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I take back what I said in the comments about the bound not shrinking.  I am convinced that one can get much tighter bounds, and that the tighter bounds will depend massively on the quadratic residues for each of the primes involved.

To set the stage, I change notation slightly.  I use $m$ for the argument to $j()$, and I use $n$ for the number of distinct prime factors of $m$.  I also insist $m > 1$.

One of the more accessible results is that $j(m) = j(k)$ if $m$ and $k$ have the same prime factors.  Further if the prime factors are all larger than $n$, then $j(m)= n+1$, so there is a certain uniformity in the analysis and results for a large and simply stated class of numbers.

(Advertisement; I am working on similar statements where $n$ is replaced by something like $\sqrt(n)$.  Email me for more detail.)

Many of the standard bounds for $j(m)$ can be expressed in terms of $n$.  When I posted my comments above, I thought something similar would be true for this version.  However, checking a few examples leads me somewhere completely different.

For $m$ a prime power, $j(m)=2$.  The same holds for the new variation if and only if $-1$ is a qr of the prime dividing $m$.  However, things change when $n>1$. $h(6)=j(6)=4$, but this does not hold for all numbers of the form $2p$. $h(m)$ can vary from 2 to 4 depending on $m$ even if $n$ is restricted to 2.

To make things interesting, I computed $h(385)$, which is bounded above by 4. There are 6 values of a mod 385 to show $j(385)=4$. To get $h(385)=4$, I had to find a square which was 4 mod 7, 9 mod 11, and 4 mod 5. The square of 47 fit the bill, but I could imagine different primes with qrs that would not nicely fit in the set {-3,-2,-1}, so $h(m)$ will not always reach 4 or higher when $n=3$; it will depend on getting qrs which are small negative numbers. For higher $n$, you may not achieve $h(m)>n$ without choosing the $n$ prime factors carefully. And this analysis is using only squarefree $m$; I do not know what happens in the general case.

I am having a few challenges showing upper bounds without having to worry about quadratic residues. This problem is a can of worms I am not ready to tackle. Certainly nothing like the uniform results involving sufficiently large prime factors will hold. One approach involves "tiling" a candidate interval with appropriate primes, and then solving $n$ many quadratic congurences simultaneously. I don't know enough about quadratic residues to see any clean looking results here. It is an interesting variation on which I welcome other viewpoints.

Gerhard "Jacobsthal's Function Is Tough Already" Paseman, 2011.07.15

I take back what I said in the comments about the bound not shrinking.  I am convinced that one can get much tighter bounds, and that the tighter bounds will depend massively on the quadratic residues for each of the primes involved.

To set the stage, I change notation slightly.  I use $m$ for the argument to $j()$, and I use $n$ for the number of distinct prime factors of $m$.  I also insist $m > 1$.

One of the more accessible results is that $j(m) = j(k)$ if $m$ and $k$ have the same prime factors.  Further if the prime factors are all larger than $n$, then $j(m)= n+1$, so there is a certain uniformity in the analysis and results for a large and simply stated class of numbers.

(Advertisement; I am working on similar statements where $n$ is replaced by something like $\sqrt(n)$.  Email me for more detail.)

Many of the standard bounds for $j(m)$ can be expressed in terms of $n$.  When I posted my comments above, I thought something similar would be true for this version.  However, checking a few examples leads me somewhere completely different.

For $m$ a prime power, $j(m)=2$.  The same holds for the new variation if and only if $-1$ is a qr of the prime dividing $m$.  However, things change when $n>1$. $h(6)=j(6)=4$, but this does not hold for all numbers of the form $2p$. $h(m)$ can vary from 2 to 4 depending on $m$ even if $n$ is restricted to 2.

To make things interesting, I computed $h(385)$, which is bounded above by 4. There are 6 values of a mod 385 to show $j(385)=4$. To get $h(385)=4$, I had to find a square which was 4 mod 7, 9 mod 11, and 4 mod 5. The square of 47 fit the bill, but I could imagine different primes with qrs that would not nicely fit in the set {-3,-2,-1}, so $h(m)$ will not always reach 4 or higher when $n=3$; it will depend on getting qrs which are small negative numbers. For higher $n$, you may not achieve $h(m)>n$ without choosing the $n$ prime factors carefully. And this analysis is using only squarefree $m$; I do not know what happens in the general case.

EDIT 2011.07.29 I let the presence of quadratic residues rattle me into thinking that $h(m)$ would depend on the multiplicity of prime factors of $m$. It does not. As is the case for $j(m)$, the value of $h(m)$ depends only on the set of prime factors of $m$, and so there is no "general case": it suffices to assume $m$ is squarefree. END EDIT 2011.07.29

I am having a few challenges showing upper bounds without having to worry about quadratic residues. This problem is a can of worms I am not ready to tackle. Certainly nothing like the uniform results involving sufficiently large prime factors will hold. One approach involves "tiling" a candidate interval with appropriate primes, and then solving $n$ many quadratic congurences simultaneously. I don't know enough about quadratic residues to see any clean looking results here. It is an interesting variation on which I welcome other viewpoints.

