Mixing
On the conjecture that, for every $n$, $lfloor e^{frac{p_{n^2}#}{p_{n^2 + 1}}}rfloor $ is a square number.
$begingroup$
I was playing around with numbers and I discovered that $$leftlfloor e^{frac{p_4#}{p_5}}rightrfloor=leftlfloor e^{frac{210}{11}}rightrfloor =13981^2,$$ with floor function $lfloor x rfloor := max{minmathbb{Z} : mleqslant x}$; $ p_n$ denotes the $n^{text{th}}$ prime number; $ $and primorial $$p_n#_c := prod_{i=1}^n (p_i + c).tag{$p_n#_0 = p_n#$}$$
Furthermore, $$leftlfloor e^{frac{p_1#}{p_2}}rightrfloor = 1^2.$$ I then made a conjecture with very few support, that $$leftlfloor e^{frac{p_{n^2}#}{p_{n^2 + 1}}}rightrfloor tag1$$ is always a square number, say $k_n^{ 2}$.
Does somebody have a big enough computer to find the value of $(k_n)_{ngeqslant3}$? Or can my conjecture be proven/disproven with a pen and paper? And perhaps, to support the potential of this not being a coincidence, every divisor of $13981$ takes the form $(13981 - 10x)$ for some $xgeqslant 0$. Maybe if this conjecture is true, $k_n > 1$ has this certain property?
Edit 1: With some computational power, I found that $k_3^{ 2}$ has $approx 176,000$ digits. It turned out, however, that I was wrong, and really, $k_3^{ 2}$ has $approx 3,340,970$ digits.
Edit 2: Expressions equivalent to $(1)$, avoiding $e$, for any base $binmathbb{N}_{>1}$:$$leftlfloor b^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(b)}}rightrfloor=k_n^{ 2}$$E.g. base $b=2$:$$leftlfloor 2^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(2)}}rightrfloor=k_n^{ 2}$$E.g. base $b=n^2$:$$leftlfloor (n^2)^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(n^2)}}rightrfloor=k_n^{ 2}$$
prime-numbers exponential-function conjectures square-numbers primorial
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
$endgroup$
|
show 14 more comments
$begingroup$
I was playing around with numbers and I discovered that $$leftlfloor e^{frac{p_4#}{p_5}}rightrfloor=leftlfloor e^{frac{210}{11}}rightrfloor =13981^2,$$ with floor function $lfloor x rfloor := max{minmathbb{Z} : mleqslant x}$; $ p_n$ denotes the $n^{text{th}}$ prime number; $ $and primorial $$p_n#_c := prod_{i=1}^n (p_i + c).tag{$p_n#_0 = p_n#$}$$
Furthermore, $$leftlfloor e^{frac{p_1#}{p_2}}rightrfloor = 1^2.$$ I then made a conjecture with very few support, that $$leftlfloor e^{frac{p_{n^2}#}{p_{n^2 + 1}}}rightrfloor tag1$$ is always a square number, say $k_n^{ 2}$.
Does somebody have a big enough computer to find the value of $(k_n)_{ngeqslant3}$? Or can my conjecture be proven/disproven with a pen and paper? And perhaps, to support the potential of this not being a coincidence, every divisor of $13981$ takes the form $(13981 - 10x)$ for some $xgeqslant 0$. Maybe if this conjecture is true, $k_n > 1$ has this certain property?
Edit 1: With some computational power, I found that $k_3^{ 2}$ has $approx 176,000$ digits. It turned out, however, that I was wrong, and really, $k_3^{ 2}$ has $approx 3,340,970$ digits.
Edit 2: Expressions equivalent to $(1)$, avoiding $e$, for any base $binmathbb{N}_{>1}$:$$leftlfloor b^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(b)}}rightrfloor=k_n^{ 2}$$E.g. base $b=2$:$$leftlfloor 2^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(2)}}rightrfloor=k_n^{ 2}$$E.g. base $b=n^2$:$$leftlfloor (n^2)^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(n^2)}}rightrfloor=k_n^{ 2}$$
prime-numbers exponential-function conjectures square-numbers primorial
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
$endgroup$
2
$begingroup$
You didn't perchance use your computer's calculator? For huge numbers, you need software like Mathematica (which I use). There are probably clever number-theoretic tests to tell whether a huge number is a square, and I faintly remember a discussion like that here in MSE.
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:02
1
$begingroup$
Yes, that's it. I just realized it considered 600 digits as "large". Yours has more than 3 million digits, so you may have a slight problem applying the algorithm. :)
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:27
2
$begingroup$
@iadvd and it is funny because I was thinking to myself before I discovered something like this, what if I just played around with numbers? and then all of a sudden, bloop there you go. I have also heard of Mills' constant, and have tried making approximations of it (because a formula for it has not been discovered, as far as I know).
