949941, 227634408, 400708894, 1033162084, 2102388551, 3093472814, 7137437912, 14133072157, 20112871792, 42387769980, 100251560595, 146971536592, 323724968937, 1003651412950, 1458252205147, 2895374552463, 7409811047825, 15404761757071, 19996463086597, 51408670348612, 119666659114170, 191206974700443, 409118905032525, 611140496167764, 2058769515153876, 4216495639600700, 6763683971478124, 9974455244496707, 30045390491869460, 44218742292676575, 138245758910846492, 199976667976342049, 525070384258266191, 1135041350219496382, 1425787542618654982, 3908372542507822062, 8993229949524469768, 17799667357578236628
]

x_values_known = np.arange(len(sequence))
sequence_decimal = [Decimal(value) for value in sequence]
spline_rep = splrep(x_values_known, sequence_decimal, k=2)
extended_x_values = np.arange(len(sequence) + 1)
predicted_next_number = splev(extended_x_values[-1], spline_rep)
predicted_next_number_hex = hex(int(predicted_next_number))
plt.plot(x_values_known, sequence_decimal, label='sequence')
plt.plot(
    extended_x_values,
    splev(extended_x_values, spline_rep),
    label='Recreated Sequence',
    linestyle='dashed'
)
plt.scatter(
    extended_x_values[-1],
    float(predicted_next_number),
    color='red',
    marker='o',
    label='Predicted Next Number'
)
plt.legend()
plt.xlabel('Index')
plt.ylabel('Value')
plt.title('Original vs. Recreated sequence with Prediction')
plt.show()
print(f"Next key: {predicted_next_number}")
print(f"Hexadecimal: {predicted_next_number_hex}")

This is the result when we feed the script the sequence of keys up to #64, in order for it to "predict" #65:

Next key: 3.0520846598475555e+19
Hex: 0x1a78fd44662532000

Is this jesus toast or could we use it to at least try and narrow down the first 2 characters of #66?

Your insights are much appreciated.

[jacky19790729]

Next key: 3.0520846598475555e+19
Hex: 0x1a78fd44662532000
Is this jesus toast or could we use it to at least try and narrow down the first 2 characters of #66?
Your insights are much appreciated.

use this code to "predict"

Hexadecimal: 0x11f774e94c1ec000         # puzzle 62     0x363d541eb611abee
Hexadecimal: 0x7d556bf6f89d2c00         # puzzle 63     0x7cce5efdaccf6808
Hexadecimal: 0xe7655f0b50acf800         # puzzle 64     0xf7051f27b09112d4
Hexadecimal: 0x1a78fd44662532000      # puzzle 65     0x1a838b13505b26867   
Hexadecimal: 0x290860e0f31602000      # puzzle 66     ?

[CY4NiDE]

use this code to "predict"

Hexadecimal: 0x11f774e94c1ec000         # puzzle 62     0x363d541eb611abee
Hexadecimal: 0x7d556bf6f89d2c00         # puzzle 63     0x7cce5efdaccf6808
Hexadecimal: 0xe7655f0b50acf800         # puzzle 64     0xf7051f27b09112d4
Hexadecimal: 0x1a78fd44662532000      # puzzle 65     0x1a838b13505b26867   
Hexadecimal: 0x290860e0f31602000      # puzzle 66     ?

Yep. Some results get kinda close, others not much.

Could accuracy increase as we append new keys to the sequence?

Can we improve this script somehow?

Code:

spline_rep = splrep(x_values_known, sequence_decimal, k=2)

I've messed with other values for k but 2 seems to yield better results.

[NotATether]

There's simply no feasible way to withdraw the funds on lower end puzzles like #66. It will be snatched up by bots. Not maybe, but it's 100% guaranteed. There will be hundreds of withdrawal transactions with varying fees all battling each other. You will simply be left in the dust.

I don't understand how it is possible for that to happen as long as the solver does not disclose the private key. I mean, #64 and #65 both have unknown keys, right?

[zahid888]

use this code to "predict"

Hexadecimal: 0x11f774e94c1ec000         # puzzle 62     0x363d541eb611abee
Hexadecimal: 0x7d556bf6f89d2c00         # puzzle 63     0x7cce5efdaccf6808
Hexadecimal: 0xe7655f0b50acf800         # puzzle 64     0xf7051f27b09112d4
Hexadecimal: 0x1a78fd44662532000      # puzzle 65     0x1a838b13505b26867   
Hexadecimal: 0x290860e0f31602000      # puzzle 66     ?

Yep. Some results get kinda close, others not much.

Could accuracy increase as we append new keys to the sequence?

