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Conjecture on triangular numbers T(n)

  1. Jan 18, 2008 #1
    Let a triangular number T(n) = n*(n+1)/2 be factored into the product A*B with A less or equal to B.
    Let gcd(x,y) be the greatest common divisor of x and y
    For each of pair (A,B) define C,D,E,F as follows

    C = (gcd(A,n+1))^2,
    D = 2*(gcd(B,n))^2,
    E = 2*(gcd(A,n))^2,
    F = (gcd(B,n+1))^2.

    As an example.
    Let n = 36 then T(n) = 666 which can be factored into 6 distinct pairs A,B. The
    six sets (A,B,C,D,E,F) are as follows

    1. (1,666,1,648, 1369,2)
    2. (2,333,1,162,1369,8)
    3. (3,222,1,72,1369,18)
    4. (6,111,1,18,1369,72)
    5. (9,74,1,8,1369,162)
    6. (18,37,1,2,1369,648)

    My conjecture is that the products (A+Ct)*(B+Dt) and (A+Et)*(B+Ft) are each triangular numbers (= T(r) = r(r+1))/2) for all integer t. Also, the above example gave a interesting result in that when the 12 results for r (= the integer part of the square root of 2T(r)) are sorted by value the difference between adjacent r values when t = 1 was one of the even factors of 666. It would seem to me that the easiest way to prove this conjecture would be to derive and prove a formula for r. I will give the corresponding r values below for t = 1
    1. 72,110
    2. 54,184
    3. 48,258
    4. 42,480
    5. 40,702
    6. 38,1368

    Also I found that r follows an arithmetic series as t varies, so it is important to find the pattern for the above values, since the difference between them and 36 is the difference value of the arithmetic series. Anyone see a pattern?
    Last edited: Jan 18, 2008
  2. jcsd
  3. Jan 19, 2008 #2
    just a silly question related with you are posting here: is there triangular prime numbers?

    I think should be but in a quick search I didn't find an example
  4. Jan 20, 2008 #3
    A formula can be derived from the (positive) solution of r^2 + r - 2*T(r) = 0, which is
    [tex]r = {{\sqrt {1 + 8 T(r)} - 1} \over 2}}[/tex]
    But I don't see how this helps to prove your conjecture.
    Last edited: Jan 20, 2008
  5. Jan 20, 2008 #4


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    As stated in the original post, any triangular number can be writen in the form n(n+1)/2. Since one of either n or n+1 must be even, that is the product of positive integers. The only way it could be prime is if one of those numbers is 1: if n= 1, then 1(2)/2= 1 which is not a prime number. If (n+1)/2= 1, then n+1= 2 so n= 1 again.

    No, there are no prime triangular numbers.
  6. Jan 20, 2008 #5

    Yes, this is obvious, I didn't see the most obvious thing !!
  7. Jan 20, 2008 #6
    Actually, T(2)=3 is prime... but the reasoning holds for higher n.
  8. Jan 20, 2008 #7
    Progress report

    While only one prime, 3, can be represented as a triangular number, all primes other than 3 can be represented as n(n+1)/2 -i where i is a positive integer less than n in one and only one way, and as another posted noted triangular numbers and square numbers are related since 8T(n) + 1 = (2n+1)^2.

    Progress report: my conjecture does indeed hold where n is even and A is a factor of n other than n. First of all I forgot to note that my formula for C,D,E,F was meant only for the case where n is even, if n is odd then make the substitution n' = -(n+1) and the formula also works. I still am working on the general conjecture.

    Proof of my simplified conjecture follows:

    The new values of A,B,C,D,E, and F can now be expressed as

    1A. A,B,1,2*(n/2A)^2,2*A^2,(n+1)^2

    The second conjectured result follows directly from the relation

    [tex]T_{(n + 2A(n+1)t)} = (A +2t*(A^2))*(B+ t*(n+1)^2)[/tex]

    The first conjectured result (using C and D) follows directly from the relation

    [tex]T_{(n + nt/A)} = (A + t)*(B + 2t*(n/2A)^2)[/tex]

    The above relations follow directly by expansion of both sides and comparison of like terms after making the substitution n(n+1)/2 = AB

    Although the above is presumably dealing with integers only, I think the same relationship hold for n and A being within the field of rationals or within the complex number field as long as A,B,C,D,E,and F are defined as in line 1A. In this case the conditions that n is odd or that A is a factor of n are not necessary as long as T(n) is defined as n(n+1)/2 regardless of the nature of n.

