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Composite numbers and squares from recursive series

  1. Mar 9, 2006 #1
    This has to do with the 2nd order recursive sequence [tex]\{...a, b,
    c ...\}[/tex] where a,b,c are any three sucessive terms and [tex]c = 6b-a + 2k[/tex].
    I found that it has the following property.
    [tex]8ab - (a+b-k)^2 = 8bc - (b+c-k)[/tex] That is eight times the product of two adjacent terms always equals the square of the difference between [tex]k[/tex] and the sum of the adjacent terms plus a number that is independent of three sucessive terms of the series are chosen.

    A related property is the following. Again it is independent of which three sucessive terms of this sequence is chosen.

    [tex]8a(a+k) - (b-3a-k)^2 = 8b(b+k) - (c-3b-k)^2[/tex]

    The proof of both is by induction
    The proof of the first relation follows:

    [tex]8ab - (a+b-k)^2 = 8bc - (b+c-k)^2[/tex]
    [tex]\quad = 8b(6b - a +2k) - (7b - a + k)^2[/tex]
    [tex]\quad = (48-49)b^2 -a^2 -k^2 +(-8 +14)ab +(16-14)bk +2a[/tex]
    [tex]\quad = 8ab -(b^2 +a^2 +k^2 +2ab -2bk -2ak)[/tex]
    [tex]\quad = 8ab -(b+a-k)^2[/tex]

    Thus the above property is proven.

    Now if [tex]ab[/tex] is a triangular number for the first two terms of a
    sequence, to make the product [tex]ab[/tex] to be a triangular number regardless of which pair of adjacent terms [tex]a \text{and} b[/tex] are in the sequence, we need to select a "k" such that [tex]8ab+1=(a+b-k)^2[/tex]

    I will leave the proof of the second part and how it relates to square triangular numbers up to the reader.
    Last edited: Mar 9, 2006
  2. jcsd
  3. Mar 19, 2006 #2
    Given the 2nd order recursive sequence [tex]\{...a, b,
    c ...\}[/tex] where a,b,c are any three sucessive terms and [tex]c = 6b-a + 2k[/tex].
    The following equations for composits always hold

    [tex]8ab - (a+b-k)^2 = 8bc - (b+c-k)^2[/tex]
    [tex]8a(a+k) - (b-3a-k)^2 = 8b(b+k) - (c-3b-k)^2[/tex]

    The proof of both is by induction
    The proof of the first relation follows:

    [tex]8ab - (a+b-k)^{2} = 8bc - (b+c-k)^{2}[/tex]
    [tex]\quad = 8b(6b - a +2k) - (7b - a + k)^2[/tex]
    [tex]\quad = (48-49)b^2 -a^2 -k^2 +(-8 +14)ab +(16-14)bk +2ak[/tex]
    [tex]\quad = 8ab -(b^2 +a^2 +k^2 +2ab -2bk -2ak)[/tex]
    [tex]\quad = 8ab -(b+a-k)^2[/tex]

    The proof of the second relation is

    [tex] 8a(a+k) -(b-3a-k)^{2} = 8b(b+k)-(c-3b-k)^{2}[/tex]
    [tex] \quad = 8b(b+k) - (3b-a+k)^2[/tex]
    [tex]\quad = (8-9)b^2 +(8-9)a^2 -k^2 +(8-6)bk +(8-6)ak +6ab[/tex]
    [tex]\quad = 8a(a+k) -b^2 -9a^2-k^2 +2bk -6ak +6ab[/tex]
    [tex]\quad = 8a(a+k) -(b -3a - k)^2[/tex]

    Now since the above relation holds for any three consecutive terms of the recursive sequence, if the product of any two terms [tex]ab[/tex] is a triangular number then there is a [tex]k[/tex] such that [tex]8ab+1=(a+b-k)^2[/tex] because it is known that 8 times a triangular number + 1 is always a square. But recursively from the above proven relationship, the product of any two adjacent terms of such a series where [tex]k[/tex] is so defined is always a triangular number.

    In the second relation let [tex]k=0[/tex] then [tex] 8a^2 = (b-3a)^2 +C [/tex] for any two adjacent terms of the recursive series. As an example the first two square triangular number are 0 and 1, since[tex] 8*0 = (1-0)^2 -1[/tex], [tex]C=-1[/tex]. Thus if b^2 is a triangular number, the next square triangular number [tex]c^2[/tex] is found by solving [tex]8b^2 =(c-3*b)^2 - 1[/tex].

    Although square triangular numbers are well known, I haven't found a prior publication of this relationship. Surely someone before me must have noticed it. I would appreciate it if someone could lead me to a site that does discuss this.
    Last edited: Mar 19, 2006
  4. Mar 19, 2006 #3


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    Square triangular numbers can be related to solutions of Pells equation [tex]x^2-2y^2=1[/tex]. If [tex]b^2=n(n+1)/2[/tex] is your square triangular number, then x=(2n+1) and y=2b is a solution to Pells, and vice versa (non-negative solutions).

    Your formula for getting the next square triangular number c^2 from a square triangular number b^2 follows from the usual way of generating the solutions to this equation, i.e. the solutions are given by powers of [tex]3+2\sqrt{2}[/tex], so to go to the next solution you just multiply by this. You get the same relation between c and b as you have.
    Last edited: Mar 19, 2006
  5. Mar 19, 2006 #4
    Thanks. I was aware of the relation to Pells equation but I dont see a easy way to get from the power of [tex]3+2\sqrt{2}[/tex] to my relation for [tex]a \text{ and }b[/tex]. Also my relation is more general since [tex]k[/tex] need not equal 0 and the difference [tex]a_{n}*(a_{n} + k)- (a_{\left(n+1\right)} -3a_{n} -k)^2[/tex] need not equal -1 although it is a constant. For instance the sequence [tex]\{ \dots 0, 3, 22, 133, \dots\} \text{ where } k = 2[/tex] has the property that [tex]a_{n} * (a_{n} + k)[/tex] is always a triangular number since the constant is -1 as is the case with square triangular numbers.

