As mentioned in the previous post, the ratio of consecutive Fibonacci numbers converges to the golden ratio. Is there a sequence whose ratios converge to the silver ratio the way ratios of Fibonacci numbers converge to the golden ratio?

(If you’re not familiar with the silver ratio, you can read more about it here.)

The Pell numbers P_n start out just like the Fibonacci numbers:

P₀ = 0
P₁ = 1.

But the recurrence relationship is slightly different:

P_n+2 = 2P_n+1 + P_n.

So the Pell numbers are 0, 1, 2, 5, 12, 29, ….

The ratios of consecutive Pell numbers converge to the silver ratio.

Metallic ratios

There are more analogs of the golden ratio, such as the bronze ratio and more that do not have names. In general the kth metallic ratio is the larger root of

x² − kx − 1 = 0.

The cases n = 1, 2, and 3 correspond to the gold, silver, and bronze ratios respectively.

The quadratic equation above is the characteristic equation of the recurrence relation

P_n+2 = kP_n+1 + P_n.

which suggests how we construct a sequence of integers such that consecutive ratios converge to the nth metallic constant.

So if we use k = 3 in the recurrence relation, we should get a sequence whose ratios converge to the bronze ratio. The results are

0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, …

The following code will print the ratios.

    def bronze(n):
        if n == 0: return 0
        if n == 1: return 1
        return 3*bronze(n-1) + bronze(n-2)

    for n in range(2, 12):
        print( bronze(n)/bronze(n-1) )

Here’s the output.

    3.0
    3.3333333333333335
    3.3
    3.303030303030303
    3.302752293577982
    3.3027777777777776
    3.302775441547519
    3.3027756557168324
    3.302775636083269
    3.3027756378831383

The results are converging to the bronze ratio

(3 + √13)/2 = 3.302775637731995.

Plastic ratio

The plastic ratio is the real root of x³ − x − 1 = 0. Following the approach above, we can construct a sequence of integers whose consecutive ratios converge to the plastic ratio with the recurrence relation

S_n+3 = S_n+1 + S_n

Let’s try this out with a little code.

    def plastic(n):
        if n < 3: return n
        return plastic(n-2) + plastic(n-3)

    for n in range(10, 20):
        print( plastic(n)/plastic(n-1) )

This prints

    1.3
    1.3076923076923077
    1.3529411764705883
    1.3043478260869565
    1.3333333333333333
    1.325
    1.320754716981132
    1.3285714285714285
    1.3225806451612903

which shows the ratios are approaching the plastic constant 1.324717957244746.

Related posts

• Gold, Silver, and Bronze Ratios
• Imaginary gold, silver, and bronze
• Plastic powers

The number of kilometers in a mile is k = 1.609344 which is close to the golden ratio φ = 1.6180334.

The ratio of consecutive Fibonacci numbers converges to φ, and so you can approximately convert miles to kilometers by multiplying by a Fibonacci number and dividing by the previous Fibonacci number. For example, you could multiply by 8 and divide by 5, or you could multiply by 13 and divide by 8.

As you start going down the Fibonacci sequence, consecutive ratios get closer to k and closer to φ. But since the ratios converge to φ, at some point the ratios get closer to φ and further from k. That means there’s an optimal Fibonacci ratio for converting miles to kilometers.

I was curious what this optimal ratio is, and it turns out to be 21/13. There we have

|k − 21/13| = 0.0060406

and so the error in the approximation is 0.375%. The error is about a third smaller than using φ as the conversion factor.

The Lucas numbers satisfy the same recurrence relation as the Fibonacci numbers, but start with L₀ = 2 and L₁ = 1. The ratio of consecutive Lucas numbers also converges to φ, and so you could also use Lucas numbers to convert miles to kilometers.

There is an optimal Lucas ratio for converting miles to kilometers for the same reasons there is an optimal Fibonacci ratio. That ratio turns out to be 29/18, and

|k − 29/18| = 0.001767

which is about 4 times more accurate than the best Fibonacci ratio.

A Mersenne number is a number of the form 2^k − 1. A Mersenne prime is a Mersenne number which is also a prime.

It turns out that if 2^k − 1 is prime then k must be prime, so Mersenne numbers have the form 2^p − 1 is prime. What about the converse? If p is prime, is 2^k − 1 also prime? No, because, for example, 2¹¹ −  1 = 2047 = 23 × 89.

If p is not just a prime but a Mersenne prime, then is 2^p − 1 a prime? Sometimes, but not always. The first counterexample is p = 8191.

There is an interesting chain of iterated Mersenne primes:

This raises the question of whether m = 2^M₁₂ − 1 is prime. Direct testing using available methods is completely out of the question. The only way we’ll ever know is if there is some theoretical result that settles the question.

Here’s an easier question. Suppose m is prime. Where would it fall on the list of Mersenne primes if conjectures about the distribution of Mersenne primes are true?

This post reports

It has been conjectured that as x increases, the number of primes p ≤ x such that 2^p – 1 is also prime is asymptotically

e^γ log x / log 2

where γ is the Euler-Mascheroni constant.

If that conjecture is true, the number of primes less than M₁₂ that are the exponents of Mersenne primes would be approximately

e^γ log M₁₂ / log 2 = 226.2.

So if m is a Mersenne prime, it may be the 226th Mersenne prime, or M_n for some n around 226, if the conjectured distribution of Mersenne primes is correct.

