0012: Article 2 (Tribonacci constant and Pi)

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A Tale of Four Constants

by Tito Piezas III

Abstract: Ramanujan-type pi formulas using the silver ratio, golden ratio, tribonacci constant, and plastic constant will be given. The constants’ commonality as being all: 1) Pisot numbers, 2) limiting ratios of special integer sequences, and 3) appearing in the metric properties of certain geometric solids will also be high-lighted.

Contents:

I. Silver Ratio

II. Golden Ratio

III. Tribonacci Constant

IV. Plastic Constant

I. Silver Ratio

As a basic introduction, recall that the silver ratio S = 1+√2 = 2.41421… is the positive root of the quadratic,

S²-2S-1 = (S-1)²-2 = 0 (eq.1)

and is the limiting ratio of the Pell-Lucas numbers x_n = {1, 3, 7, 17, 41, 99, 239,...} and Pell numbers y_n = {1, 2, 5, 12, 29, 70, 169,…}. These are the numerators and denominators of the best rational approximations to √2, being the {x, y} values of the Pell equation, x²-2y² = ±1. These sequences both have the recurrence relation S(n) = 2S(n-1) + S(n-2). Note also that the solution to the Pythagorean triple where the legs only differ by 1 such as,

21² + 20² = 29²

or, in general,

((x+1)/2)² + ((x-1)/2)² = y²

is given by solutions to x²-2y² = -1.

II. Golden Ratio

The golden ratio φ = (1+√5)/2 = 1.61803… is the positive root of,

φ²-φ-1 = 0 (eq.2)

and is best known in the context of the Fibonacci numbers {1, 1, 2, 3, 5, 8, 13,….}. This has recurrence relation F(n) = F(n-1) + F(n-2), or each term is the sum of the previous two, and φ is its limiting ratio (also for the Lucas numbers, a fact which will be relevant later).

However, the constants S and φ also appears in other contexts. Recall the well-known approximations,

e^π√43 ≈ 12³(9²-1)³ + 743.9997…

e^π√67 ≈ 12³(21²-1)³ + 743.999998…

e^π√163 ≈ 12³(231²-1)³ + 743.9999999999992…

with the last being the so-called Ramanujan constant. Not so well-known are the ff ones. Given the silver ratio S and the golden ratio φ, then,

1) e^π√6 ≈ 2⁶S⁴ + 23.8…

2) e^π√8 ≈ 2⁹S³ + 23.9…

3) e^π√16 ≈ 2^(21/2)S⁶ + 23.999…

4) e^π√22 ≈ 2⁶S¹² + 23.9999…

1) e^π√5 ≈ 2⁶φ⁶ - 24.2…

2) e^π√10 ≈ 2⁶φ¹² + 23.98…

3) e^π√15 ≈ 2¹²φ⁸ - 24.001…

4) e^π√25 ≈ 2⁶φ²⁴ - 24.00004…

The fact that the excess hovers around the integer “24” is not coincidence, as the blue numbers are the exact values of certain Dedekind eta quotients (to be given later), a modular form which involves a 24th root. All these can be used for pi formulas, though for brevity we will use only the fourth one for S and φ:

as well as,

Amazingly, for the last, the coefficients involve the four consecutive Lucas numbers (in red) as {2, 1, 3, 4, 7, 11, 18, 29, 47, 76,…}. Why that is so, I have no idea. (Update, 5/23/11) Qiaochu Yuan from MathStackExchange gave the identity for Fibonacci numbers,

F_n-1 + F_n φ = φⁿ

A related one for Lucas numbers is,

L_n φ - L_n+1 = (1/φ^n-1) √5

Hence, the last formula simplifies as,

with discriminant d = 11. This has a beautiful infinite nested radical expression,

Compare this to the one for the golden ratio given in the next section. Furthermore, analogous to the golden ratio, we have the approximation,

e^π√11 ≈ (1/T +1)²⁴ - 24.008…

where the power and excess now both involve the integer 24. Let the constant’s reciprocal be v = 1/T. The expression 1/T +1 = 1.543689… is the exact value of a special eta quotient, hence we can also use it in a pi formula. Define,

A = (1/2)(1+2v)(1+4v)

B = (1/4)(-1+2v+4v²)

C = (v+1)²⁴

h₄ = (2n)!³ / (n!⁶)

where v = 1/T, then,

1/π = Σ h₄ (An+B) / Cⁿ

from n = 0 to ∞. There is also an interesting and orderly q-continued fraction for the tribonacci constant T. Let q = -1/(e^π√11), then,

which gives the precise relationship between e^π√11 and its approximation (1/T + 1)²⁴. The general form is the Heine continued fraction. Let q = exp(2πiτ), then,

described in page 23 of W. Duke's Continued Fractions and Modular Functions where w(τ) is the eta quotient,

and related to the j-function as,

In particular, the value of the j-function involving the field Q(√-11) is exactly an integer (a power of 2), but this can be nicely expressed in terms of the irrational tribonacci constant as,

Also, as I pointed out (in 2006) to E. Weisstein of MathWorld, the tribonacci constant appears, not in the Platonic solids, but in the Archimedean solid, the snub cube,

There is certainly more to this constant than meets the eye.

Note: (9/27/10) I just learned that my observation about the tribonacci constant and the snub cube was preceded by John Sharp in his 1998 article in the Mathematical Gazette, "Have You Seen This Number?"

IV. Plastic constant

The plastic constant P = 1.32472.… is the real root of the cubic eqn,

P³-P-1 = 0 (eq.4)

Recall that the Fibonacci numbers {1, 1, 2, 3, 5, 8,...} have the recurrence relation F(n) = F(n-1) + F(n-2). This forms a square spiral,

On the other hand, the Padovan sequence {1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16,…} has P(n) = P(n-2) + P(n-3), and forms a triangular spiral,

The role of the plastic constant is, just like for the previous three constants, it is a limiting ratio, namely, for the Padovan and Perrin sequences. Furthermore, the golden ratio φ and the plastic constant P have the similar beautiful infinite nested radical representation,

The radical soln of eq. 4 is,

with discriminant d = 23. From the previous two sections, I guess one can figure out that it appears in a similar approximation,

e^π√23 ≈ 2¹²P²⁴ - 24.00007…

as well as the pi formula,

1/π = Σ h₄ (An+B) / Cⁿ

where,

A = (1/8)(5P-4)(5P+8)

B = (1/16)(9P²+4P-16)

C = 2¹²P²⁴

h₄ = (2n)!³ / (n!⁶),

from n = 0 to ∞. Just like for the tribonacci constant, there is a q-continued fraction for the plastic constant P. Let q = -1/(e^π√23), then,

which gives the relationship between e^π√23 and the approximation (√2 P)²⁴. While the tribonacci constant is implicit in the snub cube, I found that the same can be said between the plastic constant and the snub icosidodeca-dodecahedron (2006),

but surprisingly so does for the tribonacci constant T as,

4) In basic algebraic relations:

a) φ and T:

φ + 1/φ ² = 2

T + 1/T ³ = 2

b) φ and P:

1/φ + 1 = φ

1/P + 1 = P ²

c) T and P:

(1-u)³ = (2u)² , where u = (T-1)/(T+1)

(1-v)³ = (v)² , where v = 1/(P+1)

P.S. Digressing again, just a great song I found in YouTube. Hauntingly beautiful, isn't it?

-- END --

© Sept 2010, Tito Piezas III

You can email author at tpiezas@gmail.com

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