$endgroup$
– user477343
Feb 14 '18 at 9:41
1
$begingroup$
@iadvd I like the $exp(cdot)$ better but I'm sure everyone gets the idea, and that is the whole point anyway, so i'm not too fussed :)
$endgroup$
– user477343
Feb 15 '18 at 9:54
1
$begingroup$
ok I think I might be able to have crack at this one
$endgroup$
– Adam
Oct 26 '18 at 0:01
|
show 14 more comments
$begingroup$
I was playing around with numbers and I discovered that $$leftlfloor e^{frac{p_4#}{p_5}}rightrfloor=leftlfloor e^{frac{210}{11}}rightrfloor =13981^2,$$ with floor function $lfloor x rfloor := max{minmathbb{Z} : mleqslant x}$; $ p_n$ denotes the $n^{text{th}}$ prime number; $ $and primorial $$p_n#_c := prod_{i=1}^n (p_i + c).tag{$p_n#_0 = p_n#$}$$
Furthermore, $$leftlfloor e^{frac{p_1#}{p_2}}rightrfloor = 1^2.$$ I then made a conjecture with very few support, that $$leftlfloor e^{frac{p_{n^2}#}{p_{n^2 + 1}}}rightrfloor tag1$$ is always a square number, say $k_n^{ 2}$.
Does somebody have a big enough computer to find the value of $(k_n)_{ngeqslant3}$? Or can my conjecture be proven/disproven with a pen and paper? And perhaps, to support the potential of this not being a coincidence, every divisor of $13981$ takes the form $(13981 - 10x)$ for some $xgeqslant 0$. Maybe if this conjecture is true, $k_n > 1$ has this certain property?
Edit 1: With some computational power, I found that $k_3^{ 2}$ has $approx 176,000$ digits. It turned out, however, that I was wrong, and really, $k_3^{ 2}$ has $approx 3,340,970$ digits.
Edit 2: Expressions equivalent to $(1)$, avoiding $e$, for any base $binmathbb{N}_{>1}$:$$leftlfloor b^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(b)}}rightrfloor=k_n^{ 2}$$E.g. base $b=2$:$$leftlfloor 2^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(2)}}rightrfloor=k_n^{ 2}$$E.g. base $b=n^2$:$$leftlfloor (n^2)^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(n^2)}}rightrfloor=k_n^{ 2}$$
prime-numbers exponential-function conjectures square-numbers primorial
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
$endgroup$
I was playing around with numbers and I discovered that $$leftlfloor e^{frac{p_4#}{p_5}}rightrfloor=leftlfloor e^{frac{210}{11}}rightrfloor =13981^2,$$ with floor function $lfloor x rfloor := max{minmathbb{Z} : mleqslant x}$; $ p_n$ denotes the $n^{text{th}}$ prime number; $ $and primorial $$p_n#_c := prod_{i=1}^n (p_i + c).tag{$p_n#_0 = p_n#$}$$
Furthermore, $$leftlfloor e^{frac{p_1#}{p_2}}rightrfloor = 1^2.$$ I then made a conjecture with very few support, that $$leftlfloor e^{frac{p_{n^2}#}{p_{n^2 + 1}}}rightrfloor tag1$$ is always a square number, say $k_n^{ 2}$.
Does somebody have a big enough computer to find the value of $(k_n)_{ngeqslant3}$? Or can my conjecture be proven/disproven with a pen and paper? And perhaps, to support the potential of this not being a coincidence, every divisor of $13981$ takes the form $(13981 - 10x)$ for some $xgeqslant 0$. Maybe if this conjecture is true, $k_n > 1$ has this certain property?
Edit 1: With some computational power, I found that $k_3^{ 2}$ has $approx 176,000$ digits. It turned out, however, that I was wrong, and really, $k_3^{ 2}$ has $approx 3,340,970$ digits.
Edit 2: Expressions equivalent to $(1)$, avoiding $e$, for any base $binmathbb{N}_{>1}$:$$leftlfloor b^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(b)}}rightrfloor=k_n^{ 2}$$E.g. base $b=2$:$$leftlfloor 2^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(2)}}rightrfloor=k_n^{ 2}$$E.g. base $b=n^2$:$$leftlfloor (n^2)^{frac{p_{n^2}#}{p_{n^2 + 1} cdot Ln(n^2)}}rightrfloor=k_n^{ 2}$$
prime-numbers exponential-function conjectures square-numbers primorial
prime-numbers exponential-function conjectures square-numbers primorial
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
edited Feb 16 '18 at 12:52
user477343
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
asked Feb 11 '18 at 0:59
user477343user477343
3,59831243
3,59831243
2
$begingroup$
You didn't perchance use your computer's calculator? For huge numbers, you need software like Mathematica (which I use). There are probably clever number-theoretic tests to tell whether a huge number is a square, and I faintly remember a discussion like that here in MSE.
