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# User talk:Bruno Curfs

I'd like to introduce a new sequence called "Curfs numbers", based on the Friedman numbers. The definition is as follows:

Curfs numbers: numbers which allow for a nontrivial expression using all their digits once with the operations + - * / ^ and concatenation of digits (but not of results), AND allowing the expression 0^0 with value 1.

In the original definition of Friedman numbers, there is no exception for the expression 0^0, because it is not defined. The expression 0^0 arises in different contexts and may attain any value; consider the limit lim[x->0, x^(-a/log(x))) = exp(-a)] and note that both base and exponent go to zero. In most contexts, 0^0 arises as either a special case of 0^x = 0 for x>0 or as a special case of x^0 = 1 for x>0. To allow for the expression 0^0 to mean 1 is a more or less arbitrary definition but with the purpose of excluding the powers of 10 from being non-Friedman numbers. Hence, the definition of Curfs numbers including all but a finite number of powers of 10.

The question which is interesting in this context is whether there are an infinite number of integers which are non-Curfs numbers. This is an unsolved math problem.

**OBSERVATIONS**

*1. The relationship between Friedman numbers and Curfs numbers*

All Friedman numbers are Curfs numbers.

The observation is trivial, since all operations allowed to form Friedman numbers are also allowed to form Curfs numbers and therefore the Friedman numbers form a subset of the Curfs numbers.

**Note 1**. Friedman commented that he does not think that Curfs numbers are interesting, since he conjectures that all large enough integers are Curfs numbers, i.e., there are only a finite number of integers which are NOT Curfs numbers. I countered with the conjecture that there are an infinite number zero-less integers which are NOT Friedman numbers (nor Curfs numbers). The set of zero-less integers has density 0, so even if almost all numbers are Friedman numbers (or Curfs numbers), there is still an infinite number of integers which may not be Friedman numbers (nor Curfs numbers).

**Note 2**. Powers of 10 are not Friedman numbers, so there are an infinite number of integers which are not Friedman numbers. Curfs numbers eliminate these "trivial" exceptions. And the quest is on to prove that there are still an infinite number of exceptions. The smallest Curfs numbers which are not Friedman numbers are

10^8 = 10,000^(0^0+0^0) 10^9 = 1,000^(0^0+0^0+0^0) 10^10 = 100,000^(0^0+0^0)+0 10^11 = not a Curfs number, isn't that odd? 10^12 = 10,000^(0^0+0^0+0^0)+0+0 10^13 is not a Curfs number, isn't that bad luck?

10^n with n>13 are all Curfs numbers, not trivially, see below under Explanation.

: 10^15 = 100,000^(0^0+0^0+0^0)+0+0+0+0 : 10^17 = 10^[(0^0+0^0)^(0^0+0^0+0^0+0^0)+0^0]+0 : 10^19 = 10^[(0^0+0^0)^(0^0+0^0+0^0+0^0)+0^0+0^0+0^0] : 10^21 = 10^[(0^0+0^0+0^0+0^0)*(0^0+0^0+0^0+0^0+0^0)+0^0] : 10^23 = 10^[(0^0+0^0+0^0+0^0+0^0)^(0^0+0^0)-(0^0+0^0)]+0+0+0+0 : 10^25 = 10^[(0^0+0^0+0^0+0^0+0^0)^(0^0+0^0)]+0+0+0+0+0+0+0+0+0+0 : 10^27 = 10^[(0^0+0^0+0^0)^(0^0+0^0+0^0)]+0+0+0+0+0+0+0+0+0+0+0+0+0+0 : 10^29 = 10^[(0^0+0^0+0^0)^(0^0+0^0+0^0)+0^0+0^0]+0+0+0+0+0+0+0+0+0+0+0+0 : 10^31 = 1,000,000,000,000,000^(0^0+0^0)*[(0^0+0^0+0^0)^(0^0+0^0)+0^0] : 10^63 = 10^[(0^0+0^0)^(0^0+0^0+0^0+0^0+0^0)+(0^0+0^0)^(0^0+0^0+0^0+0^0)+(0^0+0^0)^(0^0+0^0+0^0)+ (0^0+0^0)^(0^0+0^0)+(0^0+0^0)+(0^0)]+0+0+0+0+0+0+0+0+0+0+0+0 :

In between the above sequence, other Curfs numbers hide, such as

10^12+x = 10,000^(0^0+0^0+0^0)+0+x (x=1..9) 10^12+x*10 = 10,000^(0^0+0^0+0^0)+x0 (x=1..9) : 10^15+x = 100,000^(0^0+0^0+0^0)+0+0+0+x (x=1..9) 10^15+x*10 = 100,000^(0^0+0^0+0^0)+0+0+x0 (x=1..9) : 10^15+x*100 = 100,000^(0^0+0^0+0^0)+0+x00 (x=1..9) : 10^15+x*1000 = 100,000^(0^0+0^0+0^0)+x000 (x=1..9) : x1*10^16 = (x0+0^0)*10^[(0^0+0^0)^(0^0+0^0+0^0+0^0)] : xy*10^26 = xy*[(0^0+0^0+0^0)^(0^0+0^0)+0^0]^[(0^0+0^0+0^0)^(0^0+0^0+0^0)-0^0] (x=1..9, y=0..9) : x1*10^27 = (10+0^0)*[(0^0+0^0+0^0)^(0^0+0^0)+0^0]^[(0^0+0^0+0^0)^(0^0+0^0+0^0)] (x=1..9) etc.

**Explanation**
For n>=63, we can use the binary expression for n to construct the exponent as a sum of powers of 2, and using 0^0 for (0^0+0^0)^0, and (0^0+0^0) for (0^0+0^0)^(0^0). This will give an expression for n using less than the available number of zeroes and proves that 63 is an upper bound for the lowest power of 10 above which all powers of 10 are Curfs numbers.

For n>=8 even, we always have an expression {10^(n/2)}^(0^0+0^0) using n/2+4 zeroes, where {x} means x written in its digits. To give expressions for n odd, we need to express an extra factor 10 using only zeroes. The smallest expression of 10 using only zeroes is (0^0+0^0+0^0)^(0^0+0^0)+0^0 using 12 zeroes, so if n/2+4+12 <= n+1 then 10^(n+1) is also a Curfs number, hence for n>=30 all powers 10^n are Curfs numbers.

For n between 13 and 30 we need special expressions to make sure, see above.

*2. Zero-less integers*

If an integer contains at most one zero, then it is a Friedman number iff it is a Curfs number.

**Proof** (scetch). One of the implications is trivial (see above). Suppose it is a Curfs number, and it uses the expression 0^0 = 1 (if not, it is already a Friedman number), then this means that one of the zeroes in this expression must come from using (other) expressions x and y with x-y = 0. The idea is that we can always eliminate the zeroes from any expression 0^0 by using a different expression for 1.

Since, if x,y<>0 then x/y = 1 and we can replace the expression 0^0 by the expression (x/y + 0), proving it is a Friedman number.

And if x=y=0, x and y being expressions, they in turn must consist of expressions a,b,c,d <>0 such that x = a-b = 0 = c-d = y. And we can eliminate the zeroes with a similar approach, e.g., replace 0^0 = (x-y)^0 = [(a-b) - (c-d)]^0 by (a/b)/(c/d) + 0. Since the number under consideration has only a finite number of digits, this recursion must end, so we can find an expression of the number not using 0^0 = 1 and therefore it is a Friedman number. QED.

Hence, we have the following corollary

If there exist an infinite number of zero-less non-Friedman numbers, then there exist an infinite number of non-Curfs numbers.

Bruno Curfs 21:25, 27 August 2013 (UTC)