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Good morning. I was motivated by a question I answered here. I wanted to find a closed form for the continued fraction

$$[2;4,6,...,\overline{2n}]=2+\frac{1}{4+\frac{1}{6+...\frac{1}{2n+\frac{1}{2n+...}}}}$$

It makes sense first to find the closed form of the repeating part; i.e., the form of $[2n;\overline{2n}]$. Let $x_n=[2n;\overline{2n}]$.

$$x_n=2n+\frac{1}{2n+\frac{1}{2n+...}}\quad \Rightarrow \quad x_n=2n+\frac{1}{x_n}\quad \Rightarrow \quad x_n^2-2nx-1=0$$

Thus, $x_n=n+\sqrt{n^2+1}$

Now representing the c.f. as $y_n=[2;4,6,...,\overline{2n}]=[2;4,6,...,2(n-1),x_n]$, I started working backwards and trying to represent as a simple rational function of variable $x_n$. So far, looking at small cases, we have that

\begin{eqnarray*}y_0&=&x_0&=&\frac{x_0+0}{0x_0+1}\\y_1&=&2+\frac{1}{x_1}&=&\frac{2x_1+1}{x_1+0}\\y_2&=&2+\frac{1}{4+\frac{1}{x_2}}&=&\frac{9x_2+2}{4x_2+1}\\y_3&=&2+\frac{1}{4+\frac{1}{6+\frac{1}{x_3}}}&=&\frac{56x_3+9}{25x_3+4} \end{eqnarray*}

So I noticed that if I represent

$$y_n=\frac{a_nx_n+b_n}{c_nx_n+d_n} \qquad T_n=\begin{pmatrix}a_n & b_n\\c_n & d_n\end{pmatrix} \qquad \vec{x}=\begin{pmatrix}x_n\\1\end{pmatrix}$$

then we can define an operation such that

$$y_n=T_n\vec{x}=\frac{a_nx_n+b_n}{c_nx_n+d_n}$$

Then, if we let

$$A_n=\begin{pmatrix}2n & 1\\1 & 0\end{pmatrix}$$

we have the following.

\begin{eqnarray*}y_1&=&A_1\vec{x_1}\\y_2&=&(A_1A_2)\vec{x_2}\\y_3&=&(A_1A_2A_3)\vec{x_3}\\...\\y_n&=&\left(\prod_{k=1}^nA_k\right)\vec{x_n} \end{eqnarray*}

Certainly from here, using a CAS, it would be easy to determine the actual value given a certain $n$. First of all, is this correct? My calculations seem right. For a true closed form, though, should I try and solve for the individual $a_i$ to $d_i$? Are there any easier approaches as well?

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    $\begingroup$ he asked the same question a few hours ago, I guess restricting to $2n+1$ prime .... math.stackexchange.com/questions/3302699/… $\endgroup$ – Will Jagy Jul 24 at 19:09
  • $\begingroup$ Interesting. I started pursuing my question yesterday on and off after answering his. Thanks for the link! $\endgroup$ – Eleven-Eleven Jul 24 at 19:14
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    $\begingroup$ Great minds think alike $\endgroup$ – Antonio Hernandez Maquivar Jul 24 at 21:05
  • $\begingroup$ Do you have any additional computed values for $a_{n}.$ You could cross-reference the sequence $\{a_n\}$ to Sloan's database. So for instance $a_n:=0,2,9,56,\ldots$ Similarly $c_n:=0,1,4,25,\ldots$ $\endgroup$ – Antonio Hernandez Maquivar Jul 24 at 21:37
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    $\begingroup$ oeis.org/A036242 is one of them $\endgroup$ – Will Jagy Jul 25 at 20:03
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This is a partial answer, some information is provided for potential answerers that need some numeric support. I cannot see a path to a closed formula. There is a small mathematical part, that puts order in the structure, a non-linear recursion relation of order two for the coefficients is extracted, and the asymmptotic is evaluated. There is also a small computer aided part with explicit data.


