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Arithmetic Series and Triangular Numbers

  1. Re: Arithmetic Series and Factors of Triangular Numbers

    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
     
  2. jcsd
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