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Triangular numbers facturable into n*(n+k)

  1. May 14, 2010 #1
    Conjecture All triangular numbers T(i) in the recursive series i(0) = 0, i(1) = k*2+1 and following the recursive relationship i(j) = 6*i(j-1) - i(j-2) + 2 can be factured into the product n*(n+k) where n and k are integers. where k = 0 these are the square triangular numbers. I know that the series relationship for square triangular numbers is well known, but has anyone before proven or made the above conjecture?
     
  2. jcsd
  3. May 15, 2010 #2
    Besides the recursive relation for the argments i of the triangular numbers the two factors differing by k each follow a closely related sequence the value n follows the sequence n(0) = 0, n(1) = k+1 and the following values are determined by the recursive relation n(i) = 6*n(i-1) - n(i-2) + 2k. For n = 2 the series is 0, 3, 22, 133, ... . Therefore the following products are triangular numbers 0*2, 3*5, 22*24, 133*135, ... k can take negative values also. E.G. for k = -2, the series is 0, -1, -10, -63, ... and the products 0*2, 1*3, 10*12, 63*65,... are also triangular numbers.
    I am interested If k can take complex values also. For instance, if k can equal "i" then (1+i)*(1+2i) = (1+2i)*(2+2i)/2 which is in the form C * (C +1)/2 where C is a complex integer).
     
    Last edited: May 15, 2010
  4. May 19, 2010 #3
    EDIT: This is NOT a counterexample if conjecture is interpreted as OP intended. It is based on a misinterpretation. See sequel.

    If [itex]i_j[/itex] satisfies [itex]i_0=0,i_1=2k+1[/itex] and [itex]i_j=6i_{j-1}-i_{j-2}+2[/itex], then for [itex]k=28[/itex], [itex]i_5=68265=T_{369}=369.370/2[/itex].

    But [itex]247(247+28)=67925<68265[/itex] and [itex]248(248+28)=68448>68265[/itex], so [itex]i_5[/itex] is not expressible as [itex]n(n+k)[/itex].
     
    Last edited: May 20, 2010
  5. May 19, 2010 #4
    I believe I said t(i) can be expressed as n*(n + k)

    t(i,k) = t(5,28) = 68265*(68265+1)/2 = 2330089245
    n = 48257
    n*(n+k) = 48257*(48257 + 28) = 2330089245

    I am sorry that I did not make myself clear that t(i) meant to use i as the argument, not as the triangular number.
     
    Last edited: May 19, 2010
  6. May 19, 2010 #5
    Sorry I hadn't realised that the [itex]i[/itex] in [itex]T(i)[/itex] was one of the [itex]i(n)[/itex]. I read it just as, "when an [itex]i(j)[/itex] is also a [itex]T(i)[/itex] for some [itex]i[/itex] then there is an [itex]n[/itex] such that [itex]T(i)=i(j)=n(n+k)[/itex] where [itex]i(1)=2k+1[/itex]".

    I thought there was probably something wrong because the cases where [itex]i(j)=T(i)[/itex] for some [itex]i[/itex] are quite rare.

    So, back to the drawing board.
     
  7. May 19, 2010 #6
    I found that this works for complex numbers also If you let the T(x) function take complex arguments. As an example, let k = 1+i and I = 2

    then

    I(0) = 0, I(1) = 3 + 2i, I(2) = (20 + 12i)

    T(I) = (21 + 12i) * (10 + 6i) = 138 + 246i

    n(0) = 0, n(1)= 2 + i , n(2) = 14 + 8i

    n*(n + k) = (14 + 8i)*(15 + 9i) = 138 + 246i
     
  8. May 19, 2010 #7
    Yeah, I need to read up on triangular numbers that are square to get a better start on proving this, but although I enjoy the diversion of looking for number patterns, especially with triangular numbers, I have just a little knowledge of number theory and thats it. If anyone can prove this for any k other than 0 or 1 which are known series, let alone proving it for all k, i would appreciate it. Also, I have arrived at a more general conjecture:

    Let T(x) be defined as x*(x+1)/2. Let C be any complex integer and let the product
    [tex] A_{0} * B_{0} = T(C)[/tex]. Let [tex]B_{0} - A_{0} = K[/tex]. Then there exists an infinite series of complex numbers [tex]D_{i}[/tex] such that [tex]T(C_i) = D_{i}*(D{i} + K)[/tex] where [tex]C_{i}) = 6C_{i-1) - C_{i-2} + 2[/tex] and [tex]D_{i} = 6D_{i-1} - D_{i-2} + 2K[/tex].
     
