In my paper Powers of \(2\) whose digits are powers of \(2\) (see also https://research.carolin-zoebelein.de/public.html#bib6), I’m discussing digits of powers of 2, and which conditions are necessary to get for them powers of 2, too.
Given be the set of powers of \(2\) by \(P_{y} = 2^{y}\), \(y \in \mathbb{N}_{0}\). It is unknown if, apart from \(P_{y=0} = 2^{0} = 1\), \(P_{y=1} = 2^{1} = 2\), \(P_{y=2} = 2^{2} = 4\), \(P_{y=3} = 2^{3} = 8\) and \(P_{y=7} = 2^{7} = 128\), there exist more \(P_{y}\)’s whose digits are powers of \(2\) (A130693 in the On-line Encyclopedia of Integer Sequences (OEIS) http://oeis.org/A130693 [Dres07]) [Well97], too.
Looking at the set of powers of \(2\)’s [Sloa], we know that a \(m\)-digit power of \(2\) by \(P_{y}\), has a periodicity of \(\varphi\left(5^{k}\right) = 4 \cdot 5^{k-1}\) for the last \(k \leq m\) digits, starting at \(2^{k}\) [YaYa64]. Taking the known periodicity of the last \(k\)-digits into account, we want to discuss properties for the last \(k^{\prime} > k\) digits, for fixed last \(k\)-digits of \(P_{y}\).
Notation. If we write \(2^{y}_{k}\), we are talking about the \(k\)’th digit (counted from right to left, starting counting by \(1\)) of \(2^{y}\), in base \(10\) representation. For step sizes we write \(d^{k + 1}_{y,k}\), meaning the step size of the \(k + 1\)-digit, starting by \(2^{y}\), with a \(k\)-digit periodicity. Furthermore, we will denote the set of all one-digit powers of \(2\) by \(\mathcal{P}_{2} := \{ 1, 2, 4, 8 \}\).
For this, at first, we also considered \(k\)-digit structures of powers of \(2\) in generally, and used the following two lemmas as starting point for our proofs in the mentioned paper.
Lemma 2.1 (\(k\)-digits structure). Let be \(P_{y} = 2^{y}\), \(y \in \mathbb{N}_{0}\), and the last \(k^{\star}\)-digits periodical with \(\varphi\left(5^{k^{\star}}\right) = 4 \cdot 5^{k^{\star} - 1}\), for all \(2^{y} \geq 2^{k^{\star}}\), \(k^{\star} \geq 2\). Then for \(2^{k + k^{\star} + \varphi\left(5^{k}\right)}\), \(k \in \left[ k^{\star}, k^{\star} + \varphi\left(5^{k^{\star} - 1}\right) - 1 \right]\), the last \(k\)-digits are given by \(2^{1 + k^{\star} + \varphi\left(5^{1}\right)}_{1} \cdot 2^{k - 1}\), with \(k - x \approx \left(1 - \log_{10}\left(2\right)\right)k - k^{\star}\log_{10}\left(2\right)\) leading zeros for \(k \geq 2\), and at least one leading zero for \(k \geq 3\).
Proof. We know, that for the last \(k\)-digits \(2^{k + k^{\star} + \varphi\left(5^{k}\right)} \sim 2^{k + k^{\star}}\), which have \(x \approx \left(k + k^{\star}\right) \log_{10}\left(2\right)\) digits. Since, we also have the periodicity \(\varphi\left(5^{k}\right)\), we directly get \(k - x \approx \left(1 - \log_{10}\left(2\right)\right)k - k^{\star}\log_{10}\left(2\right)\) for the number of leading zeros. Looking at \(0 \leq k - x\), we receive \(k \gtrsim k^{\star} \frac{\log_{10}\left(2\right)}{1 - \log_{10}\left(2\right)}\), and hence \(k \geq 2\) by the constraint \(k^{\star} \geq 2\), and for \(1 \geq k - x\), with \(k = k^{\star}\), we recive \(k \gtrsim \frac{1}{1 - 2\log_{10}\left(2\right)}\), and hence \(k \geq 3\). Finally it is easy to see, that the statement is always satisfied for \(k \geq k^{\star}\), because of \(k^{\star} \gtrsim k^{\star} \frac{\log_{10}\left(2\right)}{1 - \log_{10}\left(2\right)} \approx 0.4k^{\star}\) for \(k = k^{*}\).
Lemma 2.2 (\(k^{\star}\)-digits fixed structure). Let be \(P_{y} = 2^{y}\), \(y \in \mathbb{N}_{0}\), and the last \(k^{\star}\)-digits periodical with \(\varphi\left(5^{k^{\star}}\right) = 4 \cdot 5^{k^{\star} - 1}\), for all \(2^{y} \geq 2^{k^{\star}}\), \(k^{\star} \geq 2\). Then for \(2^{k + k^{\star} + \varphi\left(5^{k}\right)}\), \(k \in \left[ k^{\star}, k^{\star} + \varphi\left(5^{k^{\star} - 1}\right) - 1 \right]\), the last \(k + 1\) to \(k + \delta k\)-digits are fixed for at least \(\delta k = k^{\star}\) digits.
Proof. Consider \(\left( 2^{k + k^{\star} + \varphi\left(5^{k}\right)} - 2^{1 + k^{\star} + \varphi\left(5^{1}\right)}_{1} \cdot 2^{k - 1} \right) \cdot 10^{-k} \cdot 2^{\varphi\left(5^{\delta k}\right)} \approx \left( 2^{\left(k + 1\right) + k^{\star} + \varphi\left(5^{k + 1}\right)} - 2^{1 + k^{\star} + \varphi\left(5^{1}\right)}_{1} \cdot 2^{\left(k + 1\right) - 1} \right) \\ \cdot 10^{-\left(k + 1\right)} \left( 2^{k + k^{\star} + \varphi\left(5^{k}\right)} - 2^{1 + k^{\star} + \varphi\left(5^{1}\right)}_{1} \cdot 2^{k - 1} \right) \cdot 2^{\varphi\left(5^{\delta k}\right)} \approx \left( 2^{k + k^{\star} + \varphi\left(5^{k}\right)} \cdot 2^{4\varphi\left(5^{k}\right)} - 2^{1 + k^{\star} + \varphi\left(5^{1}\right)}_{1} \cdot 2^{k - 1} \right) \\ \cdot 5^{-1}\), for which we can equating the coefficients with approximation. We look at \(\varphi\left(5^{\delta k}\right) \approx 4\varphi\left(5^{k}\right)\), and receive \(\delta k \approx \lfloor \log_{5}\left(4 \cdot 5^{k}\right) \rfloor \approx \lfloor 1.86 k \rfloor \approx k\). Finally, we can conclude \(\delta k \gtrsim k^{\star}\) for \(k \in \left[ k^{\star}, k^{\star} + \varphi\left(5^{k^{\star} - 1}\right) - 1 \right]\).
References
[Dres07] Dresden, Gregory P.: A130693 - OEIS: Powers of 2 whose digits are powers of 2.
[Sloa] Sloane, N. J. A.: Table of n, 2n for n = 0..1000 - OEIS.
[Well97] Wells, David: The Penguin dictionary of curious and interesting numbers : Penguin, 1997
[YaYa64] Yaglom, AM ; Yaglom, IM: Challenging Mathematical Problems with Elementary Solutions Bd. I, Holden-Day Inc. (1964)