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Max Alekseyev
Max Alekseyev
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The answer is Yes.

The generating function for $t_n$ is $$\sum_{n\geq 0} t_n x^n = \frac14\big(\frac{1+x}{\sqrt{1-6x+x^2}}-1\big).$$ Correspondingly, $$\sum_{n\geq 0} t_n x^n \equiv \frac{1+x}{\sqrt{1+x^2}}-1 \pmod{3}.$$ It follows that for $n>0$, $$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv \binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$$$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv (-1)^{\lfloor n/2\rfloor}\binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$ from where Lucas' theorem gives the desired result.

The answer is Yes.

The generating function for $t_n$ is $$\sum_{n\geq 0} t_n x^n = \frac14\big(\frac{1+x}{\sqrt{1-6x+x^2}}-1\big).$$ Correspondingly, $$\sum_{n\geq 0} t_n x^n \equiv \frac{1+x}{\sqrt{1+x^2}}-1 \pmod{3}.$$ It follows that for $n>0$, $$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv \binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$ from where Lucas' theorem gives the desired result.

The answer is Yes.

The generating function for $t_n$ is $$\sum_{n\geq 0} t_n x^n = \frac14\big(\frac{1+x}{\sqrt{1-6x+x^2}}-1\big).$$ Correspondingly, $$\sum_{n\geq 0} t_n x^n \equiv \frac{1+x}{\sqrt{1+x^2}}-1 \pmod{3}.$$ It follows that for $n>0$, $$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv (-1)^{\lfloor n/2\rfloor}\binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$ from where Lucas' theorem gives the desired result.

Source Link
Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

The answer is Yes.

The generating function for $t_n$ is $$\sum_{n\geq 0} t_n x^n = \frac14\big(\frac{1+x}{\sqrt{1-6x+x^2}}-1\big).$$ Correspondingly, $$\sum_{n\geq 0} t_n x^n \equiv \frac{1+x}{\sqrt{1+x^2}}-1 \pmod{3}.$$ It follows that for $n>0$, $$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv \binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$ from where Lucas' theorem gives the desired result.