## Archive for the ‘sequences’ Category

### Sequences 2, Padovan and Perrin numbers

III. Padovan sequence

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}

which start with index n = 0.  Hence, $P = \frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{0}{3}$ $P^2 =\frac{1}{9}(a^2+b^2)+\frac{0}{3}(a+b)+\frac{2}{3}$ $P^3 =\frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{3}{3}$

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

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### 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 $\zeta(5)$

### The silver ratio and a continued fraction for log(2)

Define the three sequences, $C_n = \sum_{k=0}^n {\binom n k} {\binom {n+k}k} = 1, 3, 13, 63, 321,\dots$ $B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251,\dots$ $A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, \dots$

The last two are Apery numbers and have been discussed previously. The first are the central Delannoy numbers which obeys the limit, $\lim_{n \to \infty} \frac{C_{n+1}}{C_n} = \left(1+\sqrt{2}\right)^2 = 5.8284\dots$

which is the square of the silver ratio.  (The ratios for the others have already been mentioned.)  These have the recurrence relations, $n C_n = 3(2n-1)C_{n-1}-(n-1)C_{n-2}$ $n^2 B_n = (11n^2-11n+3)B_{n-1}+(n-1)^2B_{n-2}$ $n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2}$

To recall, the polynomials $P(n) =11n^2-11n+3$ and $P(n) = 34n^3-51n^2+27n-5$  generated numbers for the continued fractions of $\zeta(2), \zeta(3)$, so I was curious if $P(n) = 3(2n-1)$ could be used in an analogous manner.  It turns out, depending on what sign to use, it gives either $\log(2)$ or $\arctan(\frac{1}{3})$, $\frac{1}{2}\log(2) = \cfrac{1}{v_1 - \cfrac{1^2}{v_2 - \cfrac{2^2}{v_3 - \cfrac{3^2}{v_4-\ddots}}}}$

or, $\arctan(\frac{1}{3}) = \cfrac{1}{v_1 + \cfrac{1^2}{v_2 + \cfrac{2^2}{v_3 + \cfrac{3^2}{v_4 + \ddots}}}}$

where, starting with n = 1, $v_n = 3(2n-1)$

This can be partly demystified since one continued fraction for the natural logarithm, and arctan, is, $\frac{1}{2}\log\left(1+\frac{2x}{y}\right)= \cfrac{x}{1(x+y) - \cfrac{(1x)^2}{3(x+y) - \cfrac{(2x)^2}{5(x+y) - \cfrac{(3x)^2}{7(x+y) - \ddots}}}}$

and, $\arctan(\frac{x}{z}) = \cfrac{x}{1z + \cfrac{(1x)^2}{3z + \cfrac{(2x)^2}{5z + \cfrac{(3x)^2}{7z + \ddots}}}}$

Hence, if $x+y = z$, as with the case {x, y, z} = {1, 2, 3}, then there will be identical-looking continued fractions that differ only in the signs.  But it remains interesting how the recurrence relations of these three binomial sums are involved in the continued fractions of $\log(2), \zeta(2), \zeta(3)$.  Later, we shall see there is a recurrence relation for the cfrac of $\zeta(4)$ as well.