power(0) is (: 1 ←x :), power(1+n) is (: x.power(n, x) ←x :), for each natural n (x varies over whatever values we know how to multiply by themselves and 1). One may equally define: power(1) is the identity, power(n+m) = (: power(n,x).power(m,x) ←x :) and leave power(0) out of the discussion when no multiplicative identity is available.

Notice, for natural i, that power(power(i, 3), 2) +1 is always a multiple of power(i+1, 3). Proof:

For i = 0, power(power(0,3),2) is power(1,2) is 2 and 2+1 is 3 which is power(0+1,3); once known for i−1, so i>0 and 2.i>i, making power(power(i−1,3),2) equal to q.power(i,3)−1 for some natural q, we can cube each to find power(power(i,3),2) equal to the cube of q.power(i,3)−1;

- which comes as a series of terms, in steadily diminishing (from 3) powers of q.power(i,3);
- given 2.i>i, all terms but the last two (powers 1 and 0) are multiples of power(2.i,3) hence, in particular, of power(i+1,3);
- furthermore, the coefficient of the power(1) term is 3, making it also a multiple of power(i+1,3);
- leaving power(power(i,3),2)+1 equal to some multiple of power(i+1,3);
- QED.

Now I'm more interested in a power of 3 that is one short of a power of
2. The obvious example is 3 itself, one short of 4, but is there a bigger ?
Lest you ask why, powers of 3 provide an interesting number system, which I'd
ideally encode within the conventional binary context leaving as little slack
as possible – but just enough to (for example) distinguish keep
reading

from that was the last digit of the number

. For these
purposes, 243 = power(5,3) and 256 = power(8,2) differ by only 13, which is
pretty good: a byte can encode a bundle of 5 trits or an out-of-band
message.

So let me look at polynomials in the powers of 2 and 3; and I'll abbreviate power as just p.

- 1 = p(0,3) = p(0,2) = 3−2
- 2 = 1+1
- 3 = 1+3
- 4 = p(2,2)
- 5 = 3+2 = p(2,2)+1
- 7 = p(2,2)+3
- 8 = p(3,2)
- 9 = p(2,3) = p(3,2)+1
- 11 = p(2,3) + 2 = p(3,2) + 3
- 13 = p(2,3) + p(2,2)
- 16 = p(4,2)
- 17 = p(4,2) + 1 = p(2,3) + p(3,2)
- 19 = p(4,2) + 3
- 23 = p(3,3) − p(2,2) first −
- 27 = p(3,3)
- 29 = p(3,3) + 2
- 31 = p(3,3) + p(2,2) = p(5,2) − 1
- 32 = p(5,2)
- 37 = p(6,2) − p(3,3)
- 41 = p(5,2) + p(3,2)
- 43 = p(3,3) + p(4,2)
- 47 = p(7,2) − p(4,3)
- 53 = p(6,2) − p(2,3) −2 = 2.p(3,3) − 1 = p(4,3) −p(3,3) −1 = p(2,7)+p(2,2)
- 59 = p(5,2) + p(3,3)
- 61 = p(6,2) −3
- 64 = p(6,2)
- 67 = p(6,2) +3
- 71 = p(4,3) −2.(2+3) = p(4,3) −p(2,3) −1 = p(6,2) +p(3,2) −1
- 73 = p(6,2) + p(2,3)
- 79 = p(4,3) − 2
- 81 = p(4,3)
- 83 = p(4,3) + 2
- 89 = p(4,3) + p(3,2)
- 97 = p(4,3) + p(4,2)
- 101 = p(7,2) − p(3,3)
- 103 =

I'm also interested in which powers of 2 are one more or less than a prime. Conjecture: every prime is the sum or difference of a power of 2 and a power of 3.

- 2−1=p(0,n), 2+1=3
- p(2,2)−1 = 3, p(2,2)+1 = 3+2
- p(2,3)−1 = (3−1).(3+1) = 2.p(2,2) = p(3,2), p(2,3)+1 = 2.(p(2,2)+1)
- p(3,2)+1 = p(2,3), p(3,2)−1 = p(2,2)+3
- p(3,3)−1 = 2.(p(2,3)+p(2,2)), p(3,3)+1 = p(2,2).(p(3,2)−1)
- p(4,2)−1 = 3.(3+2), p(4,2)+1 = p(3,2)+p(2,3)
- p(5,2)−1 = p(3,3)+p(2,2)+1, p(5,2)+1 = 3.(p(3,2)+2)
- p(4,3)
- p(3,2)−1 = 7, p(3,2)+1 = 2.p(2,2) + 1 = (3−1).(3+1) + 1 = p(2,3)
- p(4,2)−1 = (p(2,2)−1).(p(2,2)+1) = 15, p(4,2)+1 = 17
- p(5,2)−1 = 31, p(5,2)+1 = 3.(p(2,3)+2)
- p(6,2)−1 = (p(3,2)−1).(p(3,2)+1) = 7.p(2,3)
- 3+1=p(2,2), p(2,2)+1=5
- p(3,2)−1=7, p(3,2)+1=p(2,3)
- p(4,2)+1=17
- p(5,2)−1=31
- p(6,2)−1 = (p(3,2)−1).(p(3,2)+1) = 7.p(2,3)
- p(9,2) = p(6,3) −p(5,3) +p(3,3) −1 = (p(3,3)−p(2,3)+1).p(3,3) −1