Can we improve this script somehow?

Code:

spline_rep = splrep(x_values_known, sequence_decimal, k=2)

I've messed with other values for k but 2 seems to yield better results.

Index 3: Predicted = 0xd, Actual = 0x8, Error = 5.0
Index 4: Predicted = 0x5, Actual = 0x15, Error = 15.83333333333334
Index 5: Predicted = 0x30, Actual = 0x31, Error = 0.2857142857142918
Index 6: Predicted = 0x5c, Actual = 0x4c, Error = 16.04901960784312
Index 7: Predicted = 0x63, Actual = 0xe0, Error = 124.75357443229606
Index 8: Predicted = 0x202, Actual = 0x1d3, Error = 47.404329004328815
Index 9: Predicted = 0x31c, Actual = 0x202, Error = 282.866702978386
Index 10: Predicted = 0x13c, Actual = 0x483, Error = 838.5322535426649
Index 11: Predicted = 0x9e5, Actual = 0xa7b, Error = 149.13061026670766
Index 12: Predicted = 0x1403, Actual = 0x1460, Error = 92.41323240818747
Index 13: Predicted = 0x2241, Actual = 0x2930, Error = 1774.144396004227
Index 14: Predicted = 0x4a1b, Actual = 0x68f3, Error = 7895.604944862127
Index 15: Predicted = 0xd8f3, Actual = 0xc936, Error = 4029.671642258196
Index 16: Predicted = 0x14745, Actual = 0x1764f, Error = 12041.382349990992
Index 17: Predicted = 0x2784f, Actual = 0x3080d, Error = 36797.025408181595
Index 18: Predicted = 0x59719, Actual = 0x5749f, Error = 8826.371450069128
Index 19: Predicted = 0x8b61a, Actual = 0xd2c55, Error = 292410.36592774664
Index 20: Predicted = 0x1af328, Actual = 0x1ba534, Error = 45579.312763758004
Index 21: Predicted = 0x30fdc8, Actual = 0x2de40f, Error = 203193.17374296952
Index 22: Predicted = 0x4360b7, Actual = 0x556e52, Error = 1183130.437051029
Index 23: Predicted = 0x955cee, Actual = 0xdc2a04, Error = 4640021.909114862
Index 24: Predicted = 0x1ce3cea, Actual = 0x1fa5ee5, Error = 2892282.0998125
Index 25: Predicted = 0x3b79f62, Actual = 0x340326e, Error = 7826676.155909941
Index 26: Predicted = 0x4992721, Actual = 0x6ac3875, Error = 34804051.331749916
Index 27: Predicted = 0xc998ee1, Actual = 0xd916ce8, Error = 16244230.842530549
Index 28: Predicted = 0x181a56c4, Actual = 0x17e2551e, Error = 3670438.391939521
Index 29: Predicted = 0x25955523, Actual = 0x3d94cd64, Error = 402618432.6683471
Index 30: Predicted = 0x82c6e33c, Actual = 0x7d4fe747, Error = 91683829.12309027
Index 31: Predicted = 0xd6239bac, Actual = 0xb862a62e, Error = 499185022.823071
Index 32: Predicted = 0xe9b22d07, Actual = 0x1a96ca8d8, Error = 3216669648.649395
Index 33: Predicted = 0x371532851, Actual = 0x34a65911d, Error = 653104948.3604355
Index 34: Predicted = 0x5949f8bc7, Actual = 0x4aed21170, Error = 3855448663.1672974
Index 35: Predicted = 0x5af449ee4, Actual = 0x9de820a7c, Error = 17972423575.53308
Index 36: Predicted = 0x1391413025, Actual = 0x1757756a93, Error = 16210213485.866455
Index 37: Predicted = 0x2dbf7270dd, Actual = 0x22382facd0, Error = 49513939981.24811
Index 38: Predicted = 0x2886559a9b, Actual = 0x4b5f8303e9, Error = 149672520013.72742
Index 39: Predicted = 0x98c8113e3b, Actual = 0xe9ae4933d6, Error = 347459810714.7615
Index 40: Predicted = 0x20b05d20b7e, Actual = 0x153869acc5b, Error = 788113342243.4756
Index 41: Predicted = 0x1696ccb8e1d, Actual = 0x2a221c58d8f, Error = 1343066079089.604
Index 42: Predicted = 0x50b26b4d29f, Actual = 0x6bd3b27c591, Error = 1864358884081.7373
Index 43: Predicted = 0xdef4cb268c7, Actual = 0xe02b35a358f, Error = 83326651591.11523
Index 44: Predicted = 0x1875de8190b7, Actual = 0x122fca143c05, Error = 6898060186802.75
Index 45: Predicted = 0x1230f00b831f, Actual = 0x2ec18388d544, Error = 31407275332132.58
Index 46: Predicted = 0x689e837c30da, Actual = 0x6cd610b53cba, Error = 4636639038432.0
Index 47: Predicted = 0xcd26aa6013f0, Actual = 0xade6d7ce3b9b, Error = 34358976174165.78
Index 48: Predicted = 0xec974b5f9e34, Actual = 0x174176b015f4d, Error = 148984356258072.8
Index 49: Predicted = 0x2d6a754d3f7ab, Actual = 0x22bd43c2e9354, Error = 187823628313687.25
Index 50: Predicted = 0x2b7ce396f3181, Actual = 0x75070a1a009d4, Error = 1293723206998098.5
Index 51: Predicted = 0x11abcd83d36e18, Actual = 0xefae164cb9e3c, Error = 757478132797404.0
Index 52: Predicted = 0x18b4f3390decf8, Actual = 0x180788e47e326c, Error = 190672196844172.0
Index 53: Predicted = 0x2258a643d0ea28, Actual = 0x236fb6d5ad1f43, Error = 306834910754076.0
Index 54: Predicted = 0x31634c7beb9390, Actual = 0x6abe1f9b67e114, Error = 1.6143936485346692e+16
Index 55: Predicted = 0xf7c9efde77f4e0, Actual = 0x9d18b63ac4ffdf, Error = 2.55276090216256e+16
Index 56: Predicted = 0xaaf00881ca1cd0, Actual = 0x1eb25c90795d61c, Error = 9.013109354212386e+16
Index 57: Predicted = 0x48bd5d0ecad0f40, Actual = 0x2c675b852189a21, Error = 1.2761382323882934e+17
Index 58: Predicted = 0x2e13f10ba558040, Actual = 0x7496cbb87cab44f, Error = 3.175539853443205e+17
Index 59: Predicted = 0x10359b6a07b1e600, Actual = 0xfc07a1825367bbe, Error = 3.2969207850953344e+16
Index 60: Predicted = 0x1c178523462fa100, Actual = 0x13c96a3742f64906, Error = 5.984454014555484e+17
Index 61: Predicted = 0x11f774e94c1ec000, Actual = 0x363d541eb611abee, Error = 2.6137405792622295e+18
Index 62: Predicted = 0x7d556bf6f89d2c00, Actual = 0x7cce5efdaccf6808, Error = 3.801338671410483e+16
Index 63: Predicted = 0xe7655f0b50acf800, Actual = 0xf7051f27b09112d4, Error = 1.1258296599663145e+18
Index 64: Predicted = 0x1a78fd44662532000, Actual = 0x1a838b13505b26867, Error = 4.753071358864998e+16

