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This infinite sequence a(n) starting: $1, 1, 2, 2, 3, 4, 5, 7, 10, 14, 20, 30, 45, 68, 104, 161...$ is the antidiagonal sums of a triangle that has several properties in common with the Pascal triangle, but is defined by a different recurrence.

http://oeis.org/A186425

Does the ratio a(n+1)/a(n) converge to the Golden Ratio?

Example: $71646979665/44294592612 = 1.617510748830995479435294...$

which is close to the golden ratio.


Edit 29.6.2012:

The number triangle as a Mathematica program is:

Clear[t]; t[n_, 1] = 1;  t[n_, k_] :=   t[n, k] = If[n >= k, Sum[t[n - i, k - 1], {i, 1, k - 1}], 0]; A = Table[Table[t[n, k], {k, 1, 12}], {n, 1, 12}]; MatrixForm[A] 

Where the recurrence in latex is:

$$\displaystyle T(n,1)=1, k>1: T(n,k) = \sum\limits_{i=1}^{k-1} T(n-i,k-1)$$

And the number triangle is:

$$\displaystyle \begin{bmatrix} 1&0&0&0&0&0&0 \\ 1&1&0&0&0&0&0 \\ 1&1&1&0&0&0&0 \\ 1&1&2&1&0&0&0 \\ 1&1&2&3&1&0&0 \\ 1&1&2&5&4&1&0 \\ 1&1&2&6&9&5&1 \end{bmatrix}$$

of which the terms $1, 1, 2, 2, 3, 4, 5, 7, 10, 14...$ are the anti-diagonal sums.


Edit 29.6.2012:

Example: a(300)/a(299) =

$26846082250877975424944195125049203782232824963599896667759960/ 16591791295818073926293326126802690704313919207844843640228293$

= $1.61803398874987500298662197019...$

Golden ratio = $1.61803398874989484820458683437...$


Edit 27.4.2013: By changing a single index "k" in the recurrence to "n" we get the Pascal triangle.

Clear[t]; t[n_, 1] = 1;  t[n_, k_] :=   t[n, k] = If[n >= k, Sum[t[n - i, k - 1], {i, 1, n - 1}], 0]; A = Table[Table[t[n, k], {k, 1, 12}], {n, 1, 12}]; MatrixForm[A] 

So in form the two programs are almost identical.

  • 2
    This is somewhat puzzling: When I go to oeis link, I see that the author is Mats Granvik and I read there: "a(n+1)/a(n) tends to the golden ratio".2012-06-29
  • 0
    Yes, I know this is shameless self promotion. But it is a conjecture. I don't know if it is true, as I don't know how to prove anything.2012-06-29
  • 1
    I am not accusing you of self-promotion or any other bad intentions. I just went to OEIS, saw the name which was familiar from MSE, and came back to recommend the OP that perhaps the right person to ask would be Mats Granvik. But, well, I found out that the OP is Mats Granvik. :-)2012-06-29
  • 16
    Is it usual for unproved conjectures to be stated as fact on the OEIS?2012-06-29
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    Rahul: I don't know. But I and another guy in the oeis usually state things as facts if the numbers match the sequences.2012-06-29
  • 1
    It would greatly improve this question if the recurrence that defines the sequence was included here. It is inconvenient to readers to be asked to study/solve a problem whose statement requires parsing OEIS comments.2012-06-29
  • 0
    hardmath: I will add the recurrence to the question.2012-06-29
  • 0
    The combinatorial interpretation for $T(n,m)$ where $m>1$: the number of postive integer sequences $a_1,\ldots,a_{m-1}$ such that $a_k\le k$ and $a_1+\cdots+a_{m-1}.2012-06-29
  • 1
    Maybe you can show that $a(n+1)-a(n)-a(n-1)$ is small?2012-06-29
  • 9
    To *state things as facts if the numbers match* is a serious departure from acceptable scientific practice. One can only hope the OEIS reacts promptly to this.2012-06-29
  • 1
    It may be a useful observation (provable by induction) that each column becomes constant $(k-1)!$ after a triangular-number $k(k-1)/2$ of smaller values.2012-06-29

0 Answers 0