Two and three – the first two primes

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

Written by Eddy.