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Is the mystery of the 66th puzzle solved with the address 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so 2832ED74F2B5E35EE 46346217550346335726
Considering that the next value is in the range from \(2^{65}\) to \(2^{66}\), this suggests that the numbers in the series increase exponentially. To determine the pattern, let's look at the ratio of consecutive terms:
1. \(\frac{51510}{49} \approx 1051.2244897959183\)
2. \(\frac{54538862}{51510} \approx 1058.8014366142497\)
3. \(\frac{42387769980}{54538862} \approx 777.2030516514994\)
4. \(\frac{51408670348612}{42387769980} \approx 1212.8184703481304\)
5. \(\frac{44218742292676575}{51408670348612} \approx 860.1417230366171\)
Looking at these ratios, we notice that they range from approximately 1050 to 1210, but do not increase steadily. This suggests that there may be a change in the rate of exponential growth. Let's try fitting an exponential function to the series and extrapolate the next value.
Let's denote the series as \(a_n\), where \(n\) is the term number. Then, \(a_n = a_{n-1} \times k\), where \(k\) is the growth factor.
Let's assume that \(k\) is approximately constant. We can then calculate \(k\) by taking the ratio of successive terms and using it to predict the next term:
1. \(k_1 \approx \frac{51510}{49}\)
2. \(k_2 \approx \frac{54538862}{51510}\)
3. \(k_3 \approx \frac{42387769980}{54538862}\)
4. \(k_4 \approx \frac{51408670348612}{42387769980}\)
5. \(k_5 \approx \frac{44218742292676575}{51408670348612}\)
Let's take the average of these ratios for approximations of \(k\), and then use it to predict the next term in the series:
\[
k \approx \frac{k_1 + k_2 + k_3 + k_4 + k_5}{5}
\]
Once we have \(k\), we can predict the next term by multiplying the last term by \(k\):
\[
\text{Next term} = a_5 \times k
\]
Let's calculate this.
Let's first calculate the average growth factor \(k\) using the given relationships:
\[
k \approx \frac{1051.2244897959183 + 1058.8014366142497 + 777.2030516514994 + 1212.8184703481304 + 860.1417230366171}{5}
\]
\[
k \approx \frac{4959.189171446415}{5}
\]
\[
k \approx 991.837834289283
\]
Now let's use this average growth factor to predict the next term of the series:
\[
\text{Next term} = 44218742292676575 \times 991.837834289283
\]
\[
\text{Next term} \approximately 4.388266766674762 \times 10^{19}
\]
So, the predicted next term in the series is approximately \(4.388266766674762 \times 10^{19}\) in the given range from \(2^{65}\) to \(2^{66}\).
Private key in hexadecimal format
000000000000000000000000000000000000000000000000000000000000000260fe8dbe1902e000
With public key - 5 minutes for bsgs sequental with keyhunt (8 threads, k=128).address:
With pubkey - 5 minutes for bsgs sequental with keyhunt (8 threads, k=128).
address: 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so pubkey: 024ee2be2d4e9f92d2f5a4a03058617dc45befe22938feed5b7a6b7282dd74cbdd
pvk: 000000000000 00000000000000000000000000000000002832ed74f2b5e35ee
rawtx: 0100000001b3138da741af259146ca2c4895bac7e0c08147679f88ce3302fb4db0584bf3 1201000 0006a47304402201d69eb8b038805e8c681d981c27f9d22b6b3c62a0fd14b0e37d904547e5931ae 02204b475fe1ef50f4c9823ba37aa9b4c90dbae59e1a4028 a61f838987dfefa9da7b0121024ee2b e2d4e9f92d2f5a4a03058617dc45befe22938feed5b7a6b7282dd74cbddfeffffff01158e662300 0000001600140d835f0298e606c9932183672c3156113aca0b2f00000000
SigScript: 47304402201d69eb8b038805e8c681d981c27f9d22b6b3c62a0fd14b0e3 7d904547e5931ae02204
b475fe1ef50f4c9823ba37aa9b4c90dbae59e1a4028a61f838987dfefa9da7b0121024ee2be2d4e 9f92d2f5a4a03058617dc45befe22938feed5b7a6b7282dd74cbdd