Another base representation investigation

[David Wilson]

I looked into another question about n-ary representations.

Call n a n-ary b-ary number if it only has n-ary digits in its b-ary 
representation. For example, 110 is a binary decimal number, because it only 
has binary digits in its base 10 representation.

In thinking about decimal b-ary numbers, it occured to me that for b > 10, 
most large numbers would tend to have b-ary digits > 9. This made me think 
that n would have a decimal b-ary representation for only a few b, maybe a 
bounded number. This made me wonder:

Question 1: What number(s) have decimal b-ary representations for the 
largest number of b >= 2?

In programming this, I quickly realized that for b > n, the b-ary 
representation of n is the single digit n. This implied that n < 10 is 
decimal b-ary for every b, while n >= 10 can be decimal b-ary only for b <= 
n. To make the question meaningful for all n, I had to restrict the question 
to b <= n, effectively excluding one-digit numbers from consideration. This 
would not affect the answer for n >= 10. So:

Question 1a: What number(s) have decimal b-ary representations for the 
largest number of 2 <= b <= n?

Now I start computing. For each n, I compute f(n) = number of 2 <= b <= n 
for which n is decimal b-ary, and record the increasing values of f(n). I 
get this:

n f(n)

2 1

3 2

4 3

5 4

6 5

7 6

8 7

9 8

10 9

11 10

12 11

13 12

14 13

15 14

16 15

17 16

18 17

19 18

20 19

22 20

24 21

26 22

28 23

30 24

33 25

36 26

39 27

42 28

45 29

51 30

56 31

57 32

66 33

69 34

72 35

81 36

84 37

104 38

105 40

144 42

177 43

225 46

324 47

441 48

576 50

729 51

1296 52

2025 53

2304 55

4761 57

11664 60

It is here that I notice that many of the larger values of n are square 
numbers, that is n = r^2. I realize that such n will have at least the two 
decimal n-ary representations 100 in base r and 10 in base r^2, and probably 
additional ones, like 20, 30, 400 or 900 in other bases, depending on r. So 
the square numbers n really have an advantage here.

To follow this lead, I decided to check out binary b-ary representations. I 
did the analogous thing for base 2, and got:

n f(n)

2 1

3 2

4 3

9 4

30 5

81 6

4096 7

Well, there are four square n in there. But 4096 = 2^12 is intriguing. It 
has binary b-ary representations for 7 bases:

b b-ary representation of 4096

2 1000000000000

4 1000000

8 10000

16 1000

64 100

4095 11

4096 10

Well there's an obvious pattern to most of these. Specifically, 4096 is a 
power of b = 2, 4, 8, 16, 64 and 4096, so 4096 has a binary b-ary 
representation in all of these bases.

Now it becomes apparent that we can find a number n with as many b-ary 
representatations as we wish. If we want k or more representations, we 
choose n = 2^m where m has k or more divisors. Since a binary b-ary number 
is an k-ary b-ary number for k >= 2, my original question 1a is answered in 
the negative, there is no number n with a largest number of decimal b-ary 
representations.



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