If we're interested in 0-lessness or more generally n-lessness of powers of 2 or squares and the like why stick to base 10? Are there infinitely many 2-less powers of 2 in base 3? While we're conjecturing how about this. Although there are only finitely many o-less powers of 2, for any n there is a power of 2 (hence infinitely many) whose first n digits are non-zero. By the way, the first n digits are in fact not just eventually periodic, they're periodic, period. N'est-ce pas? David At 07:29 PM 2/23/2005 -0700, you wrote:
Rich>My favorite effort in this direction is "2^86 is the last power of 2 >without a 0 digit". David Gale> Really? In my Automatic Ant book (page 43) I speculate that statements like "there are only a finite number of 7-less powers of 2" may be undecidable, in the Godel sense that there exists no proof of the statement nor of its negation from the standard axioms. Rich, you apparently have a proof with 7 replaced by 0. Could you elaborate?
Whoa! Nellie!
I have NOT proved 2^86 is the last power of 2 with no 0. It's a CONJECTURE. All the proof I have will fit comfortably in an arbitrarily small margin. I should have been more careful in my wording. My apologies for the confusion.
The only evidence I know of is reported in Hakmem item 57. Mike Beeler wrote a program that checked a block of low-order digits, and we ran it out to exponent 30M. Every power after 2^86 had at least one 0 digit.
---- Here are some notes I made up for power-digit puzzles.
The power-matching problem seems more interesting if we allow anagrams. Examples: 125/512, 169/196/961, 256/625, 1024/2401, 4096/9604, 1048576/5764801. Of course 169/961 is a cheat. And I need a rule for handling leading 0s.
Here's a "stats-gone-wrong" example:
Based on statistics alone, modelling numbers as random digit strings of a known size, 22nd and higher powers should all eventually have a 0 digit, since log10(10/9) = 1-.95424 = .04576 > 1/22.
However, there's a parametric exception:
(1,1)^22 = 1,22,231,1540,7315,26334,74613,170544,319770,497420,646646,705432 646646, ... ,22,1. (Pascal's Triangle)
1000001^22 = 1000022000231001540007315026334074613170544319770497420646646705432 646646497420319770170544074613026334007315001540000231000022000001 so (10^n+1)^22 is 8-less for n>=6.
Of course, (10^n)^22 is also 8-less, and if k^22 is 8-less, so is (10k)^22. This sort of example don't work for 0-less, since 0 is the padding digit. Variations on 10001 include 10002, 20001, 20003, 100379, 23400765, 1001001, 9999, 29993, 299930001, etc. Powers of these could be 8-less in a more or less ad hoc way, and could be extensible.
Is there a parametric set of 0-less squares? 777...444 seems to fail.
9999^2 = 99980001. Divide by 9 to get 3333^2 = 11108889. Close, but no cigar. [RIP WAH]. Multiply by 4 to get 6666^2 = 44435556. Bingo.
Is there any way to find 0-less powers other than scanning for them? Presumably the 0-less squares thin out but never completely disappear. It might be possible to use stats to prove density upper bounds on 0-less squares or powers of 2.
Rich rcs@cs.arizona.edu
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