### Brand-friedman_prese..

```Michael Brand
VAC29

1024 = (4-2)10
25=52
But not any power of 10.

In other bases: B-Friedman numbers


11001TWO  101TWO
10TWO

-- Erich Friedman (2000)
F (n)  x  Friedman 0  x  n
lim F (n) / n  ?
n


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,
L(b)  k  a.m.[
0] .b  Friedman
t
k


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)

10411041 = 1041 × (104+1)
LB ( s ) r
B
1
r
[ sB ]  s LB ( s )
B
1

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.
enc x1 , x2 ,, xn   x1
xn
x2


For any x and s>0 there is a number cx s.t. x
can be represented by cx repetitions of s.
For example:
if x  0,
 ss
s
s
x   
otherwise.
s 
s


x

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).
L(s)r

10
1 
r
t
L(s)r
 10 k 'L ( s ) r  0
a.[s ] .[
0] .b   a  (0  10)
 s  L(s)


0  b

10  1 

t times
k




(0,10,L(s),s,10L(s),1,10L(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 radicalfree integers.
Let G be the number of unique order
arrangements for these integers.
|span([s]r’(s-3)-C)|≥Gr’(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
r '( s 3 )C
G
by
 r ' sL ( s )

1  1  10


A density of 1 follows if Gs-3>10sL(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>BL(s)s/(s-3) to
log B k  i 1 iki / k
h
◦ 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.
□

questions?
```