power(0) is (: x->1 :), power(1+n) is (: x-> x.power(n,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) = (: x-> power(n,x).power(m,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. 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.
1 = p(0,3) = p(0,2) 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 = 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.