Gerhard "Jacobsthal's Function Is Tough Already" Paseman, 2011.07.15

I take back what I said in the comments about the bound not shrinking.  I am convinced that one can get much tighter bounds, and that the tighter bounds will depend massively on the quadratic residues for each of the primes involved.

To set the stage, I change notation slightly.  I use $m$ for the argument to $j()$, and I use $n$ for the number of distinct prime factors of $m$.  I also insist $m > 1$.

One of the more accessible results is that $j(m) = j(k)$ if $m$ and $k$ have the same prime factors.  Further if the prime factors are all larger than $n$, then $j(m)= n+1$, so there is a certain uniformity in the analysis and results for a large and simply stated class of numbers.

(Advertisement; I am working on similar statements where $n$ is replaced by something like $\sqrt(n)$.  Email me for more detail.)

Many of the standard bounds for $j(m)$ can be expressed in terms of $n$.  When I posted my comments above, I thought something similar would be true for this version.  However, checking a few examples leads me somewhere completely different.

For $m$ a prime power, $j(m)=2$.  The same holds for the new variation.  However, things change when $n>1$. $h(6)=j(6)=4$, but this does not hold for all numbers of the form $2p$. $h(m)$ can vary from 2 to 4 depending on $m$ even if $n$ is restricted to 2.

To make things interesting, I computed $h(385)$, which is bounded above by 4. There are 6 values of a mod 385 to show $j(385)=4$. To get $h(385)=4$, I had to find a square which was 4 mod 7, 9 mod 11, and 4 mod 5. The square of 47 fit the bill, but I could imagine different primes with qrs that would not nicely fit in the set {-3,-2,-1}, so $h(m)$ will not always reach 4 or higher when $n=3$; it will depend on getting qrs which are small negative numbers. For higher $n$, you may not achieve $h(m)>n$ without choosing the $n$ prime factors carefully. And this analysis is using only squarefree $m$; I do not know what happens in the general case.

I am having a few challenges showing upper bounds without having to worry about quadratic residues. This problem is a can of worms I am not ready to tackle. Certainly nothing like the uniform results involving sufficiently large prime factors will hold. One approach involves "tiling" a candidate interval with appropriate primes, and then solving $n$ many quadratic congurences simultaneously. I don't know enough about quadratic residues to see any clean looking results here. It is an interesting variation on which I welcome other viewpoints.

Gerhard "Jacobsthal's Function Is Tough Already" Paseman, 2011.07.15

I take back what I said in the comments about the bound not shrinking.  I am convinced that one can get much tighter bounds, and that the tighter bounds will depend massively on the quadratic residues for each of the primes involved.

To set the stage, I change notation slightly.  I use $m$ for the argument to $j()$, and I use $n$ for the number of distinct prime factors of $m$.  I also insist $m > 1$.

One of the more accessible results is that $j(m) = j(k)$ if $m$ and $k$ have the same prime factors.  Further if the prime factors are all larger than $n$, then $j(m)= n+1$, so there is a certain uniformity in the analysis and results for a large and simply stated class of numbers.

(Advertisement; I am working on similar statements where $n$ is replaced by something like $\sqrt(n)$.  Email me for more detail.)

Many of the standard bounds for $j(m)$ can be expressed in terms of $n$.  When I posted my comments above, I thought something similar would be true for this version.  However, checking a few examples leads me somewhere completely different.

For $m$ a prime power, $j(m)=2$.  The same holds for the new variation if and only if $-1$ is a qr of the prime dividing $m$.  However, things change when $n>1$. $h(6)=j(6)=4$, but this does not hold for all numbers of the form $2p$. $h(m)$ can vary from 2 to 4 depending on $m$ even if $n$ is restricted to 2.

To make things interesting, I computed $h(385)$, which is bounded above by 4. There are 6 values of a mod 385 to show $j(385)=4$. To get $h(385)=4$, I had to find a square which was 4 mod 7, 9 mod 11, and 4 mod 5. The square of 47 fit the bill, but I could imagine different primes with qrs that would not nicely fit in the set {-3,-2,-1}, so $h(m)$ will not always reach 4 or higher when $n=3$; it will depend on getting qrs which are small negative numbers. For higher $n$, you may not achieve $h(m)>n$ without choosing the $n$ prime factors carefully. And this analysis is using only squarefree $m$; I do not know what happens in the general case.

I am having a few challenges showing upper bounds without having to worry about quadratic residues. This problem is a can of worms I am not ready to tackle. Certainly nothing like the uniform results involving sufficiently large prime factors will hold. One approach involves "tiling" a candidate interval with appropriate primes, and then solving $n$ many quadratic congurences simultaneously. I don't know enough about quadratic residues to see any clean looking results here. It is an interesting variation on which I welcome other viewpoints.

