## Posts Tagged ‘sequences’

### Sequences 2, Padovan and Perrin numbers

Just like the golden ratio and tribonacci constant, powers of the plastic constant P can also be expressed in terms of sequences associated with it. P is a root of the equation,

$P^3=P+1$

or,

$P = \frac{1}{3}\left(\frac{27+3\sqrt{69}}{2}\right)^{1/3}+\frac{1}{3}\left(\frac{27-3\sqrt{69}}{2}\right)^{1/3}$

Define,

\begin{aligned} a & = \left(\tfrac{27+3\sqrt{69}}{2}\right)^{1/3}\\b&=\left(\tfrac{27-3\sqrt{69}}{2}\right)^{1/3}\end{aligned}

then powers of P  are,

$P^{n} = \frac{1}{9}(a^2+b^2)U_{n+1}+\frac{1}{3}(a+b)U_{n+2}+\frac{1}{3}V_n$

where U and V are the Padovan and Perrin sequences, respectively,

\begin{aligned} U_n &= 1,0,0,1,0,1,1,1,2,2,3,4,5,7,9,12,16\dots\\ V_n &=3,0,2,3,2,5,5,7,10,12,17,22,29,\dots\end{aligned}

$P = \frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{0}{3}$

$P^2 =\frac{1}{9}(a^2+b^2)+\frac{0}{3}(a+b)+\frac{2}{3}$

$P^3 =\frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{3}{3}$

and so on.  These sequences obey,

$W_n = W_{n-2} + W_{n-3}$

and their limiting ratio, of course, is P.  While the Fibonacci sequence has a nice representation as a square spiral, the Padovan is a spiral of equilateral triangles,

The Perrin sequence also has a notable feature regarding primality testing.  Let $x_1, x_2, x_3$ be the roots of,

$P^3=P+1$

then, starting with n = 0,

$V_n=x_1^n+x_2^n+x_3^n = 3,0,2,3,2,5,5,7,10,12,17,22,29,\dots$

Indexed in this manner, if n is prime, then n divides $V_n$.  For example $V_{11} = 22$.  However, there are Perrin pseudoprimes, composite numbers that pass this test, with the smallest being n = 521^2.

Lastly, like all the four limiting ratios of this family of recurrences, the plastic constant P  can be expressed in terms of the Dedekind eta function as,

\begin{aligned} P &=\frac{e^{\pi i/24}\,\eta(\tau) }{\sqrt{2}\,\eta(2\tau)}\end{aligned}

where,

$\tau=\frac{1+\sqrt{-23}}{2}$

### Sequences 1, Tribonacci numbers

In this 3-part series of posts, we’ll discuss well-known sequences with the recurrence,

$aP_{n-3} + bP_{n-2} + cP_{n-1} = P_n$

where {a, b, ccan only be zero or unity.  Aside from the Fibonacci and Lucas numbers which is a = 0, there is the Narayana sequence with b = 0, the Padovan and Perrin with c = 0, and the tribonacci has a = b = c = 1.  All four cases may then share similar properties and one of which, interestingly enough, is that their limiting ratios, a root of the following equations,

\begin{aligned} x^2 &=x+1\\ y^3 &= y^2+1\\ z^3 &= z+1\\ t^3 &= t^2+t+1\end{aligned}

can also be used to express $\zeta(3)$,  or Apery’s constant.

I. Fibonacci and Lucas numbers

Given the two roots of,

$x^2=x+1$

with $x_1 > x_2$, the larger root being the golden ratio, we get the Lucas numbers L(n) and Fibonacci numbers F(n),

\begin{aligned} L_n &= x_1^n+x_2^n = 2,1,3,4,7,11,18,29,\dots\\[2mm] F_n &= \frac{x_1^n-x_2^n}{\sqrt{5}} = 0,1,1,2,\,3,\,5,\,8,\,13,\dots\end{aligned}

(The starting index is n = 0.)  Expanding powers of the golden ratio, then for n > 0,

\begin{aligned} & {x_1}^n = \Big(\frac{1+\sqrt{5}}{2}\Big)^n = \frac{L_n+F_n\sqrt{5}}{2}\end{aligned}

We’ll see this can be generalized to powers of the tribonacci constant.

II. Tribonacci numbers

These are a generalization of the Fibonacci numbers, being,

$t_n = t_{n-1}+t_{n-2}+t_{n-3}$

Pin-Yen Lin has a nice paper involving these numbers.  First, define the following three sequences with this recurrence, but with different initial values,

\begin{aligned}S_n &=0,0,1,1,2,4,7,13,24,\dots\\U_n &=0,3,2,5,10,17,32,49,\dots\\V_n &=3,1,3,7,11,21,39,71,\dots \end{aligned}

(The starting index as usual is n = 0.)  The first and the third are recognized by the OEIS, with the first being the tribonacci numbers.  The limiting ratio for all three is the tribonacci constant, T, the real root of,

$x^3=x^2+x+1$

or,

$T = \frac{1}{3}+\frac{1}{3}(19+3\sqrt{33})^{1/3}+\frac{1}{3}(19-3\sqrt{33})^{1/3}$