Code:

import numpy as np
from scipy.interpolate import splrep, splev
from decimal import Decimal

sequence = [1, 3, 7, 8, 21, 49, 76, 224, 467, 514, 1155, 2683, 5216, 10544, 26867, 51510, 95823, 198669, 357535, 863317, 1811764, 3007503, 5598802, 14428676, 33185509, 54538862, 111949941, 227634408, 400708894, 1033162084, 2102388551, 3093472814, 7137437912, 14133072157, 20112871792, 42387769980, 100251560595, 146971536592, 323724968937, 1003651412950, 1458252205147, 2895374552463, 7409811047825, 15404761757071, 19996463086597, 51408670348612, 119666659114170, 191206974700443, 409118905032525, 611140496167764, 2058769515153876, 4216495639600700, 6763683971478124, 9974455244496707, 30045390491869460, 44218742292676575, 138245758910846492, 199976667976342049, 525070384258266191, 1135041350219496382, 1425787542618654982, 3908372542507822062, 8993229949524469768, 17799667357578236628, 30568377312064202855]
initial_points = 3
results = []
for i in range(initial_points, len(sequence)):
    x_values_known = np.arange(i)
    sequence_decimal = [Decimal(value) for value in sequence[:i]]
    spline_rep = splrep(x_values_known, sequence_decimal, k=2)
    predicted_next_number = splev(i, spline_rep)
    actual_value = sequence[i]
    predicted_hex = hex(int(predicted_next_number))
    actual_hex = hex(actual_value)
    results.append({
        'index': i,
        'predicted': predicted_hex,
        'actual': actual_hex,
        'error': abs(float(predicted_next_number) - actual_value)
    })
for result in results:
    print(f"Index {result['index']}: Predicted = {result['predicted']}, Actual = {result['actual']}, Error = {result['error']}")

Everything is useless, I have tried many predictions like this, none of them worked

[Waveilona]

Hi puzzle solvers , I am a user Waveilon also i have telegram @waveilon ,
So i am here to share with you my opinions ,
the first transaction from puzzle #66 will be transfered to this address 1EEs9PydaesijB9tpbFXRXpbMEuTHELSTH BTC
also i will post later here or tg groups  tx id which can confirm that i solved it!
I want that the puzzle was honestly paid , That's why i am here.
All we know if public key that address broadcasts anybody can take it and find the key with bsgs or kangaroo
I don't want someone interferered to my transaction.

