Comments on: Digits in powers of 2

By: Which powers of 2 do not contain the digit 7? | mathblag

Which powers of 2 do not contain the digit 7? | mathblag — Mon, 10 Dec 2012 01:31:10 +0000

[...] interest in this question was inspired by a blog post by John D. Cook. Share this:PrintEmailMoreStumbleUponGoogle [...]

By: Dave Radcliffe

Dave Radcliffe — Sun, 09 Dec 2012 07:35:05 +0000

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.

By: Visto nel Web – 54 « Ok, panico

Visto nel Web – 54 « Ok, panico — Sun, 25 Nov 2012 08:41:59 +0000

[...] Digits in powers of 2 per Bit3Lux ::: The Endeavour [...]

By: David Feuer

David Feuer — Sat, 24 Nov 2012 20:01:10 +0000

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.

By: robru

robru — Sat, 24 Nov 2012 00:06:38 +0000

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

By: John

John — Fri, 23 Nov 2012 17:38:56 +0000

lvps1000vm: Thanks for the code.

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

By: TomF

TomF — Fri, 23 Nov 2012 17:38:55 +0000

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)).

By: lvps1000vm

lvps1000vm — Fri, 23 Nov 2012 15:24:52 +0000

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.

By: lvps1000vm

lvps1000vm — Fri, 23 Nov 2012 15:21:13 +0000

In fact, 168 looks like the greatest number to miss any digit

def digits(n):
    s = set()
    while n > 0:
        s.add(n%10)
        n /= 10
        if len(s) == 10:
            break
    return s

for i in range(100, 100000):
    p = 2**i
    if len(digits(p)) < 10:
        print i, len(digits(p))
    if i % 10000 == 0:
        print "checked",i

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

By: Yves Langlois

Yves Langlois — Fri, 23 Nov 2012 13:51:23 +0000

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.