I’ve already written about the tribonacci constant before.  But I want to include how Lin found that powers of x can be expressed in terms of those three sequences. Define,

$a =\sqrt[3]{19+3\sqrt{33}}$

$b =\sqrt[3]{19-3\sqrt{33}}$

then, similar to the golden ratio,

$T^n = \frac{1}{9}(a^2+b^2)\,S_n+\frac{1}{9}(a+b)\,U_n+\frac{1}{3}\,V_n$

Hence, starting with = 1,

$T = \frac{0}{9}(a^2+b^2)+\frac{3}{9}(a+b)+\frac{1}{3}$

$T^2 = \frac{1}{9}(a^2+b^2)+\frac{2}{9}(a+b)+\frac{3}{3}$

$T^3 = \frac{1}{9}(a^2+b^2)+\frac{5}{9}(a+b)+\frac{7}{3}$

and so on.  Interesting, isn’t it, that powers of the tribonacci constant can be expressed in this manner.

There is a primality test regarding Lucas numbers: if n is a prime then $L_n-1$ is divisible by n.  For example $L_5 = 11$, minus 1, is divisible by 5.  However there are Lucas pseudoprimes, composite numbers that pass this test, with the smallest being n = 705.

The third tribonacci sequence can be formed analogously to the Lucas numbers.  Given the three roots $x_1, x_2, x_3$ of,

$x^3=x^2+x+1$

then, starting with n = 0,

$V_n = x_1^n+x_2^n+x_3^n = 3,1,3,7,11,21,39,71,\dots$

I notice that likewise, if n is prime, then $V_n-1$ is divisible by n.  But there are also tribonacci-like pseudoprimes.  The smallest is n = 182.  Steven Stadnicki was nice enough to compute the first 36.  It turns out they are relatively rarer, as there are only 21 less than $10^8$, while there are  852 Lucas pseudoprimes in the same range.

### Even powers of Fibonacci numbers

In a previous post, it was pointed out that powers of Fibonacci numbers also obey recurrence relations.  For example, it is the case that,

$F_n^2-2F_{n+1}^2-2F_{n+2}^2+F_{n+3}^2 = 0$

In general, for even kth powers, it takes k+2 consecutive Fibonacci numbers to sum up to zero.  However, using Mathematica’s LatticeReduce function which has an integer relations algorithm, I found that if reduced to k+1 terms, then it can still sum up to a constant, though it is now non-zero.  Thus,

\begin{aligned} &F_n^2-3F_{n+1}^2+F_{n+2}^2 = (-1)^{n+1}\,2\\[1.5mm] &F_n^4-4F_{n+1}^4-19F_{n+2}^4-4F_{n+3}^4+F_{n+4}^4 = 6\\[1.5mm] &F_n^6-14F_{n+1}^6-90F_{n+2}^6+350F_{n+3}^6-90F_{n+4}^6-14F_{n+5}^6+F_{n+6}^6 = (-1)^n\, 80\\[1.5mm] &F_n^8-33F_{n+1}^8-747F_{n+2}^8+3894F_{n+3}^8+16270F_{n+4}^8+3894F_{n+5}^8-747F_{n+6}^8\\&\;\;-33F_{n+7}^8+F_{n+8}^8 = 2520\end{aligned}

and so on, with k = 10 summing to $(-1)^{n+1}\,226800$.  Notice the formulas are palindromic, the same read forwards or backwards.

I was curious if this sequence of constants,

$C(2p) = 2, 6, 80, 2520, 226800, \dots$

had a generating function. Unfortunately, OEIS didn’t recognize it, so that question is unanswered for now.

Update, May 26, 2012:  Jim Cullen found a recurrence relation which is equivalent to the formula,

\begin{aligned}C(2p)&=\prod_{n=1}^p \frac{2(2n-1)(F(n))^2}{n}=\frac{(2p)!}{p!^2}\prod_{n=1}^p (F(n))^2\\&=2,6,80, 2520, 226800,53222400,\dots\end{aligned}

hence the next constant is(12) = 53222400.  The product of the first p Fibonacci numbers (n) is called a fibonorial.

### Binomial sums and the zeta function

Recall the three sequences,

$C_n = \sum_{k=0}^n {\binom n k} {\binom {n+k}k} = 1, 3, 13, 63, 321, 1683\dots$

$B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251, 11253\dots$

$A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, 819005\dots$

Equivalently,

$C_n = \,_2F_1(-n,n+1;\,1;\,-1)$

$B_n = \,_3F_2(-n,-n,n+1;\,1,1;\,1)$

$A_n = \,_4F_3(-n,-n,n+1,n+1;\,1,1,1;\,1)$

where $_pF_q$ is the generalized hypergeometric function.  Then it is known that,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\, C_n C_{n+1}} = \frac{1}{2}\log(2)\\ &\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2 B_n B_{n+1}} = \frac{1}{5}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3 A_n A_{n+1}} = \frac{1}{6}\, \zeta(3)\end{aligned}

Beautiful, aren’t they? Since the numbers increase at a near-geometric rate (for example, $\frac{A_{n+1}}{A_n} \to (1+\sqrt{2}\,)^4 \approx 33.97$  as n  goes to infinity), then the convergence is very fast.

