## Archive for May, 2012

### Borwein and Bradley’s Apery-like formulas for zeta(4n+3)

Apery gave,

\begin{aligned} \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom {2k}k}\end{aligned}

J. Borwein and D. Bradley found this can be generalized to $\zeta(4n+3)$. Define the functions,

\begin{aligned} &B(a_0)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\\ &B(a_0,a_1,a_2,\dots)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\; \sum_{p=1}^{k-1} \frac{1}{p^{a_1}}\;\sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\;\dots \end{aligned}

then,

\begin{aligned} \frac{2}{5}\,\zeta(3) &= B(3)\\ \frac{2}{5}\,\zeta(7) &= B(7)+5B(3,4)\\ \frac{2}{5}\,\zeta(11) &= B(11)+5B(7,4)-\frac{15}{2}B(3,8)+\frac{25}{2}B(3,4,4)\\ \frac{2}{5}\,\zeta(15) &= B(15)+5B(11,4)-\frac{15}{2}B(7,8) +\frac{25}{2}B(7,4,4)+\frac{130}{6}B(3,12)\\&-\frac{225}{6}B(3,8,4)+ \frac{125}{6}B(3,4,4,4) \end{aligned}

and so on.  Beautiful, aren’t they? Notice that all the $a_i$ (excepting $a_0$) are all divisible by 4. This infinite family has a generating function. Let z $\not=$ non-zero integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^3(1-z^4/k^4)}&=\frac{5}{2}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^3\;\binom {2k}k} \frac{1}{1-z^4/k^4}\prod_{j=1}^{k-1}\frac{1+4z^4/j^4}{1-z^4/j^4}\end{aligned}

On the other hand, for s = 4n+1,

\begin{aligned} 2\,\zeta(5) &= 4B(5)-5B(3,2)\\[2.5mm] 4\,\zeta(9) &= 9B(9)-5B(7,2)+20B(5,4)+45B(3,6)-25B(3,4,2)\\[2.5mm] 12\,\zeta(13) &= 28B(13)-10B(11,2)+90B(9,4)\\&+90B(7,6)-50B(7,4,2)-60B(5,8)+100B(5,4,4)\\&-310B(3,10)+75B(3,8,2)+450B(3,6,4)-125B(3,4,4,2)\end{aligned}

with this version for $\zeta(13)$ found by Jim Cullen.  There are various versions for both s = 4n+1 and 4n+3.  For example, for $\zeta(7)$, we have the relations,

\begin{aligned} 4\zeta(7) &= 8B(7,0)\,-\,5B(3,4)\,-\,8B(5,2)+5B(3,2,2)\\ 0 &= -2B(7,0) - 55B(3,4)-8B(5,2)+5B(3,2,2)\end{aligned}

Eliminating the last two terms will yield the shorter relation given by Borwein and Bradley. There is a generating function for all s = 2n+1, but none is known that is only for s = 4n+1. See Apery-Like Formulae for $\zeta(4n+3)$ for more details.

### Sequences 3, Fibonacci’s rabbits and Narayana’s cows

(Under construction)

### 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.

### Odd powers of Fibonacci numbers

The Fibonacci numbers $F_n$,

$F_n = 0, 1, 1, 2, 3, 5, 8, 13, \dots$

obey the following recurrence relations,

\begin{aligned}&F_n-F_{n-1}-F_{n-2} = 0\\[1.5mm]&F_n^2-2F_{n-1}^2-2F_{n-2}^2+F_{n-3}^2 = 0\\[1.5mm]&F_n^3-3F_{n-1}^3-6F_{n-2}^3+3F_{n-3}^3+F_{n-4}^3 = 0\\[1.5mm]&F_n^4-5F_{n-1}^4-15F_{n-2}^4+15F_{n-3}^4+5F_{n-4}^4-F_{n-5}^4 = 0\\[1.5mm]&F_n^5-8F_{n-1}^5-40F_{n-2}^5+60F_{n-3}^5+40F_{n-4}^5-8F_{n-5}^5-F_{n-6}^5 = 0\end{aligned}

and so on.  As a number triangle, the coefficients are,

\begin{aligned} &1;\; \bold{1, -1,-1}=0\\ &2;\; 1, -2, -2, \;1=0\\ &3;\; \bold{1, -3, -6, \;3, \;1}=0\\ &4;\; 1, -5, -15, 15, \;5, \;-1=0\\ &5;\; \bold{1, -8, -40, 60, 40, -8, -1}=0 \end{aligned}

See Ron Knott’s article on the fibonomials, so-called since the above is reminiscent of the binomial triangle.  However, I found another set of recurrence relations can be given as,

\begin{aligned} &F_{n-1}^2-F_{n+1}^2 = -F_{2n}\\[1.5mm] &F_{n-1}^3-F_{n}^3-F_{n+1}^3 = -F_{3n}\\[1.5mm] &F_{n-2}^4+3F_{n-1}^4-3F_{n+1}^4-F_{n+2}^4 = -6F_{4n}\\[1.5mm] &F_{n-2}^5-3F_{n-1}^5-6F_{n}^5+3F_{n+1}^5+F_{n+2}^5 = 6F_{5n}\\[1.5mm] &F_{n-3}^6+4F_{n-2}^6-20F_{n-1}^6+20F_{n+1}^6-4F_{n+2}^6-F_{n+3}^6 = -120F_{6n}\\[1.5mm] &F_{n-3}^7-8F_{n-2}^7-40F_{n-1}^7+60F_{n}^7+40F_{n+1}^7-8F_{n+2}^7-F_{n+3}^7 = -240F_{7n}\\[1.5mm]\end{aligned}

etc.  As a number triangle,

\begin{aligned} &2;\; 1, -1 =-1\\ &3;\; \bold{1, -1,-1}=-1\\ &4;\; 1, \;\;\;3, -3, \;-1=-6\\ &5;\; \bold{1, -3, -6, \;3, \;\;1}\;=\;6\\ &6;\; 1, \;\;\;4, -20, 20, -4, -1=-120\\ &7;\; \bold{1, -8, -40, 60, 40, -8, -1}=-240 \end{aligned}

Compare the two triangles.  Notice how, for odd powers, the same coefficients appear, though moved up by one odd power.  I have no explanation for the phenomenon, other than the fact that I’ve seen several instances already of a “recycled” polynomial appearing in many contexts.

### 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?