    I will concentrate on A = a factor of n+1 next
    Last edited: Jan 21, 2008
  9. Jan 22, 2008 #8
    You conjecture is true for A = n/2 and B = n+1, when n = even number, for the expression (A+Dt)(B+Ct)

    as gcd(n,n+1) = 1, then C = 1 and D = 2

    [tex](A+Dt)(B+Ct) = (n/2 + t)(n+1 + 2t) = \frac{[n+2t]*[(n+2t)+1]}{2}[/tex]

    But (A+Et)(B+Ft) is not a triangular number for all t

    gcd(A,n) = n/2 ==> [tex]2[gcd(A,n)]^2 = \frac{n^2}{2}[/tex] ==>

    ==> [tex]A+Et = \frac{n + n^2t}{2}[/tex]

    gcd(B,n+1) = n+1 ==> [tex]gcd(B,n+1)^2 = (n+1)^2[/tex] ==>

    ==> [tex]B+Ft = n + 1 + (n+1)^2t = n + n^2t + (2nt + t + 1)[/tex]

    showing that (A+Et)(B+Ft) is triangular if and ony if t = 0
    Last edited: Jan 22, 2008
  10. Jan 22, 2008 #9
    We have to test when A is not n/2, say A = kx and B = wy
  11. Jan 22, 2008 #10
    write A = kx and B = wy

    its easy to see that if k divides n/2 and x not, then x divides n+1
    same argument for B factors

    supose we take A and B such that k divides n/2, x does not, w divides n+1, y does not, so n/2 = ky and n+1 = xw ==> 2ky + 1 = xw

    gcd(A,n+1) = x ==> C = x^2
    gcd(B,n) = y ==> D = 2y^2
    gcd(A,n) = k ==> 2k^2
    gcd(B,n+1) = w ==> w^2

    now we have to find a r(r+1)/2 form
  12. Jan 22, 2008 #11
    You didn't look far enough!
    [tex](A + Et)*(B+Ft) = \frac{n+n^2t}{2} *(n+n^2t +2nt + t + 1)[/tex]

    [tex] = \frac{n+n^{2}t}{2}*((n+n^{2}t+ nt+1) + nt + t) [/tex]

    [tex] = \frac{n +n^{2}t +nt}{2}*(n+n^{2}t + nt + 1)[/tex]

    [tex] = T_{n + n^{2}t +nt}[/tex]

    What's more important you can let A be any integer, rational or complex number and still the product
    (A+Et)*(B+Ft) = m(m+1)/2 where T(n) = A*B is a triangular number and E and F are defined as in line 1A of my January 20 post and m = n + 2A(n+1)t
    Last edited: Jan 22, 2008
  13. Jan 22, 2008 #12
    Thanks that is helpful, I will use these new terms in my next response.
  14. Jan 22, 2008 #13
    For (A+Ct)*(B + Dt); r = n+2xyt
    by symmetry
    For (A+Et)*(B+Ft); r = n+2wkt
  15. Jan 23, 2008 #14
    Now put k = (a^n)/2, y = 1/b^n, x = 1/b^n and w = c^n then my equations become

    T(2ky + 2xyt) = (kx +tx^2)*(wy+2ty^2)

    T(2ky +2kwt) = (ky +2tk^2)*(wy+tw^2)

    Is it true that for n>2 this can have no solution except a = 0 since the requirement that 2ky + 1 = wx can not be met due to Fermat's theorm?
    Last edited: Jan 23, 2008
  16. Jan 23, 2008 #15
    This is not correct, you cannot put "nt" in the other side like this, and what about "t"?
  17. Jan 23, 2008 #16
    (n+n^2t)*(nt+t)=nt*(1+nt)*(n+1) = nt*(n + n^2t + nt + 1) so I was right to express it that way. Sorry for not showing all steps but I had trouble with my latex expressions.
  18. Jan 23, 2008 #17
    we have to write (A+Et)*(B+Ft) in the r(r+1)/2 form, right?

    if [tex]A+Et = \frac{n + n^2t}{2}[/tex], then [tex]n + n^2t[/tex] can be called "r"

    as[tex]B+Ft = n + (n+1)^2t + 1[/tex], and [tex]n + n^2t \neq n + (n+1)^2t[/tex], I can NOT write the product (A+Et)*(B+Ft) in the r(r+1)/2 form, since [tex] n + (n+1)^2t [/tex] is not "r",

    try to find a concrete counter example proving that (A+Et)*(B+Ft) is triangular for t > 0, I may be mistaken, but I believe that there is not such example.
  19. Jan 24, 2008 #18
    I just finished showing that r = n + n^2t + nt . You are under the false impression that r must equal (A+Et) or (B+Ft) but all I said was that r(r+1)/2 = (A+Et)*(B+Ft)
    In the example you chose r = n+n^2t +nt . Work it out and see for yourself.
  20. Jan 25, 2008 #19
    Sorry, my big mistake, pure laziness !!! Of course you're right
  21. Jan 29, 2008 #20
    Let T(r) = r(r+1)/2
    I have used this format as follows
    Let n = 2ab and n+1 = cd then the following 4 equations hold for all t
    T(n + 2act) = ac(b+ct)(d + 2at)
    T(n + 2adt) = ad(b+dt)(c + 2at)
    T(n + 2bct) = bc(a+ct)(d + 2bt)
    T(n + 2bdt) = bd(a+dt)(c + 2bt)

    Knowing that T(n) + T(n+1) = (n+1)^2, I tried letting a=c=1, b = x^n/y^n and c = z^n/y^n in the above formulas and got

    (z^n/y^n)^2 = z^n(z^n + y^n + x^n)/2y^(2n) each time by choosing t = 1/2ac etc.
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