[seqfan] Re: [math-fun] Triangular+Triangular = Factorial
Richard Guy
rkg at cpsc.ucalgary.ca
Fri Sep 10 21:29:15 CEST 2010
Just cleaning up old email, so this is a very,
very belated response, which may have been taken
care of by someone long ago. Copied to seqfans,
in case someone wants to make a sequence or
two out of it. Solutions of
x(x+1)/2 + y(y+1)/2 = z!
are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2
and the first few values of z for which there
are solutions can easily be ascertained:
(x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3),
(2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8),
(210,825,9), (760,2610,10), (1770,2030,10),
none for z = 11, 12, one for z = 13
(see Ed's solution below), none for z = 14,
two for z = 15 (see below), one for z = 16,
two for z = 17, none for z = 18, 19, 20,
two for z = 21, none for z = 22, 23,
two for z = 24, none for z = 25, 26,
eight for z = 27, one for z = 28,
two for z = 29, none for z = 30, 31,
four for z = 32, 33, sixteen for z = 34,
none for z = 35, 36, ..., 41,
two for z = 42, none for z = 43, 44, ..., 48,
sixteen for z = 49, none for z = 50, 51, 52, 53,
one for z = 54, none for z = 55, 56, ..., 65,
two for z = 66, none for z = 67,
sixteen for z = 68, ...
(E&OE, and PARI is slowing down a bit now: AND it would
take rather longer to find the actual solutions!) R.
On Tue, 20 Jan 2009, Ed Pegg
Jr wrote:
> A very belated response.
>
> Not squares, but related. Using Triangular numbers.
> 1 3 6 10 15 21
>
> T[3] = 3!
> T[3] + T[6] = 4!
> T[14]+T[5] = T[15] = 5!
> T[45] + T[89] = 7!
> T[210] + T[825] = 9!
> T[1770] + T[2030] = 10!
> T[71504] + T[85680] = 13!
> T[213384] + T[1603064] = T[299894] + T[1589154] = 15!
>
> I don't see an easy way to extend these. The density of triangular numbers seems to be sufficient for extended solutions.
>
> --Ed Pegg Jr
>
> Date: Mon, 3 Jul 2006 11:44:20 -0600 (MDT)
> From: Richard Guy <rkg at cpsc.ucalgary.ca>
> Reply-To: math-fun <math-fun at mailman.xmission.com>
> To: Number Theory List <NMBRTHRY at listserv.nodak.edu>,
> Math Fun <math-fun at mailman.xmission.com>
> Subject: [math-fun] Factorial n
>
> Presumably 0! = 1! = 0^2 + 1^2.
> 2! = 1^2 + 1^2
> 6! = 12^2 + 24^2
>
> are the only integer solutions of
>
> n! = x^2 + y^2
>
> but is there a proof? [Later:
for n > 6, n! will always contain a
prime factor of shape 4k - 1 raised to an
odd power, in fact raised to the first
power, by a suitable form of the prime
number theorem.] R.
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