Can You Dij It? #2

It’s very simple, but I’m fascinated by it. I’m talking about something I call the digit-line, or the stream of digits you get when you split numbers in a particular base into individual digits. For example, here are the numbers one to ten in bases 2 and 3:

Base = 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010…
Base = 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101…


If you turn them into digit-lines, they look like this:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0… (A030190 in the Online Encyclopedia of Integer Sequences)
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 2, 1, 0, 0, 1, 0, 1… (A003137 in the OEIS)


At the tenth digit of the two digit-lines, both digits equal zero for the first time:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0


When the binary and ternary digits are represented together, the digit-lines look like this:

(1,1), (1,2), (0,1), (1,0), (1,1), (1,1), (0,1), (0,2), (1,2), (0,0)


But in base 4, the tenth digit of the digit-line is 1. So when do all the digits of the digit-line first equal zero for bases 2 to 4? Here the early integers in those bases:

Base 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101…

Base 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002…

Base 4: 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, 130, 131, 132, 133, 200…


And here are the digits of the digit-line in bases 2 to 4 represented together:

(1,1,1), (1,2,2), (0,1,3), (1,0,1), (1,1,0), (1,1,1), (0,1,1), (0,2,1), (1,2,2), (0,0,1), (1,2,3), (1,1,2), (1,2,0), (0,2,2), (1,1,1), (1,0,2), (1,0,2), (1,1,2), (0,0,3), (0,1,3), (0,1,0), (1,0,3), (0,2,1), (0,1,3), (1,1,2), (1,0,3), (0,1,3), (1,1,1), (0,1,0), (1,1,0), (0,1,1), (1,2,0), (1,1,1), (1,2,1), (1,0,0), (0,1,2), (0,2,1), (1,1,0), (1,1,3), (0,2,1), (1,2,1), (1,2,0), (1,0,1), (1,0,1), (0,2,1), (1,0,1), (1,1,1), (1,2,2), (1,0,1), (1,2,1), (0,2,3), (0,1,1), (0,0,2), (0,2,0), (1,1,1), (0,1,2), (0,2,1), (0,1,1), (1,2,2), (1,2,2), (0,2,1), (0,0,2), (1,2,3), (0,2,1), (1,1,3), (0,2,0), (0,2,1), (1,2,3), (1,1,1), (1,0,1), (0,0,3), (1,0,2), (0,1,1), (0,0,3), (1,0,3), (0,1,2), (1,1,0), (0,0,0)

At the 78th digit, all three digits equal zero. But the 78th digit of the digit-line in base 5 is 1. So when are the digits first equal to zero in bases 2 to 5? It’s not difficult to find out, but the difficulty of the search increases fast as the bases get bigger. Here are the results up to base 13 (note that bases 11 and 12 both have zeroes at digit 103721663):

dig=0 in bases 2 to 3 at the 10th digit of the digit-line
dig=0 in bases 2 to 4 at the 78th digit of the digit-line
dig=0 in bases 2 to 5 at the 182nd digit of the digit-line
dig=0 in bases 2 to 6 at the 302nd digit of the digit-line
dig=0 in bases 2 to 7 at the 12149th digit of the digit-line
dig=0 in bases 2 to 8 at the 45243rd digit of the digit-line
dig=0 in bases 2 to 9 at the 255261st digit of the digit-line
dig=0 in bases 2 to 10 at the 8850623rd digit of the digit-line
dig=0 in bases 2 to 12 at the 103721663rd digit of the digit-line
dig=0 in bases 2 to 13 at the 807778264th digit of the digit-line


I assume that, for any base b > 2, you can find some point in the digit-line at which d = 0 for all bases 2 to b. Indeed, I assume that this happens infinitely often. But I don’t know any short-cut for finding the first digit at which this occurs.


Previously pre-posted:

Can You Dij It? #1

For Revver and Fevver

This shape reminds me of the feathers on an exotic bird:

feathers

(click or open in new window for full size)


feathers_anim

(animated version)


The shape is created by reversing the digits of a number, so you could say it involves revvers and fevvers. I discovered it when I was looking at the Halton sequence. It’s a sequence of fractions created according to a simple but interesting rule. The rule works like this: take n in base b, reverse it, and divide reverse(n) by the first power of b that is greater thann.

