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T. Amdeberhan
T. Amdeberhan
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The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity of $n$. To address Cigler's request, here is an illustration for the case even: $(2n)a_{2n}=3(4n-1)a_{2n-1}-(2n-1)a_{2n-2}$ and by induction assumption $$\nu_3(3(4n-1)a_{2n-1})=\nu_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu_3\left((2n-1)a_{2n-2})=\nu_3((2n-1)\binom{2n-2}{n-1}\right).$$ Hence$$\nu_3(3(4n-1)a_{2n-1})=\nu_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu_3((2n-1)a_{2n-2})=\nu_3\left((2n-1)\binom{2n-2}{n-1}\right).$$ So, $\nu_3(3(4n-1)a_{2n-1})>\nu_3((2n-1)a_{2n-2})$ and $\nu_3((2n)a_{2n})=\nu_3((2n-1)\binom{2n-2}{n-1})$; or, \begin{align} \nu_3(a_{2n})&=\nu_3\left(\frac{(2n-1)}{2n}\binom{2n-2}{n-1}\right) =\nu_3\left(\frac14\binom{2n}n\right)=\nu_3\left(\binom{2n}n\right) \end{align} as desired.

The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity of $n$. To address Cigler's request, here is an illustration for the case even: $(2n)a_{2n}=3(4n-1)a_{2n-1}-(2n-1)a_{2n-2}$ and by induction assumption $$\nu_3(3(4n-1)a_{2n-1})=\nu_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu_3\left((2n-1)a_{2n-2})=\nu_3((2n-1)\binom{2n-2}{n-1}\right).$$ Hence $\nu_3((2n)a_{2n})=\nu_3((2n-1)\binom{2n-2}{n-1})$ or \begin{align} \nu_3(a_{2n})&=\nu_3\left(\frac{(2n-1)}{2n}\binom{2n-2}{n-1}\right) =\nu_3\left(\frac14\binom{2n}n\right)=\nu_3\left(\binom{2n}n\right) \end{align} as desired.

The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity of $n$. To address Cigler's request, here is an illustration for the case even: $(2n)a_{2n}=3(4n-1)a_{2n-1}-(2n-1)a_{2n-2}$ and by induction assumption $$\nu_3(3(4n-1)a_{2n-1})=\nu_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu_3((2n-1)a_{2n-2})=\nu_3\left((2n-1)\binom{2n-2}{n-1}\right).$$ So, $\nu_3(3(4n-1)a_{2n-1})>\nu_3((2n-1)a_{2n-2})$ and $\nu_3((2n)a_{2n})=\nu_3((2n-1)\binom{2n-2}{n-1})$; or, \begin{align} \nu_3(a_{2n})&=\nu_3\left(\frac{(2n-1)}{2n}\binom{2n-2}{n-1}\right) =\nu_3\left(\frac14\binom{2n}n\right)=\nu_3\left(\binom{2n}n\right) \end{align} as desired.

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T. Amdeberhan
T. Amdeberhan
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The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity of $n$. To address Cigler's request, here is an illustration for the case even: $(2n)a_{2n}=3(4n-1)a_{2n-1}-(2n-1)a_{2n-2}$ and by induction assumption $$\nu_3(3(4n-1)a_{2n-1})=\nu_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu_3\left((2n-1)a_{2n-2})=\nu_3((2n-1)\binom{2n-2}{n-1}\right).$$ Hence $\nu_3((2n)a_{2n})=\nu_3((2n-1)\binom{2n-2}{n-1})$ or \begin{align} \nu_3(a_{2n})&=\nu_3\left(\frac{(2n-1)}{2n}\binom{2n-2}{n-1}\right) =\nu_3\left(\frac14\binom{2n}n\right)=\nu_3\left(\binom{2n}n\right) \end{align} as desired.

The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity $n$.

The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity of $n$. To address Cigler's request, here is an illustration for the case even: $(2n)a_{2n}=3(4n-1)a_{2n-1}-(2n-1)a_{2n-2}$ and by induction assumption $$\nu_3(3(4n-1)a_{2n-1})=\nu_3\left(9(4n-1)(2n-1)\binom{2n-2}{n-1}\right) \qquad\text{and}\qquad \nu_3\left((2n-1)a_{2n-2})=\nu_3((2n-1)\binom{2n-2}{n-1}\right).$$ Hence $\nu_3((2n)a_{2n})=\nu_3((2n-1)\binom{2n-2}{n-1})$ or \begin{align} \nu_3(a_{2n})&=\nu_3\left(\frac{(2n-1)}{2n}\binom{2n-2}{n-1}\right) =\nu_3\left(\frac14\binom{2n}n\right)=\nu_3\left(\binom{2n}n\right) \end{align} as desired.

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T. Amdeberhan
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

The $3$-adic evaluation you seek is compactly given by $$\nu_3(a_{2n})=\nu_3\left(\binom{2n}n\right) \qquad \text{and} \qquad \nu_3(a_{2n+1})=\nu_3\left(3(2n+1)\binom{2n}n\right),$$ which can be proved inductively using the well-known recurrence $$na_n=3(2n-1)a_{n-1}-(n-1)a_{n-2}$$ according to the parity $n$.