And here is the message to puzzle creator or investor who increased the reward to this puzzle
I worked on this project 6 months and I want to get my reward honestly,
Even miners has information about this puzzle they can just take my reward ,
Here is my request for you : if i find the key and someone could take my reward, could you send me my reward to this address: 1EEs9PydaesijB9tpbFXRXpbMEuTHELSTH  BTC
You can take it from like 160 puzzle !
In my opinion if i can take my reward others can search for another puzzle without afraiding that kangaroo or bsgs even miners!
I think I could correctly express my opinion
Thanx all

[averagetoaster]

I also tried to predict #66 in a similar way but the ranges I got are still too big to do anything useful. I don't have high end hardware to test the predictions either, and the ranges I got are around 40% of the total range.
Factoring in that most likely the beginning and end of the ranges have been searched already, and doing other basic assumptions, you can reduce the range down but it's still too big.
Furthermore, now I am trying my luck on #130 because why not. It is very unlikely that #66 will get in my hands (or yours) so I just gave up on that.

[saatoshi_falling]

Like Bitcoin's price, this thread pumps sometimes with interesting info for a few pages, and then other times it dumps a few pages into red.
The past few pages are pretty red.
k3ntINA finally making public his Zodiac circle .gif that cracks #66 using Adobe Illustrator.
kachev87 with no proper programing skills saves some time for the would be c++ programmer to convert his ChatGTP python code to run in full GPU mode.
joseperal finding puzzle #30 is already emptied.
Waveilona who finally solved the puzzle but is holding it hostage.

[Ovixx]

Hi puzzle solvers , I am a user Waveilon also i have telegram @waveilon ,
So i am here to share with you my opinions ,
the first transaction from puzzle #66 will be transfered to this address 1EEs9PydaesijB9tpbFXRXpbMEuTHELSTH BTC
....................................................
Here is my request for you : if i find the key and someone could take my reward, could you send me my reward to this address: 1EEs9PydaesijB9tpbFXRXpbMEuTHELSTH  BTC
You can take it from like 160 puzzle !
...................................................
Thanx all

What a good joke, I almost laughed. I have never heard anything more stupid!
There is a saying in my country, Don't sell the bear's fur from the forest!

[kachev87]

Like Bitcoin's price, this thread pumps sometimes with interesting info for a few pages, and then other times it dumps a few pages into red.
The past few pages are pretty red.
k3ntINA finally making public his Zodiac circle .gif that cracks #66 using Adobe Illustrator.
kachev87 with no proper programing skills saves some time for the would be c++ programmer to convert his ChatGTP python code to run in full GPU mode.
joseperal finding puzzle #30 is already emptied.
Waveilona who finally solved the puzzle but is holding it hostage.

You stupid or what? Are everyone here IT specialists?! I don't think so! How many self writen programs are you using for this puzzle? How many ideas you bring to the table? How many years before the second funding of the puzzle you were at this forum? So my advice is to keep your stupid opinions at home! If you have any ideas work on them don't fill the topic with our crap!

[jacky19790729]

I also tried to predict #66 in a similar way but the ranges I got are still too big to do anything useful. I don't have high end hardware to test the predictions either, and the ranges I got are around 40% of the total range.
Factoring in that most likely the beginning and end of the ranges have been searched already, and doing other basic assumptions, you can reduce the range down but it's still too big.
Furthermore, now I am trying my luck on #130 because why not. It is very unlikely that #66 will get in my hands (or yours) so I just gave up on that.

I use RTX 3090 at about 1900 M keys, but it still takes 650 years to scan all range #66

The electricity bill is too expensive and I can't continue this puzzle 

 

[kTimesG]

I don't want someone interferered to my transaction.

Don't worry, no one will mess around with your transaction.

On another note, anyone else figured out that if you superimpose the bits of the found private keys in a quantum state chromodynamic entanglement matrix, it resembles a shiba?