For example, suppose n = 6 and b = 2. In base 2, 6 = 110 and reverse(110) = 011 = 11 = 3. The first power of 2 that is greater than 6 is 2^3 or 8. Therefore, halton(6) in base 2 equals 3/8. Here is the same procedure applied to n = 1..20:

1: halton(1) = 1/10[2] → 1/2
2: halton(10) = 01/100[2] → 1/4
3: halton(11) = 11/100[2] → 3/4
4: halton(100) = 001/1000[2] → 1/8
5: halton(101) = 101/1000[2] → 5/8
6: halton(110) = 011/1000 → 3/8
7: halton(111) = 111/1000 → 7/8
8: halton(1000) = 0001/10000 → 1/16
9: halton(1001) = 1001/10000 → 9/16
10: halton(1010) = 0101/10000 → 5/16
11: halton(1011) = 1101/10000 → 13/16
12: halton(1100) = 0011/10000 → 3/16
13: halton(1101) = 1011/10000 → 11/16
14: halton(1110) = 0111/10000 → 7/16
15: halton(1111) = 1111/10000 → 15/16
16: halton(10000) = 00001/100000 → 1/32
17: halton(10001) = 10001/100000 → 17/32
18: halton(10010) = 01001/100000 → 9/32
19: halton(10011) = 11001/100000 → 25/32
20: halton(10100) = 00101/100000 → 5/32…

Note that the sequence always produces reduced fractions, i.e. fractions in their lowest possible terms. Once 1/2 has appeared, there is no 2/4, 4/8, 8/16…; once 3/4 has appeared, there is no 6/8, 12/16, 24/32…; and so on. If the fractions are represented as points in the interval [0,1], they look like this:

line1_1_2

point = 1/2


line2_1_4

point = 1/4


line3_3_4

point = 3/4


line4_1_8

point = 1/8


line5_5_8

point = 5/8


line6_3_8

point = 3/8


line7_7_8

point = 7/8


line_b2_anim

(animated line for base = 2, n = 1..63)


It’s apparent that Halton points in base 2 will evenly fill the interval [0,1]. Now compare a Halton sequence in base 3:

1: halton(1) = 1/10[3] → 1/3
2: halton(2) = 2/10[3] → 2/3
3: halton(10) = 01/100[3] → 1/9
4: halton(11) = 11/100[3] → 4/9
5: halton(12) = 21/100[3] → 7/9
6: halton(20) = 02/100 → 2/9
7: halton(21) = 12/100 → 5/9
8: halton(22) = 22/100 → 8/9
9: halton(100) = 001/1000 → 1/27
10: halton(101) = 101/1000 → 10/27
11: halton(102) = 201/1000 → 19/27
12: halton(110) = 011/1000 → 4/27
13: halton(111) = 111/1000 → 13/27
14: halton(112) = 211/1000 → 22/27
15: halton(120) = 021/1000 → 7/27
16: halton(121) = 121/1000 → 16/27
17: halton(122) = 221/1000 → 25/27
18: halton(200) = 002/1000 → 2/27
19: halton(201) = 102/1000 → 11/27
20: halton(202) = 202/1000 → 20/27
21: halton(210) = 012/1000 → 5/27
22: halton(211) = 112/1000 → 14/27
23: halton(212) = 212/1000 → 23/27
24: halton(220) = 022/1000 → 8/27
25: halton(221) = 122/1000 → 17/27
26: halton(222) = 222/1000 → 26/27
27: halton(1000) = 0001/10000 → 1/81
28: halton(1001) = 1001/10000 → 28/81
29: halton(1002) = 2001/10000 → 55/81
30: halton(1010) = 0101/10000 → 10/81

And here is an animated gif representing the Halton sequence in base 3 as points in the interval [0,1]:

line_b3_anim


Halton points in base 3 also evenly fill the interval [0,1]. What happens if you apply the Halton sequence to a two-dimensional square rather a one-dimensional line? Suppose the bottom left-hand corner of the square has the co-ordinates (0,0) and the top right-hand corner has the co-ordinates (1,1). Find points (x,y) inside the square, with x supplied by the Halton sequence in base 2 and y supplied by the Halton sequence in base 3. The square will gradually fill like this:

square1

x = 1/2, y = 1/3


square2

x = 1/4, y = 2/3


square3

x = 3/4, y = 1/9


square4

x = 1/8, y = 4/9


square5

x = 5/8, y = 7/9


square6

x = 3/8, y = 2/9


square7

x = 7/8, y = 5/9


square8

x = 1/16, y = 8/9


square9

x = 9/16, y = 1/27…


square_anim

animated square


Read full page: For Revver and Fevver

Pair on a D-String

What’s special about the binary number 10011 and the ternary number 1001120221? To answer the question, you have to see double. 10011 contains all possible pairs of numbers created from 0 and 1, just as 1001120221 contains all possible pairs created from 0, 1 and 2. And each pair appears exactly once. Now try the quaternary number 10011202130322331. That contains exactly one example of all possible pairs created from 0, 1, 2 and 3.

