It's OK, I believe, but is only epsilon of the story. The point is that it works for any value of 31. It arose from PoW 1040:

Date: Mon, 10 Oct 2005 19:19:31 -0500 From: stan wagon <wagon@macalester.edu> To: pow pow <macpow@mathforum.org> Subject: Problem 1040 Parts/Attachments: 1 Shown 17 lines Text 2 OK 9 KB Image ----------------------------------------

Problem 1040: A special sum of squares.

Find 7 (not necessarily distinct) positive integers, at least one of which is greater than 1040, such that the sum of their squares is 7 times the product (of the 7 integers).

Extra Credit: I heard this problem from Gerald Heuer (Concordia College), who asked it with 31 instead of 7 in all places. If you can solve this with 31 (or 7), let me know.

but has been widely studied. See D12 in UPINT for some info. The case k=3 is the classic Markoff equation, and the more general case is usually associated with Hurwitz. Donald Knuth has pointed out that Figure 10 in D12 in UPINT should really be rotated through 2pi/3 twice to show the 3-rotational symmetry. The binary (should it be called `ternary'?) tree of solutions presumably becomes a k-valent one, so that not only is a sequence produced but a (k-1)-dimensional array. The left (slowest growing) edge shows alternate Fibs, while the right (fastest growing) shows numbers dubiously associated with Pell [and, to quote Tom Lehrer, God knows how many between]. Incidentally, there's a classic unsolved problem here -- is Fig 10 really a tree or is there some number way out there that can be reached by two different routes? [See Dick Bumby's review MR 53 #280.] OEIS lookup is working incredibly slowly for me this morning, taking around 10 minutes a try, with a better than even chance of getting ``Internal Server Error'' (Error 500), so I haven't checked how many sequences there are with characteristic function x^2 - kx + 1 but I expect that they're there for many k, even if not for k = 31. The equation x1^2 + x2^2 + ... + xk^2 = k.x1.x2...xk is a quadratic whenever you substitute for k-1 of the variables. Since {1,1,...,1} is a solution, plug in k-1 ones and get x^2 + k-1 = kx which has the solutions x = 1 and x = k-1. Perhaps k = 4, for example, is already in (I've just got another ISE), but you not only get the solutions {1,1,a(n),a(n+1)} where a(n) is 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, ... but also {1,3,11,131}, {1,11,41,1803}, {3,11,131,17291}, etc., or something like that (doubtless lots of errors in this hand calculation). [another ISE!] AHA!! OEIS has finally divulged A001835. No surprise that there are numerous manifestations but no mention of Hurwitz or his equation. AHA again -- I didn't notice that it also found at least one more entry, but I lost it, and I'm back to ISEs. [better send this -- have waited 4 hours -- my machine SAYS that it's trying to connect/searching!] R. On Wed, 26 Oct 2005, N. J. A. Sloane wrote:

Richard, is this OK? Neil

%I A111216 %S A111216 1,30,929,28769,890910,27589441,854381761,26458245150,819351217889,25373429509409, %T A111216 785756963573790,24333092441278081,753540108716046721,23335410277756170270, %U A111216 722644178501725231649,22378634123275726010849,693015013643045781104670 %N A111216 a(n)=31*a(n-1)-a(n-2). %C A111216 Take 31 numbers consisting of 29 ones together with any two successive terms from this ssequence. This set has the property that the sum of their squares is 31 times their product. (Guy) %O A111216 0,2 %K A111216 nonn %A A111216 njas, following a suggestion from Richard Guy, Oct 26 2005