Abstract

# Abstract

I work towards a formula to predict $$2^n$$ for any $$n$$. I don’t expect to complete it, but I have found the last four digits for any $$n \in \mathbb{N}$$.

Update: I have found a relationship between the sequences that in theory should allow the calculation of any digit of $$2^n$$ by storing only $$4$$ digits...

Update II: The sequence fails for some numbers as it cycles in a non prime manner, numbers would either need to be stored, or a prime cycling sequence discovered which seems unlikely.

# Statement of Problem

I would like to be able to predict any given digit of then number $$2^n$$ for any $$n$$. The sequence begins $2,4,8,16,32,64,128,256,512,...$

We can see the end digit here will always be predictable by a sequence of period $$4$$, which is $$2,4,8,6...$$. This is nice and simple.

Assuming, there exists such a sequence for any given digit, we may attempt to find them and thier relationships with each other and the original sequence, apart from the obvious $$2^n \; \mathrm{mod} \; 10^k$$, for the $$k^{th}$$ digit.

# Finding the Sequences

So categorising the sequences $$D_k$$ for the $$k^{th}$$-least significant digit, we have $D_0 = 2,4,8,6... \;\; p:4 \;\; OEIS:A000689 \\ \\ D_1 = 0|0,0,1,3,6,2,5,1,2,4,9,9,8,6,3,7,4,8,7,5...\;\; p:20 \;\; OEIS:A160590 \\ \\ D_2 = 0,0|0,0,0,0,1,2,5,0,0,0,1,3,7,5,0,1,2,5,1,3,6,2,4,8,7,4,9,8,6,2,5,1,3,7,\\ 4,9,8,7,5,1,2,4,8,6,3,6,3,6,2,4,9,9,9,9,8,7,4,9,9,9,8,6,2,4,9,8,7,4,8,6, \\ 3,7,5,1,2,5,0,1,3,7,4,8,6,2,5,0,1,2,4,8,7,5,1,3,6,3,6,3,7,5... \;\;p:100 \;\; OEIS:-- \\ \\ D_3 = 0,0,0|... \;\; p:500 \;\; OEIS:-- \\ \\$

Where the $$|$$ denotes the starting point of the recuring sequence. We can see that the sequence $$D_k$$ is prepended by $$k$$ zeroes before the sequence with periodicity $$4\cdot5^k$$ occurs.

We may describe the first sequence by: $P*v[*(((n+3) mod 4)+1)]=D_0 \\ \mathrm{where} \\ Pv=D_0=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} 2\\4\\6\\8 \end{bmatrix}=\begin{bmatrix} 2\\4\\8\\6 \end{bmatrix}$

Asessing the occurence of each number within $$D_1$$, apart from the prepended $$0$$, we have each digit $$d\in[0,9]$$ occurring exactly twice. Therefore we may express $0,0,1,3,6,2,5,1,2,4,9,9,8,6,3,7,4,8,7,5... \\ Pv=D_1 \\ \begin{bmatrix} 1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1\\ 0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&0 \end{bmatrix} \begin{bmatrix} 0\\1\\2\\3\\4\\5\\6\\7\\8\\9\\0\\1\\2\\3\\4\\5\\6\\7\\8\\9 \end{bmatrix}$

Also anaylysing the sequences take $$D_1$$ we have $0,0,1,3,6,2,5,1,2,4,\;\;\\ 9,9,8,6,3,7,4,8,7,5...$

Where it can be seen that summing the pairs of numbers downwards makes $$9$$ for each pair. We can denote the sequence $$S^{*9*}=9,9,8,6,3,7,4,8,7,5$$ the $$9^{th}s$$ conjugate of the sequence $$S=0,0,1,3,6,2,5,1,2,4$$, such that the sum of the two is $$9,9,9,9,9,9,9,9,9,9$$

In this notation
$$D_0=S_0S_0^{*10*}$$, for $$S_0=2,4$$.
$$D_1=S_1S_1^{*9*}$$, for $$S_1=0,0,1,3,6,2,5,1,2,4$$.
$$D_2=S_2S_2^{*9*}$$, for $$S_2=0,0,0,0,1,2,5,0,0,0,1,3,7,5,0,1,2,5,1,3,6,2,4,8,7,4,9,8,6,2,5,1,3,7,4,9,8,7,5,1,2,4,8,6,3,6,3,6,2,4$$.
Which may ultimately halve the storage space for a ’lookup-method’, for quoting digits of a given $$2^n$$. It can be seen that all of the sequences tested end in $$2,4$$ and begin with $$2n$$ zero terms.

# Predictions

Thus we may set up the following rules. $\begin{array}{|c|c|} \hline If\;True & Last \; Digit \\ \hline n\;mod\;4=1 & 2 \\ n\;mod\;4=2 & 4 \\ n\;mod\;4=3 & 8 \\ n\;mod\;4=4 & 6 \\ \hline \end{array}$

$\begin{array}{|c|c|} \hline If\;True & 2^{nd} \; to \; Last \; Digit \\ \hline n-1\;mod\;20=k & D_1(k) \\ \hline \end{array}$

In general, for any digit of $$2^n$$, which is $$p$$ places away from the least significant, if $(n-p) \; mod \; 4\cdot5^p = k ,$

the $$p^{th}$$ least significant digit 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 $$2^{50000013}$$, must end in the digits $$8192$$.

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.