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:02
1
$begingroup$
Yes, that's it. I just realized it considered 600 digits as "large". Yours has more than 3 million digits, so you may have a slight problem applying the algorithm. :)
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:27
2
$begingroup$
@iadvd and it is funny because I was thinking to myself before I discovered something like this, what if I just played around with numbers? and then all of a sudden, bloop there you go. I have also heard of Mills' constant, and have tried making approximations of it (because a formula for it has not been discovered, as far as I know).
$endgroup$
– user477343
Feb 14 '18 at 9:41
1
$begingroup$
@iadvd I like the $exp(cdot)$ better but I'm sure everyone gets the idea, and that is the whole point anyway, so i'm not too fussed :)
$endgroup$
– user477343
Feb 15 '18 at 9:54
1
$begingroup$
ok I think I might be able to have crack at this one
$endgroup$
– Adam
Oct 26 '18 at 0:01
|
show 14 more comments
2
$begingroup$
You didn't perchance use your computer's calculator? For huge numbers, you need software like Mathematica (which I use). There are probably clever number-theoretic tests to tell whether a huge number is a square, and I faintly remember a discussion like that here in MSE.
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:02
1
$begingroup$
Yes, that's it. I just realized it considered 600 digits as "large". Yours has more than 3 million digits, so you may have a slight problem applying the algorithm. :)
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:27
2
$begingroup$
@iadvd and it is funny because I was thinking to myself before I discovered something like this, what if I just played around with numbers? and then all of a sudden, bloop there you go. I have also heard of Mills' constant, and have tried making approximations of it (because a formula for it has not been discovered, as far as I know).
$endgroup$
– user477343
Feb 14 '18 at 9:41
1
$begingroup$
@iadvd I like the $exp(cdot)$ better but I'm sure everyone gets the idea, and that is the whole point anyway, so i'm not too fussed :)
$endgroup$
– user477343
Feb 15 '18 at 9:54
1
$begingroup$
ok I think I might be able to have crack at this one
$endgroup$
– Adam
Oct 26 '18 at 0:01
2
2
$begingroup$
You didn't perchance use your computer's calculator? For huge numbers, you need software like Mathematica (which I use). There are probably clever number-theoretic tests to tell whether a huge number is a square, and I faintly remember a discussion like that here in MSE.
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:02
$begingroup$
You didn't perchance use your computer's calculator? For huge numbers, you need software like Mathematica (which I use). There are probably clever number-theoretic tests to tell whether a huge number is a square, and I faintly remember a discussion like that here in MSE.
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:02
1
1
$begingroup$
Yes, that's it. I just realized it considered 600 digits as "large". Yours has more than 3 million digits, so you may have a slight problem applying the algorithm. :)
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:27
$begingroup$
Yes, that's it. I just realized it considered 600 digits as "large". Yours has more than 3 million digits, so you may have a slight problem applying the algorithm. :)
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:27
2
2
$begingroup$
@iadvd and it is funny because I was thinking to myself before I discovered something like this, what if I just played around with numbers? and then all of a sudden, bloop there you go. I have also heard of Mills' constant, and have tried making approximations of it (because a formula for it has not been discovered, as far as I know).
$endgroup$
– user477343
Feb 14 '18 at 9:41
$begingroup$
@iadvd and it is funny because I was thinking to myself before I discovered something like this, what if I just played around with numbers? and then all of a sudden, bloop there you go. I have also heard of Mills' constant, and have tried making approximations of it (because a formula for it has not been discovered, as far as I know).
$endgroup$
– user477343
Feb 14 '18 at 9:41
1
1
$begingroup$
@iadvd I like the $exp(cdot)$ better but I'm sure everyone gets the idea, and that is the whole point anyway, so i'm not too fussed :)
$endgroup$
– user477343
Feb 15 '18 at 9:54
$begingroup$
@iadvd I like the $exp(cdot)$ better but I'm sure everyone gets the idea, and that is the whole point anyway, so i'm not too fussed :)
$endgroup$
– user477343
Feb 15 '18 at 9:54
1
1
$begingroup$
ok I think I might be able to have crack at this one
$endgroup$
– Adam
Oct 26 '18 at 0:01
$begingroup$
ok I think I might be able to have crack at this one
$endgroup$
– Adam
Oct 26 '18 at 0:01
|
show 14 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Conjectured "formulas" such as this are most likely the result of coincidence as one can easily find a constant $A$ and a function $f$ such that the values of
$$
g(n)=lfloor A^{f(n)} rfloor
$$
are within a given set $S={s_0, s_1, ldots}$, by knowing the density of $S$. For instance, in the case of Mills' prime-generating formula, it is known that there is always a prime in the interval $[(P-1)^3,P^3]$ for $P>1$, so there must exist a positive nonzero constant $A$ such that
$$
m(n)=lfloor A^{3^n}rfloor
$$
yields only primes.
