Bitcoin Puzzle 66

[resper234]

Decimal for puzzle 56,46,36,26,16 & 6 (reversely given below)  - increasing 3 digit. So the 66 puzzle will be in 20 digits and within 2^65 to 2^66. Also starting digit will be 4 or 5

49
51510
54538862
42387769980
51408670348612
44218742292676575




Given that the next value is within the range of \(2^{65}\) to \(2^{66}\), it suggests that the numbers in the series are increasing exponentially. To determine the pattern, let's examine 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. \(\frac{44218742292676575}{51408670348612} \approx 860.1417230366171\)

By observing these ratios, we notice that they are approximately around 1050 to 1210, but not consistently increasing. This suggests there might be a variation in the exponential growth rate. Let's try to fit 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 \(k\) is approximately constant. We can then compute \(k\) by taking the ratio of consecutive terms and use 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 to approximate \(k\), then use it to predict the next term in the series:

\[
k \approx \frac{k_1 + k_2 + k_3 + k_4 + k_5}{5}
\]

After obtaining \(k\), 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 ratios:

\[
k \approx \frac{1051.2244897959183 + 1058.8014366142497 + 777.2030516514994 + 1212.8184703481304 + 860.1417230366171}{5}
\]

\[
k \approx \frac{4959.189171446415}{5}
\]

\[
k \approx 991.837834289283
\]

Now, let's use this average growth factor to predict the next term in the series:

\[
\text{Next Term} = 44218742292676575 \times 991.837834289283
\]

\[
\text{Next Term} \approx 4.388266766674762 \times 10^{19}
\]

So, the predicted next term in the series is approximately \(4.388266766674762 \times 10^{19}\) within the given range of \(2^{65}\) to \(2^{66}\).

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