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Jose Arnaldo Bebita
Jose Arnaldo Bebita
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Here is a quick way to further refine the improved lower bound for $I(n^2)$:

Write $$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg).$$ Now use, for instance, $$q^{k+1} - \frac{1}{q^2} > q^{k+1} - 1$$ to obtain $$I(n^2)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - \frac{1}{q^2}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\Bigg(1 + \frac{q^2}{q^{k+3} - 1}\Bigg).$$

Note that $$\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{q^2}{q^{k+3} - 1}\bigg) - \bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\frac{2(q-1)}{q^{k+2} (q^{k+3} - 1)}>0$$ since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$.

In fact, this method shows that there are infinitely many ways to refine the improved lower bound for $I(n^2)$. $$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}+1}{q^{k+1}}\bigg).$$

It remains to be seen whether there is a refined (improved) lower bound that is independent of $k$ (and therefore expressed entirely in terms of $q$).

Here is a quick way to further refine the improved lower bound for $I(n^2)$:

Write $$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg).$$ Now use, for instance, $$q^{k+1} - \frac{1}{q^2} > q^{k+1} - 1$$ to obtain $$I(n^2)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - \frac{1}{q^2}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\Bigg(1 + \frac{q^2}{q^{k+3} - 1}\Bigg).$$

Note that $$\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{q^2}{q^{k+3} - 1}\bigg) - \bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\frac{2(q-1)}{q^{k+2} (q^{k+3} - 1)}>0$$ since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$.

In fact, this method shows that there are infinitely many ways to refine the improved lower bound for $I(n^2)$.

It remains to be seen whether there is a refined (improved) lower bound that is independent of $k$ (and therefore expressed entirely in terms of $q$).

Here is a quick way to further refine the improved lower bound for $I(n^2)$:

Write $$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg).$$ Now use, for instance, $$q^{k+1} - \frac{1}{q^2} > q^{k+1} - 1$$ to obtain $$I(n^2)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - \frac{1}{q^2}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\Bigg(1 + \frac{q^2}{q^{k+3} - 1}\Bigg).$$

Note that $$\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{q^2}{q^{k+3} - 1}\bigg) - \bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\frac{2(q-1)}{q^{k+2} (q^{k+3} - 1)}>0$$ since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$.

In fact, this method shows that there are infinitely many ways to refine the improved lower bound $$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}+1}{q^{k+1}}\bigg).$$

It remains to be seen whether there is a refined (improved) lower bound that is independent of $k$ (and therefore expressed entirely in terms of $q$).

Source Link
Jose Arnaldo Bebita
Jose Arnaldo Bebita
  • 982
  • 2
  • 15
  • 34

Here is a quick way to further refine the improved lower bound for $I(n^2)$:

Write $$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg).$$ Now use, for instance, $$q^{k+1} - \frac{1}{q^2} > q^{k+1} - 1$$ to obtain $$I(n^2)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - \frac{1}{q^2}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\Bigg(1 + \frac{q^2}{q^{k+3} - 1}\Bigg).$$

Note that $$\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{q^2}{q^{k+3} - 1}\bigg) - \bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\frac{2(q-1)}{q^{k+2} (q^{k+3} - 1)}>0$$ since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$.

In fact, this method shows that there are infinitely many ways to refine the improved lower bound for $I(n^2)$.

It remains to be seen whether there is a refined (improved) lower bound that is independent of $k$ (and therefore expressed entirely in terms of $q$).