Moreover, don't forget that seemingly remarkable mathematical
coincidences are easy to generate, so by "playing around" one might find formulas such as yours without any real meaning.
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Conjectured "formulas" such as this are most likely the result of coincidence as one can easily find a constant $A$ and a function $f$ such that the values of
$$
g(n)=lfloor A^{f(n)} rfloor
$$
are within a given set $S={s_0, s_1, ldots}$, by knowing the density of $S$. For instance, in the case of Mills' prime-generating formula, it is known that there is always a prime in the interval $[(P-1)^3,P^3]$ for $P>1$, so there must exist a positive nonzero constant $A$ such that
$$
m(n)=lfloor A^{3^n}rfloor
$$
yields only primes.
Moreover, don't forget that seemingly remarkable mathematical
coincidences are easy to generate, so by "playing around" one might find formulas such as yours without any real meaning.
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
$endgroup$
add a comment |
$begingroup$
Conjectured "formulas" such as this are most likely the result of coincidence as one can easily find a constant $A$ and a function $f$ such that the values of
$$
g(n)=lfloor A^{f(n)} rfloor
$$
are within a given set $S={s_0, s_1, ldots}$, by knowing the density of $S$. For instance, in the case of Mills' prime-generating formula, it is known that there is always a prime in the interval $[(P-1)^3,P^3]$ for $P>1$, so there must exist a positive nonzero constant $A$ such that
$$
m(n)=lfloor A^{3^n}rfloor
$$
yields only primes.
Moreover, don't forget that seemingly remarkable mathematical
coincidences are easy to generate, so by "playing around" one might find formulas such as yours without any real meaning.
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
$endgroup$
add a comment |
$begingroup$
Conjectured "formulas" such as this are most likely the result of coincidence as one can easily find a constant $A$ and a function $f$ such that the values of
$$
g(n)=lfloor A^{f(n)} rfloor
$$
are within a given set $S={s_0, s_1, ldots}$, by knowing the density of $S$. For instance, in the case of Mills' prime-generating formula, it is known that there is always a prime in the interval $[(P-1)^3,P^3]$ for $P>1$, so there must exist a positive nonzero constant $A$ such that
$$
m(n)=lfloor A^{3^n}rfloor
$$
yields only primes.
Moreover, don't forget that seemingly remarkable mathematical
coincidences are easy to generate, so by "playing around" one might find formulas such as yours without any real meaning.
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
$endgroup$
Conjectured "formulas" such as this are most likely the result of coincidence as one can easily find a constant $A$ and a function $f$ such that the values of
$$
g(n)=lfloor A^{f(n)} rfloor
$$
are within a given set $S={s_0, s_1, ldots}$, by knowing the density of $S$. For instance, in the case of Mills' prime-generating formula, it is known that there is always a prime in the interval $[(P-1)^3,P^3]$ for $P>1$, so there must exist a positive nonzero constant $A$ such that
$$
m(n)=lfloor A^{3^n}rfloor
$$
yields only primes.
Moreover, don't forget that seemingly remarkable mathematical
coincidences are easy to generate, so by "playing around" one might find formulas such as yours without any real meaning.
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
edited Jan 11 at 11:40
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
answered Jan 7 at 11:53
KlangenKlangen
1,72411334
1,72411334
add a comment |
add a comment |
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2
$begingroup$
You didn't perchance use your computer's calculator? For huge numbers, you need software like Mathematica (which I use). There are probably clever number-theoretic tests to tell whether a huge number is a square, and I faintly remember a discussion like that here in MSE.
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:02
1
$begingroup$
Yes, that's it. I just realized it considered 600 digits as "large". Yours has more than 3 million digits, so you may have a slight problem applying the algorithm. :)
$endgroup$
– Tito Piezas III
Feb 14 '18 at 9:27
2
$begingroup$
@iadvd and it is funny because I was thinking to myself before I discovered something like this, what if I just played around with numbers? and then all of a sudden, bloop there you go. I have also heard of Mills' constant, and have tried making approximations of it (because a formula for it has not been discovered, as far as I know).
$endgroup$
– user477343
Feb 14 '18 at 9:41
1
$begingroup$
@iadvd I like the $exp(cdot)$ better but I'm sure everyone gets the idea, and that is the whole point anyway, so i'm not too fussed :)
$endgroup$
– user477343
Feb 15 '18 at 9:54
1
$begingroup$
ok I think I might be able to have crack at this one
$endgroup$
– Adam
Oct 26 '18 at 0:01