Re: Arithmetic Series and Factors of Triangular Numbers(adsbygoogle = window.adsbygoogle || []).push({});

A+C*n, B+F*n, A+E*n, B+D*n are all arithmetic series which I define below

It is still not clear to me as why

(A+C*n)*(B+F*n) and (A+E*n)*(B+D*n) are both triangular numbers for all integers n

Can someone please visit my blog and explain it?

Perhaps someone can show somehow that

[tex]8*(A+C*n)*(B+F*n) + 1[/tex] and [tex]8*(A+E*n)*(B+D*n) + 1[/tex] are both perfect squares for all n using Mathematica or some other program.

--- In Triangular_and_Fibonacci_Numbers@yahoogroups.com, "ramsey2879"

<ramseykk2@...> wrote:

>

> I found a previously unknow property of Triangular Numbers AFAIK.

> Given that T is a triangular number having exactly M distinct ways

to

> pair the product into two factors, A*B with A </= B. For each of

these

> M distinct pairs, there are two coprime pairs of integers (C,E) and

> (E,D)where C*D = the perfect square of integral part of the square

root

> of 2*T and E*F = the next higher perfect square and each of the

> products (A+Cn)*(B+Fn) and (A+En)*(B+Dn) are triangular numbers for

all

> integer values of n.

> As an example.

> Let T = 666 which can be factored into 6 distinct pairs A,B. The

six

> sets (A,B,C,D,E,F) are as follows

>

> 1. (1,666,1,1369,2,648)

> 2. (2,333,1,1369,8,162)

> 3. (3,222,1,1369,18,72)

> 4. (6,111,1,1369,72,18)

> 5. (9,74,1,1369,162,8)

> 6. (18,37,1,1369,648,2)

>

> I don't have a proof of the general result but am working on it.

>

Still no proof, however, let ab = T(r) = r(r+1)/2, gcd(n,m) = the

greatest common divisor of m and n, then the formula for C,D,E and F

as a function of A and B is

Case 1, r is even

C = (gcd(A,r+1))^2, F = 2*(gcd(B,r))^2

E = 2*(gcd(A,r))^2, D = (gcd(B,r+1))^2

The determinant

|C F|

|E D| = (r+1)^2 - r^2 = 2r+1

Case 2 r is odd

C = 2*(gcd(A,r+1))^2, F = (gcd(B,r))^2

E = (gcd(A,r))^2, D = 2*(gcd(B,r+1))^2

The determinant

|C F|

|E D| = (r+1)^2 - r^2 = 2r+1

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# Arithmetic Series and Triangular Numbers

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