Talk:Ulam number

unreferenced computer code
Computer code without references/sources is subject to being deleted. See this discussion. Bubba73 (You talkin' to me?), 23:03, 12 August 2010 (UTC)

Overly complicated proof?
The article says this:
 * There are infinitely many Ulam numbers. To prove this, suppose, on the contrary, there are only n Ulam numbers.  Then Un&minus;1 + Un would be a sum of distinct Ulam numbers in a unique way and thus would be another Ulam number, contradicting the assumption that there are only n.
 * There are infinitely many Ulam numbers. To prove this, suppose, on the contrary, there are only n Ulam numbers.  Then Un&minus;1 + Un would be a sum of distinct Ulam numbers in a unique way and thus would be another Ulam number, contradicting the assumption that there are only n.

What if it were rephrase as follows:
 * There are infinitely many Ulam numbers. To prove this, observe that for any n,  Un&minus;1 + Un is a sum of distinct Ulam numbers in a unique way and is thus another Ulam number.
 * There are infinitely many Ulam numbers. To prove this, observe that for any n,  Un&minus;1 + Un is a sum of distinct Ulam numbers in a unique way and is thus another Ulam number.

Simpler proof, or did I miss something? Michael Hardy (talk) 02:53, 14 August 2010 (UTC)
 * But it is not necessarily true that Un&minus;1 + Un is another Ulam number. For instance, it is not true when n = 3. —David Eppstein (talk) 03:42, 14 August 2010 (UTC)
 * OK, so Un&minus;1 + Un is in some cases equal to Un&minus;2 + Un+1 and therefore would not itself be an Ulam number. But apparently the proof by contradiction still works.  (Or did I miss something here again??)  This is interesting..... could one say this is a nonconstructive proof? Michael Hardy (talk) 04:21, 14 August 2010 (UTC)
 * Are some subtler issues going on here, than mere quirks of some oddball integer sequence? Michael Hardy (talk) 04:23, 14 August 2010 (UTC)

One could write more simply without contradiction "At each step in the construction of the sequence, the set of numbers uniquely representable as a sum of members of the sequence is nonempty, because it contains Un&minus;1 + Un." —David Eppstein (talk) 04:59, 14 August 2010 (UTC)

Infinite length of sequence.
The reasoning about the infinity of Ulam numbers has deteriorated by avoiding the contradiction argument. The guy who added "citation needed" is a rank amateur and should exercise more restraint to add such comment to what is a perfectly lucid argument. What there is now is not lucid and also not formulated sufficiently sharply. The original formulation was in my not so humble opinion as a mathematician of textbook quality.

I would put back the august 14 2010 version, if I had any say in wiki. I exercise restraint, because this argument could go back and forth indefinitely, which is worse than living with the article at hand. So I leave it to the "powers-that-be". The best would be if the author would be convinced to revert his edit.

80.100.243.19 (talk) 22:23, 11 November 2010 (UTC)
 * The larger point of the change away from a proof by contradiction was that the supposed proof by contradiction was wrong. It claimed that, if there were only n Ulam numbers, then Un + Un &minus; 1 would also be a Ulam number, a contradiction. But, while Un + Un &minus; 1 would clearly have a unique representation as a sum of two Ulam numbers in this case (one half of the definition of the Ulam numbers) there's no reason to believe that it would be the smallest uniquely representable number, or that it would remain uniquely representable once other numbers are added to the sequence (the other half of the definition). It's not an important mistake as it's easily fixed, but it is a mistake. So reversion to the old version would also be a mistake. —David Eppstein (talk) 23:02, 11 November 2010 (UTC)


 * The proof needs to be verifyable and from a reliable source, otherwise it is original research, and not allowed in Wikipedia. Bubba73 You talkin' to me? 04:01, 12 November 2010 (UTC)

Is there any formulars for Ulam numbers?
As there are closed functions for Fibonacci numbers like F(x)=((1+sqrt 5)/2)^x-((-1)^x/((1+sqrt 5)/2)^x)))/sqrt 5, I was wondering if there where any closed (finite) functions that generate the xth Ulam number? This artical doesn't seem to even have any infinite ones... Robo37 (talk) 13:20, 30 July 2011 (UTC)


 * No, they are too irregular. Bubba73 You talkin' to me? 14:56, 30 July 2011 (UTC)


 * Well, there could be something analogous to Formula for primes but it wouldn't be practical, and not a closed form. Bubba73 You talkin' to me? 17:48, 10 October 2011 (UTC)

Restored external ink
I restored an external link to a paper on viXra. This was removed by Headbomb with justification WP:RS. Other external links could equally well have been removed for the same reason and the same applies to all links to arXiv including the one on this article but these were left. The paper linked was of interest and has been reviewed by Don Knuth in another link which was allowed to remain. This suggests that the editor removed the link because of a personal dislike of viXra rather than because if the credibility of the paper itself. Other links removed by the same editor on other pages may be equally suspect Weburbia (talk) 08:11, 27 September 2016 (UTC)
 * It really would be the exception to the viXra stuff if this stays. arXiv papers aren't peer reviewed either, but most are allowed to stay because they tend to be written by experts and the arxiv is at the very least moderated (although the 'general mathematics' is in general, to be rarely trusted given it's a sort of 'shove all' for awful submissions).
 * However, if Knuth mentions this paper by Gibbs favourably (and he does), then that's good enough for me and can stay per WP:USEBYOTHERS. I'll add a comment next to the link, however. Headbomb {talk / contribs / physics / books} 09:52, 27 September 2016 (UTC)


 * I think that it will probably be mentioned in the next edition of The Art of Computer Programming, vol 4A. Bubba73 You talkin' to me? 23:55, 27 September 2016 (UTC)