Talk:Friedman number

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The most reliable source I can find for the name Friedman number is Sloane's OEIS, A036057 and A080035. PrimeFan 16:59, 25 Aug 2004 (UTC)

Is there a better word than "lame"? Perhaps "trivial" would be more appropriate, or at least sound more like an encyclopedic/mathematical term. Confusing Manifestation 15:55, 30 January 2006 (UTC)

maybe, but I don't think its "trivial." That 12 is a divisor of 12, that's trivial, though valid. That 12 is a Friedman number cause 12 = (12) that's invalid, not trivial. Numerao 22:28, 30 January 2006 (UTC)
But if you did allow those sorts of things, then it would become trivial because then all integers would be Friedman numbers. So in other words, the rules for forming a Friedman number are defined to avoid trivial solutions. (Similar to how Fermat's Last Theorem is defined to avoid the trivial cases of n=2 or (x,y,z)=(1,0,1).) Confusing Manifestation 15:43, 31 January 2006 (UTC)
While there's a certain informal charm to using the word "lame," this might be one case where the word "trivial" makes more sense. I've changed the article thus. Anton Mravcek 21:26, 31 January 2006 (UTC)

On the smallest repdigits[edit]

Hi, am I missing something here, or is it simply straightforward to go ahead and check this claim about the smallest repdigits number which is also a Friedman number? It seems to me that there is an algorithm which will determine whether or not a number is Friedman in finite time, as there are only finitely many ways to combine things, and then just run it on the repdigit numbers. --Deville (Talk) 01:01, 30 March 2006 (UTC)

I don't think you're missing anything. There is an algorithm that will check whether an integer is a Friedman number in finite time, and efficiently too, but to my knowledge, Friedman's students haven't published it.
The only other algorithm I can think of is the one mentioned in the article, of checking a number against each possible expression. To implement that algorithm, it would first be necessary to catalog all possible Friedman expressions up to 8 digits.
It would still run in finite time, but wouldn't be for the impatient, even if we limit it to the repdigits. PrimeFan 16:11, 30 March 2006 (UTC)

About nice Friedman numbers under 10000[edit]

Why was the revision of Nov 20, 2006 by reverted? The statement than "all the expressions for nice Friedman numbers less than 10000 involve addition and subtraction" is indeed false, because 2^5*9^2 = 2592 < 10000, and 2^5*9^2 does not involve addition or subtraction. (talk) 14:14, 21 January 2011 (UTC)

You are right. It was an error to revert it. Maybe the editor misinterpreted the edit summary and didn't examine the actual edit. I have removed the false statement. PrimeHunter (talk) 21:09, 21 January 2011 (UTC)

Update statistics please[edit]

Since a couple of days, a lot more Friedman numbers have come to light by yours sincerely. These numbers are published on Friedman's webpage (see There are many more to be discovered by those who put their intention on it. Anyway, it seems I discovered the first 10-digit pandigital Friedman numbers, but there are quite a lot so called vampire numbers (they factor using the original digits), which de facto are Friedman numbers. (Bcurfs (talk) 09:06, 14 August 2013 (UTC))

I have updated the count to say "Erich Friedman's website shows around 100 zeroless pandigital Friedman numbers as of August 2013."[1] Friedman doesn't give the count and I don't think they should be counted by Wikipedia editors when we don't even have a source saying these are all the known examples. The "vampire number" link in the definition on Friedman's page [2] goes to a former url [3] of my website (the webhost closed). The working url is [4]. All true 10-digit vampire numbers had already been computed when I recomputed them in 2002 before computing all with 12 and 14 digits. This means many 10-digit pandigital Friedman numbers had been computed. I don't know whether anyone had pointed out the pandigital numbers among the vampire numbers but I don't think that deserves discoverer credit. PrimeHunter (talk) 13:20, 14 August 2013 (UTC)

Curfs numbers[edit]

Tentative there is an extension of the definition of Friedman numbers called Curfs numbers: those numbers allowing an expression with their own digits following the same rules as Friedman numbers, with the addition that we allow 0^0 = 1.

Note. This definition does not introduce any new symbols, but uses an "arbitrary" definition of 0^0, an expression which strictly spoken is undefined.

Surely, this allows for powers of 10 to be Friedman numbers. All powers of the form 10^n are Curfs numbers for n>=63. I can provide a proof. The number 63 is an upper bound for the lowest such n, which can be expressed with less than n zeroes, as is shown in the following example.

10^63 = 0+0+0+[(0^0+0^0+0^0)^(0^0+0^0)+1]^[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)]

However, much smaller Curfs numbers exist, such as:

10^8 = 10,000^(0^0+0^0) --smallest-- and
10^9 = 1,000^(0^0+0^0+0^0)
10^10 = 100,000^(0^0+0^0)+0
10^11 = ???
10^12 = 1,000,000^(0^0+0^0)+0+0


In a private exchange, Friedman relates that he does not like Curfs numbers, because he conjectures that

(C1) "Only a finite number of integers are NOT Curfs numbers." - (Friedman)

Supposedly, the extra rule makes it too easy for a number to be a Curfs number. The real reason comes through in his confident additional conjecture that

(C2) "All large enough numbers without a 0 are Friedman numbers." - (Friedman)

However, both conjectures turn out to be not trivial and far from proven. The relation to Curfs numbers makes more sense when one realizes that

If an integer does not contain a 0, then it is a Friedman number iff it is a Curfs number.