But there’s something more: in each case, the number is the smallest possible number with that property. As the bases get higher, that gets less obvious. In quinary, or base 5, the smallest number containing all possible pairs is 10011202130314042232433441. The digits look increasingly random. And what about base 10? There are 100 possible pairs of numbers created from the digits 0 to 9, starting with 00, 01, 02… and ending with …97, 98, 99. To accommodate 100 pairs, the all-pair number in base 10 has to be 101 digits long. It’s a string of digits, so let’s call it a d-string:

1, 0, 0, 1, 1, 2, 0, 2, 1, 3, 0, 3, 1, 4, 0, 4, 1, 5, 0, 5, 1, 6, 0, 6, 1, 7, 0, 7, 1, 8, 0, 8, 1, 9, 0, 9, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 6, 7, 6, 8, 6, 9, 7, 7, 8, 7, 9, 8, 8, 9, 9, 1

Again, the digits look increasingly random. They aren’t: they’re strictly determined. The d-string is in harmony. As the digits are generated from the left, they impose restrictions on the digits that appear later. It might appear that you could shift larger digits to the right and make the number smaller, but if you do that you no longer meet the conditions and the d-string collapses into dischord.

Now examine d-strings containing all possible triplets created from the digits of bases 2, 3 and 4:

1, 0, 0, 0, 1, 0, 1, 1, 1, 0 in base 2 = 558 in base 10

1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 1, 1, 2, 0, 1, 2, 1, 2, 2, 0, 2, 2, 2, 1, 0 in base 3 = 23203495920756 in base 10

1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 1, 1, 2, 0, 1, 2, 1, 1, 3, 0, 1, 3, 1, 2, 2, 0, 2, 2, 1, 2, 3, 0, 2, 3, 1, 3, 2, 0, 3, 2, 1, 3, 3, 0, 3, 3, 2, 2, 2, 3, 2, 3, 3, 3, 1, 0 in base 4 = 1366872334420014346556556812432766057460 in base 10

Note that there are 8 possible triplets in base 2, so the all-triplet number has to be 10 digits long. In base 10, there are 1000 possible triplets, so the all-triplet number has to be 1002 digits long. Here it is:

