Benedict Irwin edited Predictions.tex  over 9 years ago

Commit id: 9b9c6e0b7dc051cb129340d108cb16567734245e

deletions | additions      

       

In general for any digit of $2^n$, having $q$ digits,  which is $p$ places away from the least significant, if \begin{equation} (n-p) \; mod \; 4\cdot5^p = k , \; p\in[0,...], p\in[0,q-1],  \end{equation}  where $k$ will take ranges from $0$ to $4\cdot5^p-1$. Then $4\cdot5^p-1$,  the $p^{th}$ least significant digit is $D_p(k)$, that is the $k^{th}$ term in the series $D_p$. Thus by knowing that D_0(1)=2,D_1(12)=9,D_2(11)=1,D_3(10)=8. We can tell that the number 2^{513}, 2^{50000013},  must end in the digits 8192.Which according to a check is true!  However, the last part of the problem is how to generate the series $D_p$ for any given $p$.  Well we can tell that each has $4\cdot5^p$ terms, but can be stored as $2\cdot5^p$ terms as the last half of them are the $9$ conjugate of the first half. We also know that the first $2p$ terms are zeroes. and there is a chance that the last two digits of the $S_p$ series are always $2,4$, making the last two digits of the $S^{*9*}$ series $7,5$. The odd revelaing of snippets of information is interesting, it would allow the solution for a very slight number of problems automatically, if there existed such an $n$ that $2^n$ just happened to be soluble for each digit based of of this information alone. However it also indicates there could be more information yet to be revealed. A closed form would be quite spectacular, as it would be much easier to write such numbers as $2^{123456789}$ or even larger, (or at least specify a given digit upon request). Although from what has been found alone we can also tell that, for example $2^{5000013}$ must end in $8192$, also, and so forth.