So, Curfs numbers and Friedman numbers coincide when they do not contain a digit 0. The "difficulty" would be to prove that all Curfs numbers without a 0 are Friedman numbers, since all Friedman numbers are automatically Curfs numbers. The proof that a Curfs number without a digit 0 is a Friedman number is almost trivial and rests on the observation that if a number requires an expression of the form "0^0" (if not, it is certainly a Friedman number), but the 0 is not part of its digits, then it requires two additional expressions for each of these 0's, and they must be of the form x-y = 0, with x and y possibly different expressions using at least one digit unequal to 0. This would imply, however, that "0^0" (being equal to 1) can also be expressed with less digits, namely with x/y (and the digits used in the expression of the other 0 can either be used to multiply by an expression x/y = 1 or to find an altogether easier expression of the number in question). Hence, if a number without digit 0 is a Curfs number, it does in fact not require the use of the expression "0^0", and is therefore a Friedman number as well.

On the other hand, I conjecture that

(C3) The integers without a 0 which are NOT Friedman numbers form an infinite set. - (Curfs)

This would makes the Friedman numbers even more interesting, apparently, and hence it makes this conjecture worth investigating. Since the numbers under consideration here do not have a 0, the numbers in this infinite set are NOT Curfs numbers either! So, if my conjecture (C3) is true, contradicting both Friedman's conjectures (C1) and (C2), then the numbers without a 0 which are NOT Curfs numbers form an infinite set as well, and it would make them interesting. It's clear that more research is needed to decide the matter and help would be appreciated.

I realize that it may be quite difficult to prove of very large numbers without 0 that they are NOT Friedman / Curfs numbers, since we have to eliminate all possible expressions of its digits.

As of yet, seemingly supporting Friedman's conjecture, his website features a proof by Michael Brand showing that the limit of the fraction F(N)/N of the number F(N) of Friedman numbers <= N tends to 1 for N going to infinity. (This proof is not published in a math proceedings and needs peer review.) This is a way of saying that the density of Friedman numbers equals 1.

Seemingly supporting my conjecture is the observation that numbers which use less than 10 different digits have density 0 . . . but they form an infinite set with possible candidates for not being a Friedman / Curfs number!

For example, does it make the question easier to prove by considering only integers with all of its digits from a set of just two digits not including 0? (Bcurfs (talk) 09:06, 14 August 2013 (UTC)) Modified Bcurfs (talk) 00:57, 29 October 2013 (UTC)

If Curfs numbers have not been covered by Friedman or reliable sources then they don't belong in Wikipedia articles per Wikipedia:No original research. PrimeHunter (talk) 13:22, 14 August 2013 (UTC)
Point taken, hence I put it here in Talk for now. Considering, Friedman may or may not cover Curfs numbers in the future. I'm going to try and persuade him to "cover" them. On the other hand, if I could get an article published myself, would that then qualify the subject with a reference to the article? Bcurfs (talk) 00:57, 29 October 2013 (UTC)

But 0^0=0/0 (talk) 11:20, 12 July 2014 (UTC)

Well, technically, 0^0 is not defined, and neither is 0/0, so they are not equal (that could only hold if they are both defined). Your assertion that they are equal may stem from the observation that for any x<>0, and n>0, we have (x^n)/(x^n) = x/x = x^(n-n) = x^0 = 1, and so with substitution of x=0 we would get equality of 0^0 = 0/0. (Note that by the same token, 0^0 = 0/0 = 1.) However, the equality ceases to hold, as both expressions cease to mean anything. Therefore, to be more specific, I propose to replace the binary operation ^ by & with the following definition: x&y = {x^y if x.y <> 0; 1 if x.y = 0}. Of course, with this definition, 1 = 0&0 <> 0/0, because the former is defined, and the latter isn't. But once we have introduced &, to avoid having to actually write a different sign for this operator, we can just as well use ^ with the understanding that 0^0 should be interpreted as 0&0, i.e., as 1. And that is what I have been doing. Bcurfs (talk) 10:59, 2 November 2015 (UTC)

Curfs numbers (2)[edit]

My meanderings above about 10^63 were a bit obscure, even to me :-), so I have come up with some improvements. I'd like to find a proof of the following

Conjecture. All numbers of the form 10^n are Curfs numbers except for n in {1,2,3,4,5,6,7,11,13}.

We have

Lemma 1. For even exponents, 10^(2.k) is a Curfs number for k>=6.

Proof. The following expression proves Lemma 1:

10^(12 + 2.k) = 100^(((0^0 + 0^0).(0^0 + 0^0 + 0^0)) + (0^0 + ... <k terms> ... + 0^0)) for k>=0.


Lemma 2.

For odd exponents, 10^(2.k + 1) is a Curfs number for k>=8.

Proof. The following expression proves Lemma 2:

10^(17 + 2.k) = 10^((0^0 + 0^0).((0^0 + 0^0)^(0^0 + 0^0 + 0^0) + (0^0 + ... <k terms> ... + 0^0)) + 0^0) for k>=0.


Given the earlier expressions for powers of 10 (see above), and the feasibility of proving that 10^n with n<8 is not a Curfs number by listing all possible expressions with less than 8 zeroes, this leaves only the cases for the exponent n in {11,13,15}.

As it turns out 10^15 is a Curfs number as well, and can be expressed in at least four ways:

10^15 = 100,000^(0^0 + 0^0 + 0^0) +
10^15 = 1,000^(0^0 + 0^0 + 0^0 + 0^0 + 0^0) + 0.0
10^15 = 10^((0^0 + 0^0)^(0^0 + 0^0 + 0^0 + 0^0) - 0^0)
10^15 = 10^((0^0 + 0^0 + 0^0 + 0^0)^(0^0 + 0^0) - 0^0)

So, the conjecture is reduced to the conjecture that 10^11 and 10^13 aren't Curfs numbers. Any proofs are welcome.

Bcurfs (talk) 10:59, 2 November 2015 (UTC)