We also have the nice evaluations,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right) \\ &\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\ &\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = -\frac{19}{3}\,\zeta(5) +\frac{2}{3}\,\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot 3^2}\left(\zeta(4, \tfrac{1}{3})-\zeta(4,\tfrac{2}{3}) \right)\\ &\sum_{n=1}^\infty \frac{1}{n^6\,\binom{2n}n} = \;\;?\\ &\sum_{n=1}^\infty \frac{1}{n^7\,\binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot 3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6,\tfrac{2}{3})\right)\\ \end{aligned}

with the Riemann zeta function $\zeta(s)$ and the more general Hurwitz zeta function $\zeta(s,a)$,

\begin{aligned} &\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\\&\zeta(s,a) = \sum_{n=0}^\infty \frac{1}{(n+a)^s} \end{aligned}

respectively.  (Note that for $a = 1$, the Hurwitz reduces into the Riemann.)  The expression for p = 5 in the paper here used Dirichlet L-functions, but a poster from mathstackexchange gave it in terms of the Hurwitz zeta.  The one for p = 7 is from Mathworld’s article on central binomial coefficients.

However, none are known for p > 7 (as well as p = 6).  Based on odd p, it is easy to assume that the next has the form,

\begin{aligned} & a_0\sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n} = a_1\,\zeta(9) +a_2\,\zeta(2)\zeta(7)+a_3\, \zeta(3)\zeta(6)+a_4\, \zeta(4)\zeta(5)+a_5\,\pi\sqrt{3}\,H_8\\ \end{aligned}

where,

$H_8 = \left(\zeta(8,\tfrac{1}{3})-\zeta(8,\tfrac{2}{3})\right)$

and the six $a_i$ are integers.  One can use Mathematica’s LatticeReduce function (which employs an integer relations algorithm) to find them, if any exists. Unfortunately, it didn’t find any exact relation, nor for analogous forms for prime p = 11 or 13.  Either my old version of Mathematica is just not powerful enough, or odd p > 7 do not have analogous forms to the ones above.

Can you find the next in the family?

### Zudilin’s continued fraction for Zeta(4)

Euler proved the following general continued fraction formula,

$\frac{1}{u_1}+\frac{1}{u_2}+\frac{1}{u_3} = \cfrac{1}{u_1 - \cfrac{u_1^2}{u_1+u_2 - \cfrac{u_2^2}{u_2+u_3 - \cfrac{u_3^2}{u_3+u_4 - \ddots}}}}$

which automatically gives a representation for the Riemann zeta function $\zeta(s)$.  However, the convergence is rather slow.  Apery found a much faster version for $\zeta(3)$ and proved its irrationality.  The status for other odd s, however, remains open.  While it has already been proved irrational for all even s, Wadim Zudilin nonetheless found an interesting faster version for $\zeta(4)$.

First, consider the following known binomial sums,

$\sum_{n=1}^\infty \frac{1}{n\,\binom{2n}n} = \frac{1}{3\sqrt{3}}\,\pi$

$\sum_{n=1}^\infty \frac{1}{n^2\,\binom{2n}n} = \frac{1}{3}\,\zeta(2)$

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3\,\binom{2n}n} = \frac{2}{5}\,\zeta(3)$

$\sum_{n=1}^\infty \frac{1}{n^4\,\binom{2n}n} = \frac{17}{36}\,\zeta(4)$

This nice pattern stops at s = 4.  (There are sums for other s, but they do not have this succinct form.)  It is suggestive then that $\zeta(4)$ belongs to the same family and may share certain similarities.

Zudilim found that,

$\zeta(4) = \cfrac{13}{p_0 + \cfrac{1^8\, q_1}{p_1+ \cfrac{2^8\,q_2}{p_2+\cfrac{3^8\,q_3}{p_3 + \ddots}}}}$

where, starting with n = 0,

$p_n = 3(2n+1)(3n^2+3n+1)(15n^2+15n+4) = 12, 2142, 26790, 142968,\dots$

$q_n = 3(3n-1)(3n+1) = -3, 24, 105, 240, 429\dots$

(Note that the cfrac does not use $q_0$.)  Like Apery’s accelerated version for $\zeta(2)$, the partial denominators use addition, in contrast to Euler’s form which has subtraction.  Also, like its siblings, it obeys a recurrence relation,

$(n+1)^5U_{n+1}=p_n U_n+n^3q_nU_{n-1}$

Starting with $U_0=1, U_1=12$, one gets the sequence,

$U_n = 1, 12, 804, 88680, \dots$

which is not yet in the OEIS.  In Theorem 4, Zudilin mentions they are positive rationals, but some Mathematica experimentation will show that apparently all the $U_i$ are integral.  (Proof anyone?)  Furthermore, they have the limit,

$\lim_{n \to \infty} \frac{U_{n+1}}{U_n} = (3+2\sqrt{3})^3$

For more details, refer to Zudilin’s An Apery-like difference equation for Catalan’s constant.  Now if only someone will do something more about $\zeta(5)$