Does the base 10 expansion of 2^*n* always contain the digit 7 if *n* is large enough?

As of 1994, this was an open question (page 196 here). I don’t know whether this has since been resolved.

The following Python code suggests that the conjecture may be true for *n* ≥ 72.

def digits(n): s = set() while n > 0: s.add(n%10) n /= 10 return s for i in range(71, 10000): p = 2**i if 7 not in digits(p): print i, p

**Update**: It appears that 2^*n* contains every digit for *n* > 168. See this comment.

**Related post**: Open question turned into exercise

Why specifically the conjecture is about only 7?

5 doesn’t seem to be there either.

9 is not present for i=108 but always after until 10000.

In fact, 168 looks like the greatest number to miss

anydigitI conjecture: maybe 7 is interesting because the rest are all proven.

Oh, the previous comment lost the identation. I was just checking that all digits are present and not just 7. After 168 doesn’t appear to return more results.

Implementation detail: this really only needs four lines:

for i in xrange(71, 10000):

p = 2**i

if ‘7’ not in str(p):

print i, p

If you really need the SET of digits in x: set(str(x)).

lvps1000vm: Thanks for the code.

I edited your comment to change your “code” tag to a “pre” tag and that restored your indentation.

TomF: You beat me to it. Not only is str() simpler than digits(), it is significantly faster, too. Here are some benchmarks:

$ time python digits.py

71 2361183241434822606848

real 0m48.748s

user 0m48.712s

sys 0m0.000s

$ time python str.py

71 2361183241434822606848

`real 0m0.575s`

user 0m0.564s

sys 0m0.008s

I think you can get much faster code by giving up on using built-in (binary) numbers altogether. Just do all the arithmetic in (binary coded) decimal. Instead of calculating 2**n each time, store an array of, say, 40000 digits, and double it on each iteration, scanning for sevens as you do. I believe this should be much faster than doing an expensive binary-to-decimal conversion each time.

Note: there are all sorts of tricks I don’t know to make BCD arithmetic faster, if you need.

I suggest that you keep only the last 60 digits or so, by setting N = (N*2) % (10**60) after each iteration. If there is no 7 in the last 60 digits, then check the last 200 digits by using pow(2, n, 10**200). If that also fails then try the last 400 digits, etc. Only compute the entire decimal expansion as a last resort. You can check exponents into the billions with this approach.