Gerhard "Jacobsthal's Function Is Tough Already" Paseman, 2011.07.15

I take back what I said in the comments about the bound not shrinking.  I am convinced that one can get much tighter boinds, and that the tighter bounds will depend massively on the quadratic residues for each of the primes involved.

To set the stage, I change notation slightly.  I use $m$ for the argument to $j()$, and I use $n$ for the number of distinct prime factors of $m$.  I also insist $m > 1$.

One of the more accessible results is that $j(m) = j(k)$ if $m$ and $k$ have the same prime factors.  Further if the prime factors are all larger than $n$, then $j(m)= n+1$, so there is a certain uniformity in the analysis and results for a large and simply stated class of numbers.

(Advertisement; I am working on similar statements where $n$ is replaced by something like $\sqrt(n)$.  Email me for more detail.)

Many of the standard bounds for $j(n)$ can be expressed in terms of $n$.  When I posted my comments above, I thought something similar would be true for this version.  However, checking a few examples leads me somewhere completely different.

For $m$ a prime power, $j(m)=2$.  The same holds for the new variation.  However, things change when $n>1$. $h(6)=j(6)=4$, but this does not hold for all numbers of the form $2p$. $h(m)$ can vary from 2 to 4 depending on $m$ even if $n$ is restricted to 2.

To make things interesting, I computed $h(385)$, which is bounded above by 4. There are 6 values of a mod 385 to show $j(385)=4$. To get $h(385)=4$, I had to find a square which was 4 mod 7, 9 mod 11, and 4 mod 5. The square of 47 fit the bill, but I could imagine different primes with qrs that would not nicely fit in the set {-3,-2,-1}, so $h(m)$ will not always reach 4 or higher when $n=3$; it will depend on getting qrs which are small negative numbers. For higher $n$, you may not achieve $h(n)>n$ without choosing the $n$ prime factors carefully. And this analysis is using only squarefree $m$; I do not know what happens in the general case.

I am having a few challenges showing upper bounds without having to worry about quadratic residues. This problem is a can of worms I am not ready to tackle. Certainly nothing like the uniform results involving sufficiently large prime factors will hold. One approach involves "tiling" a candidate interval with appropriate primes, and then solving $n$ many quadratic congurences simultaneously. I don't know enough about quadratic residues to see any clean looking results here. It is an interesting variation on which I welcome other viewpoints.

Gerhard "Jacobsthal's Function Is Tough Already" Paseman, 2011.07.15

I take back what I said in the comments about the bound not shrinking.  I am convinced that one can get much tighter bounds, and that the tighter bounds will depend massively on the quadratic residues for each of the primes involved.

To set the stage, I change notation slightly.  I use $m$ for the argument to $j()$, and I use $n$ for the number of distinct prime factors of $m$.  I also insist $m > 1$.

One of the more accessible results is that $j(m) = j(k)$ if $m$ and $k$ have the same prime factors.  Further if the prime factors are all larger than $n$, then $j(m)= n+1$, so there is a certain uniformity in the analysis and results for a large and simply stated class of numbers.

(Advertisement; I am working on similar statements where $n$ is replaced by something like $\sqrt(n)$.  Email me for more detail.)

Many of the standard bounds for $j(m)$ can be expressed in terms of $n$.  When I posted my comments above, I thought something similar would be true for this version.  However, checking a few examples leads me somewhere completely different.

For $m$ a prime power, $j(m)=2$.  The same holds for the new variation.  However, things change when $n>1$. $h(6)=j(6)=4$, but this does not hold for all numbers of the form $2p$. $h(m)$ can vary from 2 to 4 depending on $m$ even if $n$ is restricted to 2.

To make things interesting, I computed $h(385)$, which is bounded above by 4. There are 6 values of a mod 385 to show $j(385)=4$. To get $h(385)=4$, I had to find a square which was 4 mod 7, 9 mod 11, and 4 mod 5. The square of 47 fit the bill, but I could imagine different primes with qrs that would not nicely fit in the set {-3,-2,-1}, so $h(m)$ will not always reach 4 or higher when $n=3$; it will depend on getting qrs which are small negative numbers. For higher $n$, you may not achieve $h(m)>n$ without choosing the $n$ prime factors carefully. And this analysis is using only squarefree $m$; I do not know what happens in the general case.

I am having a few challenges showing upper bounds without having to worry about quadratic residues. This problem is a can of worms I am not ready to tackle. Certainly nothing like the uniform results involving sufficiently large prime factors will hold. One approach involves "tiling" a candidate interval with appropriate primes, and then solving $n$ many quadratic congurences simultaneously. I don't know enough about quadratic residues to see any clean looking results here. It is an interesting variation on which I welcome other viewpoints.

Gerhard "Jacobsthal's Function Is Tough Already" Paseman, 2011.07.15