Presentation on theme: "Michael Brand VAC29. 1024 = (4-2) 10 25=5 2 But not any power of 10. In other bases: B-Friedman numbers."— Presentation transcript:
Michael Brand VAC29
1024 = (4-2) 10 25=5 2 But not any power of 10. In other bases: B-Friedman numbers
-- Erich Friedman (2000) We prove =1 Not just for Friedman numbers, but also for B-Friedman numbers in any base B.
Step 1: We show that there exist Friedman infixes. ◦ ∃(m,k), s.t. ∀a, b, Step 2: We show a systematic way of constructing Friedman infixes with many different k that all share a constant l=L(m) and all satisfy l|k. Step 3: We show that the number of possible k grows with l enough to conclude a density of 1.
“Yields a falsehood when appended to its own quotation” yields a falsehood when appended to its own quotation. -- W.v.O. Quine quines (Bratley and Milo, 1972) = 1041 × ( )
We need a method to unambiguously encode an entire tuple in a single number. ◦ We don’t need every single integer tuple to be representable, but we do need a sufficiently dense set. ◦ We do need to minimize the number of extraneous digits used. Solution: encode the tuples of radical-free integers.
For any x and s>0 there is a number c x s.t. x can be represented by c x repetitions of s. For example:
A Friedman infix: m = [s] r = [s] r’s, for any s>3 and any large enough r’, with a suitable k = k’L(s)r = k’L(m). (0,10,L(s),s,10 L(s),1,10 L(s) -1,10,L(s)) all, together, in the span of some constant, C, repetitions of s. r ∈ span([s] r’ ) by s+...+s. We need 3 of them. This leaves k’ ∈ span([s] r-(C+3r’) ) = span([s] r’(s-3)-C ).
m = [s] r’s. k’ ∈ span([s] r’(s-3)-C ). We want to choose a good s. Chooses to be the concatenation of radical- free integers. Let G be the number of unique order arrangements for these integers. |span([s] r’(s-3)-C )|≥G r’(s-3)-C.
By using the infix m=[s] r’s, we constrain only L(m) =r’sL(s) digits. By choosing k=k’L(m), we ensure the constraints are independent for different k. The density is therefore bounded from below by A density of 1 follows if G s-3 >10 sL(s), because we can take an arbitrarily large r’.
In base 10: ◦ taking s to be the concatenation of all radical-free integers of up to h=3 digits, repeated d=13 times. For a general base B, we need to pick h=h(B) and d=d(B).
Using a large enough d, one can use Stirling’s formula to reduce G>B L(s)s/(s-3) to ◦ Shorter than minimum message length! For a large enough h, the radical-free integers’ expectation converges to that of all integers in the same range, which satisfies the property. □