Series with only positive integer coefficients

Question

I want to show that the coefficients of the series $$ f(x) = \sqrt[4]{\frac{1+4x}{1-4x}} = 1 + 2x + 2x^2 + 12x^3 + \cdots $$ are only positive integers. Since differentiating $f(x)$ shows $\left(1 - 16x^2\right)f'(x) = 2f(x)$, we can compare the coefficients of $x^{n+1}$ on both sides to get the following recurrence relation for the coefficient $a_n$ of $x^n$ in $f(x)$: $$ (n+2)a_{n+2} = 2a_{n+1} + 16na_n $$ Now it is easy to see by induction that $a_n$ is positive. However, I'm not sure how to approach showing that it is an integer. Is there a way to inductively show that the right-hand side of the recurrence divided by $\left( n+2 \right)$, or can we show that it's an integer in a different way? Any comment or idea would be greatly appreciated.

Answer 1

Define a power series with

$$ ((1+4x)/(1-4x))^{1/4} = 1 + a_1x + a_2x^2 + a_3x^3 \dots. \tag1$$

The question asks for a proof that the coefficients $\,\{a_n\}\,$ are an integer sequence. One proof uses the following theorem.

Suppose that $\,k>0\,$ is an integer and

$$ A(x) := \sum_{n=1}^\infty a_n x^n,\;\; B(x) := \sum_{n=1}^\infty b_n x^n,\;\; 1+k\,A(x) = \sqrt[k]{1+k\,B(k\,x)}. \tag2 $$

Then $\,a_n\,$ for all $\,n>0\,$ is a polynomial in $\,b_n\,$ and $\,\{a_i\}_{i=1}^{n-1}\,$ with integer coefficients. Thus, if all of the $\,\{b_n\}\,$ are integers, then the same holds for $\,\{a_n\}\,$ by induction.

The first few coefficients are

$$ a_1 = b_1,\; a_2 = b_2k - a_1^2{k\choose 2},\; a_3 = b_3k^2 -\Big(2a_1a_2 {k\choose 2} + a_1^3k {k\choose 3} \Big),\\ a_4 = b_4k^3 -\Big((2a_1a_3+a_2^2) {k\choose 2} +3a_1^2a_2k{k\choose 3} + a_1^4k^2{k\choose 4}\Big). \tag3 $$

The proof follows from comparing coefficients of

$$ (1 + k\,A(x))^k = 1 + k\,B(k\,x). \tag4 $$

Apply this theorem to the particular case of

$$ B(x) = (1+x)/(1-x) = 1+2(x+x^2+x^3+\dots) \tag5 $$

with $\,k=2\,$ twice to get the requested proof.

Answer 2

Let $$ B(x)=\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^{\infty}{\binom{2n}{n}x^n}, \quad C(x)=\frac{1-\sqrt{1-4x}}{2x}=\sum_{n=0}^{\infty}{\frac{1}{n+1}\binom{2n}{n}x^n}. $$ Note that $$ \frac{C(x)+C(-x)}{2}=B(x)\left(1-2xC(4x^2)\right)=B(-x)\left(1+2xC(4x^2)\right)=\sqrt{C(4x^2)} \ . $$ Therefore, $$ \begin{split} \sqrt[4]{\frac{1+4x}{1-4x}}&=\sqrt{\frac{B(x)}{B(-x)}}=\sqrt{\frac{1+2xC(4x^2)}{1-2xC(4x^2)}}\\ &=\sqrt{\frac{1+2x}{1-2x}}\circ\left(xC(4x^2)\right)\\ &=\frac{1+2x}{\sqrt{1-4x^2}}\circ\left(xC(4x^2)\right)\\ &=\left((1+2x)B(x^2)\right)\circ\left(xC(4x^2)\right). \end{split} $$ Thus, our series is composition of two series, one with positive integer coefficients, the other of order $1$ with nonnegative integer coefficients, so the composition also has positive integer coefficients.

Note also that $(1+2x)B(x^2)$ is the generating function for A063886 and $xC(4x^2)$ is the generating function for the sequence A151403 interleaved with $0$'s and starting with a $0$.

Update: To look at it another way, we have $$ \begin{split} (1+2x)B(x^2)&=\sqrt{\frac{1+2x}{1-2x}}=\sqrt[4]{\frac{1+4x+4x^2}{1-4x+4x^2}}\\ &=\sqrt[4]{\frac{1+4\frac{x}{1+4x^2}}{1-4\frac{x}{1+4x^2}}}=\sqrt[4]{\frac{1+4x}{1-4x}}\circ\frac{x}{1+4x^2}, \end{split} $$ so $$ \sqrt[4]{\frac{1+4x}{1-4x}}=(1+2x)B(x^2)\circ\left(\frac{x}{1+4x^2}\right)^{\left\langle-1\right\rangle}=\left((1+2x)B(x^2)\right)\circ\left(xC(4x^2)\right), $$ since $$ xC(4x^2)=x(1+4x^2C(4x^2)^2)=x(1+4(xC(4x^2))^2). $$