It’s such a simple thing: repeatedly doubling a number: 1, 2, 4, 8, 16, 32, 61, 128… And yet it yields such riches, reminiscent of DNA or a literary text:

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

2^8 = 256

2^9 = 512

2^10 = 1024

2^20 = 1048576

2^30 = 1073741824

2^40 = 1099511627776

2^50 = 1125899906842624

2^60 = 1152921504606846976

2^70 = 1180591620717411303424

2^80 = 1208925819614629174706176

2^90 = 1237940039285380274899124224

2^100 = 1267650600228229401496703205376

2^200 = 1606938044258990275541962092341162602522202993782792835301376

Although, by Benford’s law*, 1 is the commonest leading digit, do all numbers eventually appear as the leading digits of some power of 2? I conjecture that they do. indeed, I conjecture that they do infinitely often. If the function first(n) returns the power of 2 whose leading digits are the same as the digits of n, then:

first(1) = 2^0 = 1

first(2) = 2^1 = 2

first(3) = 2^5 = 32

first(4) = 2^2 = 4

first(5) = 2^9 = 512

first(6) = 2^6 = 64

first(7) = 2^46 = 70368744177664

first(8) = 2^3 = 8

first(9) = 2^53 = 9007199254740992

first(10) = 2^10 = 1024

And I conjecture that this is true of all bases except bases that are powers of 2, like 2, 4, 8, 16 and so on. A related question is whether the leading digits of any 2^n are the same as the digits of n. Yes:

2^6 = 64

2^10 = 1024

2^1542 = 1.54259995… * 10^464

2^77075 = 7.70754024… * 10^23201

2^113939 = 1.13939932… * 10^34299

2^1122772 = 1.12277217… * 10^337988

That looks like a look of calculation, but there’s a simple way to cut it down: restrict the leading digits. Eventually they will lose accuracy, because the missing digits are generating carries. With four leading digits, this happens:

1: 0001

2: 0002

4: 0004

8: 0008

16: 0016

32: 0032

64: 0064

128: 0128

256: 0256

512: 0512

1024: 1024

2048: 2048

4096: 4096

8192: 8192

16384: 1638…

32768: 3276…

65536: 6552…

But working with only fifteen leading digits, you can find that 1122772 = the leading digits of 2^1122772, which has 337989 digits when calculated in full.

Previously pre-posted (please peruse):

*Not Zipf’s law, as I originally said.