your sequence from Oct 12

[Richard Guy]

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 at macalester.edu>
> To: pow pow <macpow at 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