Well, one summer, I decided to tackle the problem. Not having any great knowledge of number theory, I used a more brute force approach. Note that for greater comprehensibility, I have broken the resulting formula up into several stages, but it would not be difficult to put it back together into one vast formula:

{e is prime:} PRIME(e) :=

Eb:Ec:((b+2)*(c+2) = e)

{if e is a prime, true iff a is a power of e:} PPOWER(a,e) :=

Ab:Ac:((b*c = a) -> ((b = 1) or (Ed: (e*d) = b)))

{if e is a prime, and a is a power of e, true iff d is the

(log_e a)th digit (counting from the right, starting with 0) in the base-e expansion of n:}

DIG(a,e,d,n) :=

(d < e) & Eb:Ec:((c < a) & (n = (b*e*a) + (d*a) + c))

{if e is prime, and a is a power of e, true iff n has exactly

(log_e a) digits in its base-e expansion (0 is counted as having 0 digits:}

LENGTH(e,a,n):=

(n < a) & Ab:((PPOWER(b,e) & (b < a)) -> (b <= n))

POWER_OF_TEN(x):=

Ee:(PRIME(e) & (e > x) & En:Ea:(LENGTH(e,a,n) & DIG(1,e,1,n) & Ai:Aj:Au:( (PPOWER(u,e) & ((e*u) < a) & DIG(u,e,i,n) & DIG(e*u,e,j,n)) -> j = (10 * i) ) & Eu:(PPOWER(u,e) & (e*u = a) & DIG(u,e,x,n)) ) )

The basic idea is that you are asserting that for some prime e greater than x, we can find a number n which, when expressed in base-e notation, will have 1 in its units place, 10 in its e's place, 100 in its (e^2)'s place, and in general have the "digit" in each place be 10 times greater than the one to its right, such that the leftmost digit is our x.

To translate into Hofstadter's notation, of course, we must use horse-shoes instead of ->'s, big carets instead of &'s, letters a through e (followed by however many ''s) exclusively, and so forth. (We must also replace <'s and <= with appropriate expansions, and substitute in for our capitalized formula abbreviations.) This is left as an exercise to the reader.

Kevin Wald wald2@husc.harvard.edu