[mochi86_]

Like Bitcoin's price, this thread pumps sometimes with interesting info for a few pages, and then other times it dumps a few pages into red.
The past few pages are pretty red.
k3ntINA finally making public his Zodiac circle .gif that cracks #66 using Adobe Illustrator.
kachev87 with no proper programing skills saves some time for the would be c++ programmer to convert his ChatGTP python code to run in full GPU mode.
joseperal finding puzzle #30 is already emptied.
Waveilona who finally solved the puzzle but is holding it hostage.

I like how your user is the exact opposite of saatoshi_rising of whom I believe is the puzzle creator(?). Sometimes I like to be delusional and think certain people use their alter ego on this forum (not saying people actually do, but may be a possibility lol ).

[citb0in]

I don't understand how it is possible for that to happen as long as the solver does not disclose the private key. I mean, #64 and #65 both have unknown keys, right?

No, you're wrong. The private keys of #64 and #65 are known:

#64 = F7051F27B09112D4
#65 = 1A838B13505B26867

As soon as someone makes an outgoing transaction of #66 the pubkey will be revealed. With the revealed pubkey it's a matter of seconds to get the private key and then you can replace the still-ongoing unconfirmed transaction and forward the coins of #66 to any other address.

Please understand that I cannot publish here the key for

#66 =

I'm still preparing the payout

[Airfin Same]

importing private key to electrum wallet then send to your address is it enough? or some bots can double spend tx

someone pls  advise how to avoid such attack i read that the second of the public key of key 66 67 is revealed looters can double spend tx and steal those coins

suppose someone lucky find the private key of the puzzle what he should do what wallet is more secure to use to send those coin

what do you mean payout!

can you give me answer here, thank you

[CY4NiDE]

Everything is useless, I have tried many predictions like this, none of them worked

I also tried to predict #66 in a similar way but the ranges I got are still too big to do anything useful. I don't have high end hardware to test the predictions either, and the ranges I got are around 40% of the total range.
Factoring in that most likely the beginning and end of the ranges have been searched already, and doing other basic assumptions, you can reduce the range down but it's still too big.
Furthermore, now I am trying my luck on #130 because why not. It is very unlikely that #66 will get in my hands (or yours) so I just gave up on that.

Exactly. I'm also long done with #66.

I'm currently trying my luck with #130 too cause even brute forcing it could be done in less time than #66.

I shared this not with the intent of actually predicting the correct key, but to maybe narrow down the range.

Still, my rig would take one year to search from 290860e0f31602000:291fff00000000000.

Way more if 290860e0f31602000 is used as mid point instead of starting point.

Anyways, sorry if the idea is dumb, and thanks for your answers. 

[ccinet]

Everything is useless, I have tried many predictions like this, none of them worked

I also tried to predict #66 in a similar way but the ranges I got are still too big to do anything useful. I don't have high end hardware to test the predictions either, and the ranges I got are around 40% of the total range.
Factoring in that most likely the beginning and end of the ranges have been searched already, and doing other basic assumptions, you can reduce the range down but it's still too big.
Furthermore, now I am trying my luck on #130 because why not. It is very unlikely that #66 will get in my hands (or yours) so I just gave up on that.

Exactly. I'm also long done with #66.

I'm currently trying my luck with #130 too cause even brute forcing it could be done in less time than #66.

I shared this not with the intent of actually predicting the correct key, but to maybe narrow down the range.

Still, my rig would take one year to search from 290860e0f31602000:291fff00000000000.

Way more if 290860e0f31602000 is used as mid point instead of starting point.

Anyways, sorry if the idea is dumb, and thanks for your answers. 

One year would be wonderful, whoever could solve 66 in one year!

[CY4NiDE]

One year would be wonderful, whoever could solve 66 in one year!

I'm not saying that the key is there tho.

What I meant is that my rig would take one year to search that portion of the range.

The key might not be there at all.

Personally I don't feel like wasting a year to find out.

But who knows, maybe someone with better hardware, able to do that range in much less time, might be willing to check it out for themselves.

[zahid888]

There's simply no feasible way to withdraw the funds on lower end puzzles like #66. It will be snatched up by bots. Not maybe, but it's 100% guaranteed. There will be hundreds of withdrawal transactions with varying fees all battling each other. You will simply be left in the dust.

I don't understand how it is possible for that to happen as long as the solver does not disclose the private key. I mean, #64 and #65 both have unknown keys, right?

I'm surprised why you are asking such weird questions these days, do you really not know anything about these things, or maybe you are someone else who is running the "@NotATether" account. Anyway, if you mean to ask who solved puzzles 64 and 65, let me tell you, puzzle 64 was solved by some unknown person and puzzle 65 was not solved but swept by the creator himself, I want to tell the creator to put it back in puzzle 65 or in my wallet