As I’ve said before, I love the way that numbers can come in many different guises. For example, take the number 21. It comes in all these guises:

21 = 10101 in base 2 = 210 in base 3 = 111 in b4 = 41 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21 in b10 = 1A in b11 = 19 in b12 = 18 in b13 = 17 in b14 = 16 in b15 = 15 in b16 = 14 in b17 = 13 in b18 = 12 in b19 = 11 in b20 = 10 in b21

But I’ve not chosen 21 at random. If you sum the 1s in the representations of 21 in bases 2 to 21, look what you get:

21 = 10101 in base 2 = 210 in base 3 = 111 in b4 = 41 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21 in b10 = 1A in b11 = 19 in b12 = 18 in b13 = 17 in b14 = 16 in b15 = 15 in b16 = 14 in b17 = 13 in b18 = 12 in b19 = 11 in b20 = 10 in b21

21 = 1

_{s=1}01_{s=2}01_{s=3}in base 2 = 21_{s=4}0 in base 3 = 111_{s=7}in b4 = 41_{s=8}in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21_{s=9}in b10 = 1_{s=10}A in b11 = 1_{s=11}9 in b12 = 1_{s=12}8 in b13 = 1_{s=13}7 in b14 = 1_{s=14}6 in b15 = 1_{s=15}5 in b16 = 1_{s=16}4 in b17 = 1_{s=17}3 in b18 = 1_{s=18}2 in b19 = 11_{s=20}in b20 = 1_{s=21}0 in b21

In other words, 21 = digcount(21,dig=1,base=2..21). But n = digcount(n,dig,b=2..n) doesn’t happen for any other digit and doesn’t happen often with 1:

3 = digcount(3,d=1,b=2..3) = 11 in b2 = 10 in b3

4 = digcount(4,d=1,b=2..4) = 100 in b2 = 11 in b3 = 10 in b4

6 = digcount(6,d=1,b=2..6) = 110 in b2 = 20 in b3 = 12 in b4 = 11 in b5 = 10 in b6

10 = digcount(10,d=1) = 1010 in b2 = 101 in b3 = 22 in b4 = 20 in b5 = 14 in b6 = 13 in b7 = 12 in b8 = 11 in b9 = 10 in b10

15 = digcount(15,d=1) = 1111 in b2 = 120 in b3 = 33 in b4 = 30 in b5 = 23 in b6 = 21 in b7 = 17 in b8 = 16 in b9 = 15 in b10 = 14 in b11 = 13 in b12 = 12 in b13 = 11 in b14 = 10 in b15

21 = digcount(21,d=1) = 10101 in b2 = 210 in b3 = 111 in b4 = 41 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21 in b10 = 1A in b11 = 19 in b12 = 18 in b13 = 17 in b14 = 16 in b15 = 15 in b16 = 14 in b17 = 13 in b18 = 12 in b19 = 11 in b20 = 10 in b21

After that, the digcount(n,d=1,b=2..n) → n/2 (see “Digital Dissection” for further discussion). But I decided to look for the first n where digcount(n,dig,b=2..n) = 666:

digcount(1270,1) = 666

digcount(3770,2) = 666

digcount(7667,3) = 666

digcount(12184,4) = 666

digcount(18845,5) = 666

digcount(25806,6) = 666

digcount(34195,7) = 666

digcount(43352,8) = 666

digcount(54693,9) = 666

It doesn’t stop there, of course. You can carry on for ever, looking for digcount(n,A) = 666, digcount(n,B) = 666, digcount(n,C) = 666, where A = 10, B = 11 and C=12, and so on. But it doesn’t start there, either. What about digcount(n,0) = 666? That isn’t easy to find, because 0 usually occurs far less often than other digits in the representation of n. Here are the integers setting records for digcount(n,0,b=2..n):

2 → digcount(2,0) = 1 ← 2= 10 in base 2

4 → digcount(4,0) = 3; ← 4 = 100 in base 2, 11 in base 3, 10 in base 4

8 → digcount(8,0) = 5 ← 8 = 1000 in base 2, 22 in base 3, 20 in base 4, 13 in base 5, 12 in base 6, 11 in base 7, 10 in base 8

12 → digcount(12,0) = 6

16 → digcount(16,0) = 8

18 → digcount(18,0) = 9

32 → digcount(32,0) = 11

36 → digcount(36,0) = 13

64 → digcount(64,0) = 15

72 → digcount(72,0) = 18

128 → digcount(128,0) = 20

144 → digcount(144,0) = 24

252 → digcount(252,0) = 25

264 → digcount(264,0) = 27

288 → digcount(288,0) = 29

360 → digcount(360,0) = 30

504 → digcount(504,0) = 33

540 → digcount(540,0) = 36

720 → digcount(720,0) = 40

900 → digcount(900,0) = 42

1080 → digcount(1080,0) = 47

1680 → digcount(1680,0) = 48

1800 → digcount(1800,0) = 53

2160 → digcount(2160,0) = 56

2520 → digcount(2520,0) = 61

3600 → digcount(3600,0) = 64

4320 → digcount(4320,0) = 66

So what is the first n for which digcount(n,0) = 666? Watch this space.