# Proving existence of powers of 10 in a sequence

I have been given the sequence of numbers

$$a(n) = 2^{2n-1}-n$$

and I want to know if any of its members is a power of 10 (other than 1). The only thing I've figured out is that the $$n$$ for such a number must be even and not a multiple of five. By searching, I know that there are no powers of 10 in the first 10000 terms, but I don't know how to prove that there are none, if that is the case.

• might be done with inequalities. When $n \equiv 8 \pmod {10}$ so that $a(n)$ is, at least, divisible by $10,$ the impressive thing is how low the exponents of $2$ and $5$ are. 8 = 2^3 3^2 5 7 13 /// 18 = 2 5^2 7^2 53 107 2473 /// 28 = 2^2 5 839 910229 2358887 /// 38 = 2 3 5 43801 cdot mbox{BIG} /// 48 = 2^4 5 cdot mbox{BIG} /// 58 = 2 5 7 cdot mbox{BIG} /// 68 = 2^2 3 5^2 29 23789 36313 202067 cdot mbox{BIG} /// 78 = 2 5 cdot mbox{BIG} /// 88 = 2^3 5 389 1399 cdot mbox{BIG} /// Jan 22 at 3:48
• Distance of closest power of ten to $2^n$ is tabulated at oeis.org/A334588 Jan 22 at 6:47

## 1 Answer

Consider that, for some integer $$n \ge 2$$, there's a positive integer $$k$$ where

$$2^{2n-1} - n = 10^k = 2^k\times 5^k \tag{1}\label{eq1A}$$

Since $$2n - 1 \gt k$$ so $$2^k \mid 2^{2n-1}$$, then with $$2^k \mid 10^k$$, we also have $$n = 2^{k}m$$ for some integer $$m$$. Dividing both sides of \eqref{eq1A} by $$2^{k}$$ gives

$$2^{2^{k+1}m - 1 - k} - m = 5^k \tag{2}\label{eq2A}$$

No $$n \le 12$$ works, so consider only $$n \gt 12$$. With $$2^{23} - 12 = 8\,388\,596$$, this requires $$k \ge 7$$. We therefore then also get

\begin{aligned} 2^{2^{k+1}m - 1 - k} - m & \ge 2^{2^{k+1} - 1 - k} - 1 \\ & \gt 2^{2^k} \\ & = 16^{2^{k-2}} \\ & \gt 16^{k} \\ & \gt 5^{k} \end{aligned}\tag{3}\label{eq3A}

However, this contradicts \eqref{eq2A}, showing there are no solutions to \eqref{eq1A} for $$n \ge 2$$.

• Your nice (+1) observation $2^k\mid n$ is the key. It may be simpler to proceed by observing that $n\ge 2^k$ already implies $2^{2n-1}-n\ge 2^{2\cdot2^k-1}-2^{k}>10^k$ whenever $k\ge2$. Jan 22 at 7:49
• Great. Could this be generalized to sequences $A^{An-1}-n$, or is the divisibility absolutely necessary? Jan 22 at 13:46
• @SuspiciousGarbage I don't see how the technique could be applied to general sequences $A^{An-1}-n$ since the divisibility is necessary to be able to show that the value of $n$ grows too fast compared to the exponent of the RHS power, e.g., $10^k$ in this case. Nonetheless, even where divisibility is not available, there might be other methods which could be applied to handle specific other sequences of the form $A^{An-1}-n$, e.g., perhaps involving modulo values. Jan 22 at 18:25