  9. May 19, 2010 #8
    As an example with integers starting with K = 5, I = 3, A = 1, B = 6, we can set I(0) = -4,
    I(1) = 3, D(0) = D(1) = 1. Then I(2), I(3), ... = 24, 143, ...; and D(2), D(3) ... = 15, 99, ...;
     
  10. Jun 3, 2010 #9
    I think I found a proof of a more comprehensive theorm than my conjecture:

    Please check for errors

    Theorm: For any given triangular number [tex]T(n_{0}) =n*(n+1)/2[/tex] and any given number [tex]N_{0}[/tex], the difference [tex]T(n_{i}) - N_{i}*(N_{i} + K)[/tex] is always constant for any constant K where the series [tex]N_{i}[/tex] is found by letting [tex]N_{1} = 3*N_{0} + K +1 +2*n_{0}[/tex] with the recursion [tex]N_{i} = 6*N_{i-1} - N_{i-2} + 2K[/tex] and letting [tex]n_{1} = 4*N_{0} +2K + 1 +3*n_{0}[/tex] for which the recursive relation is [tex]n_{i} = 6*n_{i-1} - n_{i-2} + 2[/tex].

    I have the proof and will supply it latter.
     
    Last edited: Jun 3, 2010
  11. Jun 3, 2010 #10
    My proof is as follows:

    I gave in my last post only the terms for [tex]N_{1}[/tex] and [tex]n_{1}[/tex] as function of [tex]N_{0}[/tex] and [tex]n_{0}[/tex] as well as the recursive formula for [tex]N_{i}[/tex] and [tex]n_{i}[/tex]. Apart from the recursive formula, all of the above can be directly verified by doing the math and also one can simarly verify using alternative to N_1 and n_1 terms which follow N_0 and n_0 respectively, the terms preceding the given terms : [tex]N_{-1} = 3*N_{0} + K -1 -2*n_{0}[/tex] and [tex]n_{-1} = 1 +3*n_{0}-4*N_{0}-2K [/tex].

    Once you have verified that for each of the three pairs of terms (N{-1}, n{-1}), (N{0}, n{0}) and (N{1},n{1}) that N{i}*(N{i}+K) - n{i}*(n{i} +1)/2 is the same and verified that each set of 3 terms follows the assigned recursive relation; the next step is to verify that the same formulas based upon N{0} and n{0} for the terms N{-1} and n{-1} will give N{0} and n{0} when based upon the pair (N{1},n{1}). This is shown as follows.

    [tex]N_{0} = 3*N_{1} + K -1 - 2*n_{1}[/tex]
    [tex] = 3*(3*N_{0} + K + 1 + 2*n_{0}) + K - 1 -2*(4*N_{0} + 2K + 1 + 3*n_{0})[/tex]
    [tex] = (9-8)N_{0} + (3+1-4)K + (3-1-2) + (6 - 6)n_{0} = N_{0}[/tex]


    [tex]n_{0} = 1 + 3*n_{1}-4*N_{1} -2K[/tex]
    [tex]=-4(3N_{0} + K + 1 + 2n_{0}) -2K +1 +3(4N_{0} + 2K + 1 + 3n_{0})[/tex]
    [tex] = (12-12)N_{0} +(4+2-6)K +(-4 +1-2) + (-8+9)n_{0} = n_{0} [/tex]




    Does this prove my conjecture?
     
    Last edited: Jun 4, 2010
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