Benedict Irwin edited Predictions.tex  over 9 years ago

Commit id: 757678857e7381c8925fdce8484d19d0a0372079

deletions | additions      

       

(n-p) \; mod \; 4\cdot5^p = k , \;\; p\in[0,\infty],  \end{equation}  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^{50000013}, $2^{50000013}$,  must end in the digits 8192. $8192$.  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.