Fibonacci Again

In 1970, John Dixon showed that the number of steps in the Euclidean algorithm for two positive integers  is less than or equal to 2.078[log a+1] where a is the larger of the two positive integers. 
It would seem that, roughly once a year, I return to Fibonacci, perhaps because I love playing with numbers.

Plotting known Fibonacci number intervals at stable values of  phi against their corresponding difference values yielded an interesting and perhaps comparative result to Dixon’s. The equation, which isn’t very clear on the graph, reads y=2.078ln(x)+1, which seemed curious…
What fun. 

Especially as I was looking for a trivial solution for the Schwarzchild radius of a black hole under a nonlinear gravitational field, and Euclid only dealt with integers.

I have no idea what this means, if anything at all.

2 thoughts on “Fibonacci Again

  1. “What fun?” (Only if you have an abacus for a brain and a passion for solitude.)Good thing no one asks me about Fibonacci, or Euclid, or Pythagoras.

    My own relationship with numbers is conflicted and is made workable only by figuring out which brainiac/calculator/google app I need to provide the answer to whatever mathematical dilemma is currently troubling me.

    And I don't think you even know what a “trivial solution” is. lol


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