Triangular numbers, T(An +B), that equal (Cn+D)*(En+F) for all n

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Discussion Overview

The discussion revolves around the exploration of triangular numbers and their representation through interdependent arithmetic sequences. Participants investigate the equation T(A*n+B)=(C*n+D)*(E*n+F) for all integer n, where T(x) represents the triangular number function. The scope includes theoretical exploration and the search for families of solutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a set of equations to generate interdependent arithmetic sequences A*n + B, C*n + D, and E*n + F, claiming they solve the equation T(A*n+B)=(C*n+D)*(E*n+F) for all integer n.
  • Another participant introduces additional families of solutions, suggesting an interesting interrelation between the new solutions and those previously mentioned.
  • There is a claim that the findings may have implications for congruences, although this is not elaborated upon in detail.
  • A participant expresses frustration, questioning the clarity and significance of the original claims, suggesting they may not be sophisticated enough for a more advanced audience.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus. While some present new findings and propose further exploration, others challenge the clarity and relevance of the discussion, indicating a divide in understanding and engagement.

Contextual Notes

Some assumptions about the relationships between the sequences and the implications of the findings remain unexamined. The mathematical steps leading to the proposed solutions are not fully detailed, leaving potential gaps in understanding.

Who May Find This Useful

This discussion may be of interest to those exploring number theory, particularly in the context of triangular and figurate numbers, as well as individuals looking for methods to generate arithmetic sequences with specific properties.

ramsey2879
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I found that interdependent arithmetic sequences:
A*n + B, C*n + D, and E*n +F solving the
formula T(A*n+B)=(C*n+D)*(E*n+F), for all integer n can be
generated by the equations
A=(2m+1)*(2m+2)
B=(2m+2)*2m
C=+/- (2m+2)*(m+1)
D=+/- (2m+1)*(m+1)
E=+/- (2m+1)*(2m+1)
F=+/- (2m+1)*2m
I tried to find an example not generated by these formulas but could not. I believe that this finding has many implications with congruences.
 
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More on Triangular Numbers and arithmetic sequences

I found two more solution families to T(An+B)=(Cn+D)*(En+F) for all n where T(x) = x*(x+1)/2. Plus there is an interesting interrelation between the solutions. Besides the solutions for A,B,C,D,E and F below there is the solution:
A'=8m*(2m+1)
B'=4m*(3m+1)
C'=(2m+1)*(4m+2)
D'=+/- (2m+1)*(3m+2)
E'=+/- 16m^2
F'=+/- 2m*(6m+1)
Interestingly the above solution and the previous solution merge to form the third solution as follows:
A"=A'
B"=B
C"=C'
D"=D
E"=E'
F"=F
I am further searching for more solution sets.
ramsey2879 said:
I found that interdependent arithmetic sequences:
A*n + B, C*n + D, and E*n +F solving the
formula T(A*n+B)=(C*n+D)*(E*n+F), for all integer n can be
generated by the equations
A=(2m+1)*(2m+2)
B=(2m+2)*2m
C=+/- (2m+2)*(m+1)
D=+/- (2m+1)*(m+1)
E=+/- (2m+1)*(2m+1)
F=+/- (2m+1)*2m
I tried to find an example not generated by these formulas but could not. I believe that this finding has many implications with congruences.
 
Group created on Triangular and Figurate Numbers

I posted my further findings and some links at this group which I initiated:
http://groups.yahoo.com/group/Figurate_Numbers/
I present there a general method to find other families of solutions for values of A-F. I have more to post and intend to do so on this new group in the future
 
Do you realize that you are getting no responses because no one can understand what you are talking about?

What is your point? You start by creating sequences have a certain relationship and then show that the relation can be written in a number of different ways. If you in 9th or 10th grade then, Okay, pretty good. If you are older than that, all that should be trivial to you.
 

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