We’ve discovered a dozen Mersenne primes since the turn of the century and we’re up to 51 discovered so far. We’re probably not going to get up to the 226th Mersenne prime, if there even is a 226th Mersenne prime, any time soon.

I saw someone point out recently that

10! = 7! × 5! × 3! × 1!

Are there more examples like this?

What would you call the pattern on the right? I don’t think there’s a standard name, but here’s why I think it should be called double super factorial or super double factorial.

Super factorial

The factorial of a positive number n is the product of the positive numbers up to and including n. The super factorial of n is the product of the factorials of the positive numbers up to and including n. So, for example, 7 super factorial would be

7! × 6! × 5! × 4! × 3! × 2! × 1!

Double factorial

The double factorial of a positive number n is the product of all the positive numbers up to n with the same parity of n. So, for example, the double factorial of 7 would be

7!! = 7 × 5 × 3 × 1.

Double superfactorial

The pattern at the top of the post is like super factorial, but it only includes odd terms, so it’s like a cross between super factorial and double factorial, hence double super factorial.

Denote the double super factorial of n as dsf(n), the product of the factorials of all numbers up to n with the same parity as n. That is,

dsf(n) = n! × (n − 2)! × (n − 4)! × … × 1

where the 1 at the end is 1! if n is odd and 0! if n is even. In this notation, the observation at the top of the post is

10! = dsf(7).

Super double factorial

We can see by re-arranging terms that a double super factorial is also a super double factorial. For example, look at

dsf(7) = 7! × 5! × 3! × 1!

If we separate out the first term in each factorial we have

(7 × 5 × 3 × 1)(6! × 4! × 2!) = 7!! dsf(6)

We can keep going and show in general that

dsf(n) = n!! × (n − 1)!! × (n − 2)!! … × 1

We could call the right hand side super double factorial, sdf(n). Just as a super factorial is a product of factorials, a super double factorials is a product of double factorials. Therefore

dsf(n) = sdf(n).

Factorials that equal double super factorials

Are there more solutions to

n! = dsf(m).

besides n = 10 and m = 7? Yes, here are some.

0! = dsf(0)
1! = dsf(1)
2! = dsf(2)
3! = dsf(3)
6! = dsf(5)

There are no solutions to

n! = dsf(m)

if n > 10. Here’s a sketch of a proof.

Bertrand’s postulate says that for n > 1 there is always a prime p between n and 2n. Now p divides (2n)! but p cannot divide dsf(n) because dsf(n) only has factors less than or equal to n.

If we can show that for some N, n > N implies (2n)! < dsf(n) then there are no solutions to

n! = dsf(m)

for n > 2N because there is a prime p between N and 2N that divides the left side but not the right. In fact N = 12. We can show empirically there are no solutions for n = 11 up to 24, and the proof shows there are no solutions for n > 24.

I saw a hand-drawn version of the diagram above yesterday and noticed that the points were too evenly distributed. That got me to thinking: is there any objective way to say that this famous diagram is in some sense complete? If you were to make a diagram with more points, what would they be?

Simple numbers

The numbers on the diagram are all simple. Once we’re more precise about what it means for these numbers to be “simple,” we can answer the questions above.

The angles in the diagram are all rational parts of a circle, that is, rational multiples of 2π. For the rest of the post, I’ll say “rational angle” to mean a rational multiple of 2π.

The sines and cosines all involve only one square root, i.e. no nested roots. A more useful way to express this is that all the values are the roots of a quadratic polynomial with integer coefficients.

Completeness

Could we add more rational angles whose sines and cosines are roots of quadratics? Maybe the chart would be too cluttered to put in a textbook, but would it be possible in principle? Could there be some chart analogous to the one above that has, for example, (1 + √7)/5 as one of the labels?

The angles in the common unit circle diagram, integer multiples of π/4 and π/6, are the only rational angles with sines and cosines that are roots of a quadratic polynomial with integer coefficients. That is, these are the only rational angles that have sines and cosines that are algebraic of degree 2. In that sense the diagram is complete.

The number (1 + √7)/5 is algebraic of degree 2 [1] but isn’t on our exhaustive list of possible algebraic values of degree 2. So even if you were to try numbers of the form pπ/q for very large integers p and q, you’ll never get a sine or cosine equal to (1 + √7)/5.

In 1933 Lehmer [2] showed how to classify all rational angles whose sines or cosines are algebraic of given degree. His theorem proves that the only rational angles whose sine is algebraic of degree 2 are integer multiples of π/4 and π/6.

Interestingly, there is another rational angle whose cosine is algebraic of degree 2:

cos(π/5) = (1 + √5)/4

So we could extend the unit circle diagram to include multiples of π/5, but only the cosine would be algebraic of degree 2. The sines are more complicated. For example,

sin(π/5) = √(5/8 + √(5)/8)

which is algebraic of degree 4.

Higher degrees

There are no rational angles whose sine is algebraic of degree 3, so going up to degree 3 wouldn’t help.

If we go up to degree 4 then we could add multiples of π/5, π/8, and π/12. These all have sines and cosines that are algebraic of degree 4.

Related posts

• How medieval astronomers made trig tables
• Sines of integers
• Systematically solving trig equations

[1] (1 + √7)/5  is a root of 25x² − 10x = 6 = 0.

[2] D. H. Lehmer. A Note on Trigonometric Algebraic Numbers. The American Mathematical Monthly , March 1933, Vol. 40, No. 3, pp. 165–166