An invertible $2\times 2$ matrix $T$ with entries $a,b,c,d$ induces a momographic / Möbius transformation / action on $\Bbb C$ without pole given by $$ \underbrace{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} }_{=T}z :=\frac {az+b}{cz+d}\ . $$ Personally, i prefer to write this action as $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \left[\frac z1\right] := \left[\frac {az+b}{cz+d}\right]\ , $$ so that the action can be simpler seen with bare eyes, knowing that $2\times 2$ matrices act on $2\times 1$ vectors.

(The question was implicitly asking, if a formula has action character, yes.)

From this, we have using $$ \begin{aligned} A(k) &= \begin{bmatrix} k & 1\\ 1 & 0 \end{bmatrix} \ ,\\[2mm] T(n) &= A(2)\; A(4)\; \dots\; A(2n) \\ &= \begin{bmatrix} 2 & 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 4 & 1\\ 1 & 0 \end{bmatrix} \dots \begin{bmatrix} 2n & 1\\ 1 & 0 \end{bmatrix} \\ &= \begin{bmatrix} t_n & t_{n-1}\\ u_n & u_{n-1} \end{bmatrix} \ , \end{aligned} $$ where the last form can be shown inductively.

The sequences $(t_n)$, $(u_n)$ are satisfying the parallel relations $$ \begin{aligned} t_n &= 2n \;t_{n-1} + t_{n-2}\ , \qquad & t_0 &=1\ , & t_{-1} &= 0\ ,\\ u_n &= 2n \; u_{n-1} + u_{n-2}\ , \qquad & u_0 &=0\ , & t_{-1} &= 1\ . \end{aligned} $$ These follow from $$ \begin{bmatrix} t_n & t_{n-1}\\ u_n & u_{n-1} \end{bmatrix} = \begin{bmatrix} t_{n-1} & t_{n-2} \\ u_{n-1} & u_{n-2} \end{bmatrix} \begin{bmatrix} 2n & 1\\ 1 & 0 \end{bmatrix} \ ,\ %T_1 = \begin{bmatrix} %2 & 1\\ 1 & 0 %\end{bmatrix}\ , %\ T_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\ . $$ An easy lower estimation is found by "forgetting the $t_{n-2}$ in the recursion", so we get for instance $t_n\ge (2n)!!=2^n\cdot n!$. So we have a rapid increasing sequence.


I could not find a closed form for the sequences $(t_n)$, $(u_n)$, appearing as entries in $(T(n))_{n\ge 0}$. And i also could not use the ideas of solving the differential equation $$ y'' = 2x\;y' + y $$ to get a "discrete idea" to proceed in our case. Note that the above differential equation can be used to get an asymptotic of $t(n)$. Using sage again for a quick typing:

sage: x = var('x'); y = function('y')(x);
....: desolve(diff(y,x,x) - 2*x*diff(y,x) - y, y, contrib_ode=True)
....: 
[y(x) == (1/2)^(1/4)
        *(-x^2)^(1/4)*_K2*x*bessel_I(-1/4, -1/2*x^2)
        *e^(1/2*x^2)/((-1/2*I*x^2)^(1/4)*sqrt(-x)) 
        +
        _K1*x*bessel_Y(-1/4, -1/2*I*x^2)*e^(1/2*x^2)/sqrt(-x)]

sage: latex(_)

(Result y(x) = ... was rearrange manually.) $$ \left[y\left(x\right) = \frac{\left(\frac{1}{2}\right)^{\frac{1}{4}} \left(-x^{2}\right)^{\frac{1}{4}} K_{2} x I_{-\frac{1}{4}}(-\frac{1}{2} \, x^{2}) e^{\left(\frac{1}{2} \, x^{2}\right)}}{\left(-\frac{1}{2} i \, x^{2}\right)^{\frac{1}{4}} \sqrt{-x}} + \frac{K_{1} x Y_{-\frac{1}{4}}(-\frac{1}{2} i \, x^{2}) e^{\left(\frac{1}{2} \, x^{2}\right)}}{\sqrt{-x}}\right] $$

Discretely, i am expecting a similar complexity. (As a human, i am completely ignoring all algebraic operations, and look at the word Bessel, this is a complicate function, research projects may be started if some starting idea is present. Else...)

