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The Gould sequence A001316 is defined as follows:

$$G(n)=\sum _{i=0}^n \binom{n}{i} \bmod 2$$

However, according to Wikipedia, it is also equal to the numerator of the ratio $\frac{2^n}{n!}$ in lowest terms, which leads to the following expression for $G(n)$:

$$G(n)=\frac{2^n}{gcd(2^n,n!)}$$

By evaluating the first values of $n$ for both formulations ($0\leq n\leq 15$), they seem to be equal:

{{1, 1}, {2, 2}, {2, 2}, {4, 4}, {2, 2}, {4, 4}, {4, 4}, {8, 8}, {2, 2}, {4, 4}, {4, 4}, {8, 8}, {4, 4}, {8, 8}, {8, 8}, {16, 16}}

So, is the last expression for $G(n)$ valid to represent the sequence? Is there any way to prove an equivalence between both formulations?

Cardstdani
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1 Answers1

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The $G(n)$ is essentially counting how many of the binomial coefficients $\binom{n}{i}$ are odd. But these are the coefficients of $(1+x)^n$, so we just need to look at non-zero coefficients of $(1+x)^n\bmod 2$. Let $n=\sum_i 2^{a_i}$ be $n$'s binary expansion, then $$ (1+x)^n=\prod_i(1+x)^{2^{a_i}}. $$ Now prove by induction that $(1+x)^{2^k}\equiv 1+x^{2^k}\pmod 2$ for $k\geq 0$. Hence $$ (1+x)^n\equiv \prod_i(1+x^{2^{a_i}})=\sum_s x^s\mod 2 $$ where the last sum goes over all integers whose binary $1$s are subsets of those of $n$. So, if $n$ has $k$ binary $1$'s, then $(1+x)^n \bmod 2$ has exactly $2^k$ non-zero terms, i.e. $G(n)=2^k$. Here $k$ is also called Hamming weight of $n$, say $k=H(n)$.

Your other formula can be written as $G(n)=2^{n-v_2(n!)}$, so all that remains to be proven is that number of $1$'s in binary expansion of $n$ is $n-v_2(n!)$. See for example Is there an equation for the amount of ones in a binary number? or Number of $1$'s in a binary number.

So overall $$ G(n)=2^{H(n)}=2^{n-v_2(n!)}. $$

Sil
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