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

  1. Apr 2, 2005 #1
    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
    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.
  2. jcsd
  3. Apr 10, 2005 #2
    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:
    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:
    I am further searching for more solution sets.
  4. Apr 30, 2005 #3
    Group created on Triangular and Figurate Numbers

    I posted my further findings and some links at this group which I initiated:
    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
  5. May 1, 2005 #4


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    Science Advisor

    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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