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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.

share|improve this question
1  
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". –  Martin Sleziak Jun 29 '12 at 10:11
15  
Is it usual for unproved conjectures to be stated as fact on the OEIS? –  Rahul Jun 29 '12 at 10:22
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. –  hardmath Jun 29 '12 at 10:27
1  
Maybe you can show that $a(n+1)-a(n)-a(n-1)$ is small? –  Gerry Myerson Jun 29 '12 at 12:17
8  
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. –  Did Jun 29 '12 at 12:28

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