Heuristically there should be infinitely many numbers with this property, and they should be about as common as square numbers (to within a constant multiplicative factor). In particular, let S be the string: "1248163264128256512[...]" obtained by concatenating all of the powers of 2. For a given integer n, let f(n) be the position of the first occurrence of the decimal representation of n as a substring of S. If we instead use a random string of uniform iid digits in place of S, then f(n) will follow a geometric distribution with mean 10^len(str(n)), which lies between n and 10 n. If we assume independence of f(n) and f(n+1) (i.e. we generate a different random string each time), then we can find the probability that f(n) and f(n+1) belong to the same power of 2; it is not too difficult to show that this is asymptotically proportional to 1/sqrt(n). But of course this is just a heuristic argument rather than a mathematical proof; finding the latter is incredibly difficult. Best wishes, Adam P. Goucher
Sent: Sunday, December 23, 2018 at 4:14 PM From: "Keith F. Lynch" <kfl@KeithLynch.net> To: math-fun@mailman.xmission.com Subject: [math-fun] New Year's oddity
Lots of things will change with the new year in a few days. But there's one thing that won't: The power of two in which the year first appears.
As we all know, 2^212 is 6582018229284824168619876730229402019930943462534319453394436096. What you may not have noticed, not being as perceptive as me (i.e. having a life) is that it contains both 2018 and 2019 as substrings. And that it's the *first* power of two to contain *either* of those numbers as substrings.
When, if ever, is the last time this happened? When, if ever, will be the next time? What's the expected density of this sequence? And should I add it to OEIS?
Of course the question can be extended to other radices than base ten, and to other powers than those of two.
A problem I haven't solved is which is the first power of two to start with 2018, and which is the first to start with 2019. It's easy to prove that there must be a solution, not just for 2018 and 2019, but for any positive integer.
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