1, 0, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 0, 4, 0, 0, 4, 1, 0, 5, 0, 0, 5, 1, 0, 6, 0, 0, 6, 1, 0, 7, 0, 0, 7, 1, 0, 8, 0, 0, 8, 1, 0, 9, 0, 0, 9, 1, 1, 1, 2, 0, 1, 2, 1, 1, 3, 0, 1, 3, 1, 1, 4, 0, 1, 4, 1, 1, 5, 0, 1, 5, 1, 1, 6, 0, 1, 6, 1, 1, 7, 0, 1, 7, 1, 1, 8, 0, 1, 8, 1, 1, 9, 0, 1, 9, 1, 2, 2, 0, 2, 2, 1, 2, 3, 0, 2, 3, 1, 2, 4, 0, 2, 4, 1, 2, 5, 0, 2, 5, 1, 2, 6, 0, 2, 6, 1, 2, 7, 0, 2, 7, 1, 2, 8, 0, 2, 8, 1, 2, 9, 0, 2, 9, 1, 3, 2, 0, 3, 2, 1, 3, 3, 0, 3, 3, 1, 3, 4, 0, 3, 4, 1, 3, 5, 0, 3, 5, 1, 3, 6, 0, 3, 6, 1, 3, 7, 0, 3, 7, 1, 3, 8, 0, 3, 8, 1, 3, 9, 0, 3, 9, 1, 4, 2, 0, 4, 2, 1, 4, 3, 0, 4, 3, 1, 4, 4, 0, 4, 4, 1, 4, 5, 0, 4, 5, 1, 4, 6, 0, 4, 6, 1, 4, 7, 0, 4, 7, 1, 4, 8, 0, 4, 8, 1, 4, 9, 0, 4, 9, 1, 5, 2, 0, 5, 2, 1, 5, 3, 0, 5, 3, 1, 5, 4, 0, 5, 4, 1, 5, 5, 0, 5, 5, 1, 5, 6, 0, 5, 6, 1, 5, 7, 0, 5, 7, 1, 5, 8, 0, 5, 8, 1, 5, 9, 0, 5, 9, 1, 6, 2, 0, 6, 2, 1, 6, 3, 0, 6, 3, 1, 6, 4, 0, 6, 4, 1, 6, 5, 0, 6, 5, 1, 6, 6, 0, 6, 6, 1, 6, 7, 0, 6, 7, 1, 6, 8, 0, 6, 8, 1, 6, 9, 0, 6, 9, 1, 7, 2, 0, 7, 2, 1, 7, 3, 0, 7, 3, 1, 7, 4, 0, 7, 4, 1, 7, 5, 0, 7, 5, 1, 7, 6, 0, 7, 6, 1, 7, 7, 0, 7, 7, 1, 7, 8, 0, 7, 8, 1, 7, 9, 0, 7, 9, 1, 8, 2, 0, 8, 2, 1, 8, 3, 0, 8, 3, 1, 8, 4, 0, 8, 4, 1, 8, 5, 0, 8, 5, 1, 8, 6, 0, 8, 6, 1, 8, 7, 0, 8, 7, 1, 8, 8, 0, 8, 8, 1, 8, 9, 0, 8, 9, 1, 9, 2, 0, 9, 2, 1, 9, 3, 0, 9, 3, 1, 9, 4, 0, 9, 4, 1, 9, 5, 0, 9, 5, 1, 9, 6, 0, 9, 6, 1, 9, 7, 0, 9, 7, 1, 9, 8, 0, 9, 8, 1, 9, 9, 0, 9, 9, 2, 2, 2, 3, 2, 2, 4, 2, 2, 5, 2, 2, 6, 2, 2, 7, 2, 2, 8, 2, 2, 9, 2, 3, 3, 2, 3, 4, 2, 3, 5, 2, 3, 6, 2, 3, 7, 2, 3, 8, 2, 3, 9, 2, 4, 3, 2, 4, 4, 2, 4, 5, 2, 4, 6, 2, 4, 7, 2, 4, 8, 2, 4, 9, 2, 5, 3, 2, 5, 4, 2, 5, 5, 2, 5, 6, 2, 5, 7, 2, 5, 8, 2, 5, 9, 2, 6, 3, 2, 6, 4, 2, 6, 5, 2, 6, 6, 2, 6, 7, 2, 6, 8, 2, 6, 9, 2, 7, 3, 2, 7, 4, 2, 7, 5, 2, 7, 6, 2, 7, 7, 2, 7, 8, 2, 7, 9, 2, 8, 3, 2, 8, 4, 2, 8, 5, 2, 8, 6, 2, 8, 7, 2, 8, 8, 2, 8, 9, 2, 9, 3, 2, 9, 4, 2, 9, 5, 2, 9, 6, 2, 9, 7, 2, 9, 8, 2, 9, 9, 3, 3, 3, 4, 3, 3, 5, 3, 3, 6, 3, 3, 7, 3, 3, 8, 3, 3, 9, 3, 4, 4, 3, 4, 5, 3, 4, 6, 3, 4, 7, 3, 4, 8, 3, 4, 9, 3, 5, 4, 3, 5, 5, 3, 5, 6, 3, 5, 7, 3, 5, 8, 3, 5, 9, 3, 6, 4, 3, 6, 5, 3, 6, 6, 3, 6, 7, 3, 6, 8, 3, 6, 9, 3, 7, 4, 3, 7, 5, 3, 7, 6, 3, 7, 7, 3, 7, 8, 3, 7, 9, 3, 8, 4, 3, 8, 5, 3, 8, 6, 3, 8, 7, 3, 8, 8, 3, 8, 9, 3, 9, 4, 3, 9, 5, 3, 9, 6, 3, 9, 7, 3, 9, 8, 3, 9, 9, 4, 4, 4, 5, 4, 4, 6, 4, 4, 7, 4, 4, 8, 4, 4, 9, 4, 5, 5, 4, 5, 6, 4, 5, 7, 4, 5, 8, 4, 5, 9, 4, 6, 5, 4, 6, 6, 4, 6, 7, 4, 6, 8, 4, 6, 9, 4, 7, 5, 4, 7, 6, 4, 7, 7, 4, 7, 8, 4, 7, 9, 4, 8, 5, 4, 8, 6, 4, 8, 7, 4, 8, 8, 4, 8, 9, 4, 9, 5, 4, 9, 6, 4, 9, 7, 4, 9, 8, 4, 9, 9, 5, 5, 5, 6, 5, 5, 7, 5, 5, 8, 5, 5, 9, 5, 6, 6, 5, 6, 7, 5, 6, 8, 5, 6, 9, 5, 7, 6, 5, 7, 7, 5, 7, 8, 5, 7, 9, 5, 8, 6, 5, 8, 7, 5, 8, 8, 5, 8, 9, 5, 9, 6, 5, 9, 7, 5, 9, 8, 5, 9, 9, 6, 6, 6, 7, 6, 6, 8, 6, 6, 9, 6, 7, 7, 6, 7, 8, 6, 7, 9, 6, 8, 7, 6, 8, 8, 6, 8, 9, 6, 9, 7, 6, 9, 8, 6, 9, 9, 7, 7, 7, 8, 7, 7, 9, 7, 8, 8, 7, 8, 9, 7, 9, 8, 7, 9, 9, 8, 8, 8, 9, 8, 9, 9, 9, 1, 0