    I even have a greater generalization.

    where [tex]k = -2[/tex] the following relation also holds:

    for every pair of adjacent terms [tex]a \ \text{ and }\ b[/tex],
    [tex]\ 8(a+n)(b+n) = (a+b + 2 - 4n)^2 + C \text{ where } C \text{is a constant dependent on n and}\ n [/tex] is any value including a complex value.
    Last edited: Mar 19, 2006
  6. Mar 19, 2006 #5


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    It's a straightforward computation. If your square triangular number is b^2=n(n+1)/2, then [tex]n=(-1+\sqrt{1+8b^2})/2[/tex] so [tex]x=\sqrt{1+8b^2}[/tex] and [tex]y=2b[/tex]. Multiply [tex]\sqrt{1+8b^2}+2b\sqrt{2}[/tex] by [tex]3+2\sqrt{2}[/tex] to get [tex]blah+(2\sqrt{1+8b^2}+6b)\sqrt{2}[/tex], so your next square triangle is at c^2 where [tex]c=\sqrt{1+8b^2}+3b[/tex]
    Last edited: Mar 19, 2006
  7. Mar 20, 2006 #6
    Ok that works and is a little more straight forward than my formula, thanks.
    Now to resolve my formula
    [tex]8b^2+1 = (c-3b)^2 = c^2 -6cb +9b^2[/tex]
    [tex] 1 = c^2 -6cb + b^2 \Rightarrow c*(6b-c) = b^2 - 1[/tex]
    I am lost as to where to go from here
    But [tex]c = 6b -a \Rightarrow b^2-c*a = 1[/tex]
    [tex] c = \frac{b^2-1}{a}[/tex]
    Substituting c = 6b-a

    [tex] 1 = \frac{b^2-1}{6ba-a^2}[/tex]

    How about my formula
    [tex]8(a + m)(b + m) = (a + b + 2 - 4m)^2 + C[/tex] where [tex]a[/tex] and [tex]b[/tex] are two adjacent terms in the recursive sequence defined by [tex]a_{n} = 6a_{\left(n-1\right)} - a_{\left(n-2\right)} - 4[/tex], [tex]C[/tex] is a constant dependent on m, and [tex]m[/tex] is any value?
  8. Mar 24, 2006 #7
    Relation to Pathagorean triples found.
    My friend wrote:
    "The values of [tex]c[/tex] in [tex]a^2 + b^2 = c^2[/tex] where [tex]|b - a| = 1[/tex] and gcd(a,b)=1 are generated from this recurrence relation:
    [tex]c(1)=5[/tex], [tex]c(2)=29[/tex], [tex]c(n) = 6*c(n-1) - c(n-2) [/tex]
    ( see http://www.research.att.com/~njas/sequences/A001653 [Broken] )
    What I find interesting is that apparently all Pythagorean triples can be generated with such recurrence relations. I am far from being a mathematician, and so I cannot prove any of this.
    It is sufficient to look at primitive triples where gcd(a,b)=1.
    The possible values of the difference |b - a| in primitive triples are 1,7,17,23,31,... see http://www.research.att.com/~njas/sequences/A058529 [Broken] .
    When |b-a| = 7, we have :
    c(1)=13, c(2)=17, c(3)=73, c(4)=97, c(n) = 6*c(n-2) - c(n-4)
    The values of a will be generated by:
    a(1)=5, a(2)=8, a(3)=48, a(4)=65, a(n) = 6*a(n-2) - a(n-4) + 14

    When |b-a| = 17, we have :
    c(1)=25, c(2)=53, c(3)=137, c(4)=305, c(n) = 6*c(n-2) - c(n-4)
    a(1)=7, a(2)=28, a(3)=88, a(4)=207, a(n) = 6*a(n-2) - a(n-4) + 34
    When |b-a| = 23, we have :
    c(1)=37, c(2)=65, c(3)=205, c(4)=373, c(n) = 6*c(n-2) - c(n-4)
    a(1)=12, a(2)=33, a(3)=133, a(4)=252, a(n) = 6*a(n-2) - a(n-4) + 46
    and so on...
    Does anyone know of an explanation please?
    Andras Erszegi".

    It follows from the relation [tex]8(a(n))^2 - (a(n+1)-3*a(n))^2 = C[/tex] where [tex]C[/tex] is a constant for recursive series of the form [tex]a(n) = 6a(n-1)-a(n-2)[/tex] that I found. If you note that if any two adjacent terms of the Pell-like series {a,b,2b+a,5b+2a, … 2*a(n-1) + a(n-2) } are selected as y and x of the Pathagorean triples: x^2 + y^2, x^2-y^2, and 2xy then the value for [tex](x^{2} - y^{2} - 2xy)^{2}[/tex] is a constant. Moreover, You will find that [tex]8{(a^{2} + b^{2})}^{2}-{({(2b+a)}^{2} + b^{2} \text{ minus }3(a^2+b^2))}^{2} = 4{(b^{2} - a^{2} \text{ minus } 2ab)}^{2}[/tex], i.e. the constant. Thus the series of c's for a constant absolute value of |[tex]x^2-y^2\text{ minus }2xy[/tex]| is in the form of [tex]a(n) = 6a(n-1)-a(n-2)[/tex] .
    Note that you some times have more than one such series for a constant value of [tex]{(a^{2}-b^{2}\text{ minus }2ab)}^{2}[/tex] .
    Last edited by a moderator: May 2, 2017
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