The polynomials have 1 positive integer real root and 2 complex roots
As set up with x(x+1)(x+2)=x^3+3x^2 +2x=720 the real root is 8
For some others
p._990 2 3 1
┌─┬────────────────────────┐
│1│_6j8.60233 _6j_8.60233 9│
└─┴────────────────────────┘
p._6 2 3 1
┌─┬────────────────────────┐
│1│_2j1.41421 _2j_1.41421 1│
└─┴────────────────────────┘
p._24 2 3 1
┌─┬────────────────────────────┐
│1│_2.5j2.39792 _2.5j_2.39792 2│
└─┴────────────────────────────┘
for non special numbers the real root is negative and non-integer
p.719 2 3 1
┌─┬───────────────────────────────────────┐
│1│_9.99587 3.49793j7.7262 3.49793j_7.7262│
└─┴───────────────────────────────────────┘
p.23 2 3 1
┌─┬───────────────────────────────────────────┐
│1│_3.96101 0.480507j2.36129 0.480507j_2.36129│
└─┴───────────────────────────────────────────┘
p. _721 2 3 1
┌─┬──────────────────────────────────────────┐
│1│_5.50207j7.73342 _5.50207j_7.73342 8.00413│
└─┴──────────────────────────────────────────┘
With the second polynomial, x(x+1)(x-1) the positive real root is the
middle number
as it should be.
p. _720 _1 0 1
┌─┬────────────────────────────┐
│1│9 _4.5j7.72981 _4.5j_7.72981│
└─┴────────────────────────────┘
note that in both cases the value of the polynomial is negative at x=0
whether there is a case with 3 real roots -I don't know
possibly writing the second case as x^3 =n+x helps (given a real integer
x >1 find the
then for n>1
x x^3 n
2 8 6 for 11*2*3
3 27 24 for 2*3*4
4 64 60 for 3*4*5
9 512 720 for 8*9*10
using x^3=n-3x^2-2x works but is more cumbersome
x x^3 -(3x^2-2x) n
1 1 _5 6 for 1*2*3
8 512 _208 720 for 8*9*10
p. works more easily in either case - skim off the complex roots.
Don Kelly
On 08/08/2015 8:39 PM, Don Guinn wrote:
Sorry I'm dense. I understand that. But the first polynomial is not
remotely related to the second other than the gcd of both is x-8. Why?
On Aug 8, 2015 8:08 PM, "'Pascal Jasmin' via Programming" <
[email protected]> wrote:
(x-1)x(x+1) or x(x+1)(x+2) and their expansions are algebraic solutions to
special products. All the solutions posted used one of these 2 forms.
(your example uses the 2nd's expansion)
----- Original Message -----
From: Don Guinn <[email protected]>
To: Programming forum <[email protected]>
Cc:
Sent: Saturday, August 8, 2015 9:53 PM
Subject: Re: [Jprogramming] Detecting special products
Using polynomials to solve this problem is really neat, but why does it
work? I can't find anything on the internet explaining why it works.
p._720 2 3 1
+-+----------------------------+
|1|_5.5j7.72981 _5.5j_7.72981 8|
+-+----------------------------+
p.1;-8 9 10
720 242 27 1
Obviously the polynomial for the roots is quite different from that used
above. There has to be a proof somewhere.
On Fri, Aug 7, 2015 at 9:47 PM, Don Kelly <[email protected]> wrote:
correction : 'will be the <last> number of the sequence'
in any case for a special number the real root will be an integer.
On 06/08/2015 4:06 PM, Don Kelly wrote:
also try
0=1|( <1; 2) {>p.|.1 3 2,-]720
1_
0=1|( <1; 2) {>p.|.1 3 2,-]719
0
note that
(<1;2) {>p._y 2 3 1
will return an integer for a special number and this will be the first
number of the sequence
based on x*x+1)*(x+2) -y=0 giving an integral root
p._720 2 3 1
┌─┬────────────────────────────┐
│1│_5.5j7.72981 _5.5j_7.72981 8│
└─┴────────────────────────────┘
a modification to x(x^2-1) gives an integral real root for the middle
root
p. _720 _1 0 1
┌─┬────────────────────────────┐
│1│9 _4.5j7.72981 _4.5j_7.72981│
└─┴────────────────────────────┘
puts the real root first rather than last
As for finding a list of these special numbers-the "stope " function
works well
f=:3^!.1~[:i.]
f 10
0 6 24 60 120 210 336 504 720 990
Don Kelly
On 06/08/2015 1:02 PM, Roger Hui wrote:
g=: = (^&3 - ])@:(3&(>.@%:))
g */"1 (10^40 60 80x)+/0 1 2
1 1 1
On Thu, Aug 6, 2015 at 11:23 AM, 'Pascal Jasmin' via Programming <
[email protected]> wrote:
(] = (^&3 - ])@:(1r3 >.@^~ ])) 990
----- Original Message -----
From: Kip Murray <[email protected]>
To: "[email protected]" <[email protected]>
Cc:
Sent: Thursday, August 6, 2015 1:58 PM
Subject: [Jprogramming] Detecting special products
The number 720 is special: it is 8*9*10 the product of three
successive non-negative integers. The first few special numbers are
*/"1 [ 0 1 2 +"1 0 i. 10
0 6 24 60 120 210 336 504 720 990
Write a verb test that tests whether a non-negative integer is
special:
test 720
1
test 0
1
test 721
0
--Kip Murray
--
Sent from Gmail Mobile
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