Consider the quadruplet number in base 10. There are 10000 possible quadruplets, so the all-quadruplet number is 10003 digits long. And so on. In general, the “all n-tuplet” number in base b contains b^n n-tuplets and is (b^n + n-1) digits long. If b = 10 and n = 4, the d-string starts like this:

1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 0, 0, 0, 2, 1, 0, 0, 3, 0, 0, 0, 3, 1, 0, 0, 4, 0, 0, 0, 4, 1, 0, 0, 5, 0, 0, 0, 5, 1, 0, 0, 6, 0, 0, 0, 6, 1, 0, 0, 7, 0, 0, 0, 7, 1, 0, 0, 8, 0, 0, 0, 8, 1, 0, 0, 9, 0, 0, 0, 9, 1, 0, 1, 0, 1, 1, 1, 0, 1, 2, 0, 0, 1, 2, 1, 0, 1, 3, 0, 0, 1, 3, 1, 0, 1, 4, 0, 0, 1, 4, 1, 0, 1, 5, 0, 0, 1, 5, 1, 0, 1, 6, 0, 0, 1, 6, 1, 0, 1, 7, 0, 0, 1, 7, 1, 0, 1, 8, 0, 0, 1, 8, 1, 0, 1, 9, 0, 0, 1, 9, 1, 0, 2, 0, 1, 0, 2, 1, 1, 0, 2, 2, 0, 0, 2, 2, 1, 0, 2, 3, 0, 0, 2, 3, 1, 0, 2, 4, 0, 0, 2, 4, 1, 0, 2, 5, 0, 0, 2, 5, 1, 0, 2, 6…

What about when n = 100? Now the d-string is ungraspably huge – too big to fit in the known universe. But it starts with 1 followed by a hundred 0s and every digit after that is entirely determined. Perhaps there’s a simple way to calculate any given digit, given its position in the d-string. Either way, what is the ontological status of the d-string for n=100? Does it exist in some Platonic realm of number, independent of physical reality?

Some would say that it does, just like √2 or π or e. I disagree. I don’t believe in a Platonic realm. If the universe or multiverse ceased to exist, numbers and mathematics in general would also cease to exist. But this isn’t to say that mathematics depends on physical reality. It doesn’t. Nor does physical reality depend on mathematics. Rather, physical reality necessarily embodies mathematics, which might be defined as “entity in interrelation”. Humans have invented small-m mathematics, a symbolic way of expressing the physical embodiment of big-m mathematics.

But small-m mathematics is actually more powerful and far-ranging, because it increases the number, range and power of entities and their interaction. Where are √2 and π in physical reality? Nowhere. You could say that early mathematicians saw their shadows, cast from a Platonic realm, and deduced their existence in that realm, but that’s a metaphor. Do all events, like avalanches or thunderstorms, exist in some Platonic realm before they are realized? No, they arise as physical entities interact according to laws of physics. In a more abstract way, √2 and π arise as entities of another kind interact according to laws of logic: the concepts of a square and its diagonal, of a circle and its diameter.