I gave up also because this is only the first part in the computations of some "(real) surds" of a special shape that have a period of length one when written as continued fractions.


Here is a good place to insert some numerical data. We use sage. Please ignore this part if it is disturbing, some comments wanted explicit numerical data, here it is.

sage: def A(k):    return matrix(ZZ, 2, 2, [k,1, 1,0])
sage: def T(n): return prod( [ A(k) for k in range(2,n+2,2) ] )
sage: T(2)
[2 1]
[1 0]
sage: T(4)
[9 2]
[4 1]
sage: T(6)
[56  9]
[25  4]
sage: T(8)
[457  56]
[204  25]
sage: def t(n):
....:     if n == 0: return 1
....:     if n == -1: return 0
....:     return 2*n*t(n-1) + t(n-2)
....: 
sage: [ t(n) for n in [0..10] ]
[1, 2, 9, 56, 457, 4626, 55969, 788192, 12667041, 228794930, 4588565641]
sage: [ 2^n * factorial(n) for n in [0..10] ]
[1, 2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200]
sage: def u(n):
....:     if n == 0: return 0
....:     if n == -1: return 1
....:     return 2*n*u(n-1) + u(n-2)
....: 
sage: [ u(n) for n in [0..10] ]
[0, 1, 4, 25, 204, 2065, 24984, 351841, 5654440, 102131761, 2048289660]

We can also check the continued fractions representations (easily and structurally, so let us do this). Again, please ignore if a theoretical interest predominates.

sage: def showme(n):
....:     K.<a> = QuadraticField((n+2)^2+1)
....:     print "n = %s and a satisfies a^2 = %s" % (n, a^2)
....:     M = T(n) * matrix( 2,1, [ n+2+a, 1] )
....:     ML = M.list()
....:     ML0, ML1 = ML
....:     print "Consider val = ( %s ) divided by ( %s )" % (ML0, ML1)
....:     val = ML0/ML1
....:     print( "val = %s\nhas the continued fraction %s\n"
....:            % (val, continued_fraction(val) ) )
....:     
sage: for n in range(2, 18, 2):    showme(n)
n = 2 and a satisfies a^2 = 17
Consider val = ( 2*a + 9 ) divided by ( a + 4 )
val = a - 2
has the continued fraction [2; (8)*]

n = 4 and a satisfies a^2 = 37
Consider val = ( 9*a + 56 ) divided by ( 4*a + 25 )
val = 1/33*a + 68/33
has the continued fraction [2; 4, (12)*]

n = 6 and a satisfies a^2 = 65
Consider val = ( 56*a + 457 ) divided by ( 25*a + 204 )
val = -1/991*a + 2228/991
has the continued fraction [2; 4, 6, (16)*]

n = 8 and a satisfies a^2 = 101
Consider val = ( 457*a + 4626 ) divided by ( 204*a + 2065 )
val = 1/61009*a + 136662/61009
has the continued fraction [2; 4, 6, 8, (20)*]

n = 10 and a satisfies a^2 = 145
Consider val = ( 4626*a + 55969 ) divided by ( 2065*a + 24984 )
val = -1/5887631*a + 13189446/5887631
has the continued fraction [2; 4, 6, 8, 10, (24)*]

n = 12 and a satisfies a^2 = 197
Consider val = ( 55969*a + 788192 ) divided by ( 24984*a + 351841 )
val = 1/824638849*a + 1847350760/824638849
has the continued fraction [2; 4, 6, 8, 10, 12, (28)*]

n = 14 and a satisfies a^2 = 257
Consider val = ( 788192*a + 12667041 ) divided by ( 351841*a + 5654440 )
val = -1/158124768383*a + 354230113736/158124768383
has the continued fraction [2; 4, 6, 8, 10, 12, 14, (32)*]

n = 16 and a satisfies a^2 = 325
Consider val = ( 12667041*a + 228794930 ) divided by ( 5654440*a + 102131761 )
val = 1/39771798041121*a + 89096532358730/39771798041121
has the continued fraction [2; 4, 6, 8, 10, 12, 14, 16, (36)*]

sage: 

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