Problems & Puzzles: Puzzles

Puzzle 1184  Floor[n GoldenRatio] =n+ Floor[n/GoldenRatio] for all n=>1. Sebastián Martín Ruiz, sent the following puzzle: Q) Prove that Floor[n GoldenRatio] =n+ Floor[n/GoldenRatio] for all n=>1, or find a counterexample. Golden Ratio=(1+Sqrt(5))/2

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From August 10 to 16m 2024 contributions came from Michael Branicky, Vicente F. Izquierdo, Emmanuel Vantieghem, Adam Stinchcombe

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Michael wrote:

This is true using the fact that g = 1 + 1/g and the properties of floor, where g is the Golden Ratio.

In particular, floor(n*g) = floor(n*(1 + 1/g)) = floor(n + n/g) = n + floor(n/g). #

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Vicente wrote

We know that  n * GR = n + n  / GR, That may be the reason for equality.

Also: n * (1-GR) = n + n / (1-GR)

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Emmanuel wrote:

The assertion is true.

Indeed, if  t =(1+Sqrt[5]) / 2  then we have trivially :

 : t - 1 = 1 / t 

and hence :
   n t = n + n (t - 1) = n + n / t.

Thus :
   Floor[n t] = Floor[n + n / t] = n + Floor[n / t], QED.

 

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Adam wrote:

Integer sums move in and out of the floor function freely: for every real number x and for every integer n, floor(n+x) = n + floor(x).  The defining property of the golden ratio is   1 +1/phi = phi, so n+n/phi = n*phi and then floor(n*phi) = floor(n+n/phi) = n +floor(n/phi).

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