Factors of triangular numbers forming arithmetic series

ramsey2879

For every pair (a,b) of factors that equal the trangular number m(m+1)/2 there are two distinct sets of pairs (c,d) that form a determinant equal to 2m+1 such that the products (a+cn)*(b+dn) = a triangular number for all n. Is this a previously known fact and how can it be prooved?

Quote by ramsey2879 in the topic "New Conjecture"

For instance, the triangular number T(37) has 12 factors which yields the [tex](a,b)[/tex] pairs 1,666; 2,333; 3,222; 6,111; 9,74 and 18,37. The respective sets of [tex]c,d[/tex] pairs in determinant format are

[tex]\left| \begin{smallmatrix}
1 & 648\\ 2 & 1369
\end{smallmatrix}\right|[/tex]

[tex]\left| \begin{smallmatrix}
1 & 162\\ 8 & 1369
\end{smallmatrix}\right|[/tex]

[tex]\left| \begin{smallmatrix}
1 & 72\\ 18 & 1369
\end{smallmatrix}\right|[/tex]

[tex]\left| \begin{smallmatrix}
1 & 18\\ 72 & 1369
\end{smallmatrix}\right|[/tex]

[tex]\left| \begin{smallmatrix}
1 & 8\\ 162 & 1369
\end{smallmatrix}\right|[/tex]

[tex]\left| \begin{smallmatrix}
1 & 2\\ 648 & 1369
\end{smallmatrix}\right|[/tex]

Each determinant equals [tex]37^2 - 36^2[/tex] and each of (1 + n)*(666+648n), (1 + 2n)*(666+1369n), (2+n)*(333+162n), ... (18+n)*(37+2n), (18+648n)*(37+1369n) are each triangular numbers for all integer n. In short if a given triangular number [a*b] has M factors, there are M different sets of products (a' + cn)(b' + dn) where a'b' = ab, c is prime to d and the products remain as triangular numbers for all integer n.

Hurkyl

T(n) = n²/2 + n/2

T(kn + m) = (kn+m)²/2 + (kn+m)/2
= (k²/2)n² + (km)n + (m²)/2 + (k/2)n + m/2
= (k²/2)n² + (km+ k/2)n + (m² + m)/2
= (k²/2)n² + (km + k/2)n + T(m)

(a+cn)(b+dn) = (cd)n² + (ad + bc)n + ab

Why look at the (kn+m)-th triangular number? Because I wanted the most general expression that was quadratic in n. (And later realized that one of my coefficients was your m)

ramsey2879

Quote by Hurkyl

T(n) = n²/2 + n/2

T(kn + m) = (kn+m)²/2 + (kn+m)/2
= (k²/2)n² + (km)n + (m²)/2 + (k/2)n + m/2
= (k²/2)n² + (km+ k/2)n + (m² + m)/2
= (k²/2)n² + (km + k/2)n + T(m)

(a+cn)(b+dn) = (cd)n² + (ad + bc)n + ab

Why look at the (kn+m)-th triangular number? Because I wanted the most general expression that was quadratic in n. (And later realized that one of my coefficients was your m)

This is a helpful. I did determined the following formula for c and d that fits both the data and [tex]cd = k^2/2[/tex]
Lets separate this into two parts m odd and m even
For m even
[tex]c_1 =(gcd(m+1,a))^{2}[/tex]
[tex]d_1 =(gcd(m,b))^{2}*2[/tex]
[tex]c_2 = (gcd(m,a))^{2}*2[/tex]
[tex]d_2 =(gcd(m+1,b))^{2}[/tex]

For m odd
[tex]c_1 = (gcd(m+1,a))^{2}*2[/tex]
[tex]d_1 =(gcd(m,b))^{2}[/tex]
[tex]c_2 =(gcd(m,a))^{2}[/tex]
[tex]d_2 =(gcd(m+1,b))^{2}*2[/tex]

Either way [tex]cd = k^2/2[/tex] fits.

Now to show that the data fits km + k/2 = ad + bc also with my equations. Granted that this is no proof but it could rule out my equations if it didn't. m = 36 m+1 = 37

1. (1 + n)*(666+648n)
gcd(666,36)=18; 2*1*18^2 = 648 -> k = 36
n=1 -> 2*1314 = T(36+36)
n=1 -> 3*1962 = T(72+36)
36*36-36/2 = 1*648 + 666*1
1314=1314

2. (1 + 2n)*(666+1369n)
gcd(666,37)= 37, 2*1*37^2 = 74^2/2 -> k = 74
n=1 -> 3*2035 = T(74+36)
n=2 -> 5*3404 = T(148+36)
74*36+74/2 = 2701 = 1369 + 2*666

3. (2+n)*(333+162n)

, ...

11. (18+n)*(37+2n)

12. (18+648n)*(37+1369n)

ramsey2879

Quote by Hurkyl

T(n) = n²/2 + n/2

T(kn + m) = (kn+m)²/2 + (kn+m)/2
= (k²/2)n² + (km)n + (m²)/2 + (k/2)n + m/2
= (k²/2)n² + (km+ k/2)n + (m² + m)/2
= (k²/2)n² + (km + k/2)n + T(m)

(a+cn)(b+dn) = (cd)n² + (ad + bc)n + ab

I check my data over and over and two principles remain.
1. Although k can take any integer value in the first set of equations, there are only a finite number, i.e., [tex]\Upsilon(T(m))[/tex] of k values for which the diophantine equation set below has a solution in integers.
[tex]ab = T(m)[/tex]
[tex] cd = k^2/2[/tex]
[tex]ad + bc = km + k/2[/tex]
2. The solution for each k value is given by my equations for c and d as a function of a,b,m where [tex](a,b) \iff (m,k)[/tex] .
[tex]\Upsilon(T(m))[/tex] equals the number of divisors of T(m).
I am confident here that no counter example can be found.

ramsey2879

Quote by ramsey2879

I check my data over and over and two principles remain.
1. Although k can take any integer value in the first set of equations, there are only a finite number, i.e., [tex]\Upsilon(T(m))[/tex] of k values for which the diophantine equation set below has a solution in integers.
[tex]ab = T(m)[/tex]
[tex] cd = k^2/2[/tex]
[tex]ad + bc = km + k/2[/tex]
2. The solution for each k value is given by my equations for c and d as a function of a,b,m where [tex](a,b) \iff (m,k)[/tex] .
[tex]\Upsilon(T(m))[/tex] equals the number of divisors of T(m).
I am confident here that no counter example can be found.

I have a proof for this statement. Anyone interested?

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Hurkyl Recognitions: Gold Member Science Advisor Staff Emeritus	T(n) = n²/2 + n/2 T(kn + m) = (kn+m)²/2 + (kn+m)/2 = (k²/2)n² + (km)n + (m²)/2 + (k/2)n + m/2 = (k²/2)n² + (km+ k/2)n + (m² + m)/2 = (k²/2)n² + (km + k/2)n + T(m) (a+cn)(b+dn) = (cd)n² + (ad + bc)n + ab Why look at the (kn+m)-th triangular number? Because I wanted the most general expression that was quadratic in n. (And later realized that one of my coefficients was your m)