The d-strings discussed above arise from the interaction of simpler concepts: the finite set of digits in a base and the ways in which they can be combined. Platonism is unnecessary: the arc and spray of a fountain are explained by the pressure of the water, the design of the pipes, the arrangement of the nozzles, not by reference to an eternal archetype of water and spray. In small-m mathematics, there are an infinite number of fountains, because small-m mathematics opens a door to a big-U universe, infinitely larger and richer than the small-u universe of physical reality.

N-route

In maths, one thing leads to another. I wondered whether, in a spiral of integers, any number was equal to the digit-sum of the numbers on the route traced by moving to the origin first horizontally, then vertically. To illustrate the procedure, here is a 9×9 integer spiral containing 81 numbers:

| 65 | 64 | 63 | 62 | 61 | 60 | 59 | 58 | 57 |
| 66 | 37 | 36 | 35 | 34 | 33 | 32 | 31 | 56 |
| 67 | 38 | 17 | 16 | 15 | 14 | 13 | 30 | 55 |
| 68 | 39 | 18 | 05 | 04 | 03 | 12 | 29 | 54 |
| 69 | 40 | 19 | 06 | 01 | 02 | 11 | 28 | 53 |
| 70 | 41 | 20 | 07 | 08 | 09 | 10 | 27 | 52 |
| 71 | 42 | 21 | 22 | 23 | 24 | 25 | 26 | 51 |
| 72 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |
| 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 | 81 |

Take the number 21, which is three places across and up from the bottom left corner of the spiral. The route to the origin contains the numbers 21, 22, 23, 8 and 1, because first you move right two places, then up two places. And 21 is what I call a route number, because 21 = 3 + 4 + 5 + 8 + 1 = digitsum(21) + digitsum(22) + digitsum(23) + digitsum(8) + digitsum(1). Beside the trivial case of 1, there are two more route numbers in the spiral:

58 = 13 + 14 + 6 + 7 + 7 + 6 + 4 + 1 = digitsum(58) + digitsum(59) + digitsum(60) + digitsum(61) + digitsum(34) + digitsum(15) + digitsum(4) + digitsum(1).

74 = 11 + 12 + 13 + 14 + 10 + 5 + 8 + 1 = digitsum(74) + digitsum(75) + digitsum(76) + digitsum(77) + digitsum(46) + digitsum(23) + digitsum(8) + digitsum(1).

Then I wondered about other possible routes to the origin. Think of the origin as one corner of a rectangle and the number being tested as the diagonal corner. Suppose that you always move away from the starting corner, that is, you always move up or right (or up and left, and so on, depending on where the corners lie). In a x by y rectangle, how many routes are there between the diagonal corners under those conditions?

It’s an interesting question, but first I’ve looked at the simpler case of an n by n square. You can encode each route as a binary number, with 0 representing a vertical move and 1 representing a horizontal move. The problem then becomes equivalent to finding the number of distinct ways you can arrange equal numbers of 1s and 0s. If you use this method, you’ll discover that there are two routes across the 2×2 square, corresponding to the binary numbers 01 and 10:

2x2

Across the 3×3 square, there are six routes, corresponding to the binary numbers 0011, 0101, 0110, 1001, 1010 and 1100:

3x3

Across the 4×4 square, there are twenty routes:
4x4

(Please open in new window if it fails to animate)

(Please open in new window if it fails to animate)

Across the 5×5 square, there are 70 routes:

5x5

(Please open in new window etc)

(Please open in new window etc)

Across the 6×6 and 7×7 squares, there are 252 and 924 routes:

6x6

7x7

After that, the routes quickly increase in number. This is the list for n = 1 to 14:

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600… (see A000984 at the Online Encyclopedia of Integer Sequences)

After that you can vary the conditions. What if you can move not just vertically and horizontally, but diagonally, i.e. vertically and horizontally at the same time? Now you can encode the route with a ternary number, or number in base 3, with 0 representing a vertical move, 1 a horizontal move and 2 a diagonal move. As before, there is one route across a 1×1 square, but there are three across a 2×2, corresponding to the ternary numbers 01, 2 and 10:

3x3t

There are 13 routes across a 3×3 square, corresponding to the ternary numbers 0011, 201, 021, 22, 0101, 210, 1001, 120, 012, 102, 0110, 1010, 1100:

4x4t

And what about cubes, hypercubes and higher?