Question relating to a number theoretic function












2












$begingroup$


Question; For $a geq 2$, Let $M(a) = $ the largest prime factor of $a$ i.e $M(6) = 3, M(9) = 3, M(69) = 23$ are there infinitely many $n$ so that
$M(n-2) < M(n) < M(n+2)$ ?



One can show that for infinitely many $k$ we have $M(k-1) < M(k) < M(k+1)$ by considering the following lemma;



Lemma; If $M(y) > M(y-1),$ $exists b in mathbb{N} cup {0}$ so that $M(y^{2^b} - 1) < M(y^{2^b}) < M(y^{2^b} +1)$



Proof of Lemma; If one has $M(y-1) < M(y) < M(y+1)$ then we are done, if not then we have $M(y^2-1) < M(y^2)$, If one has $M(y^2-1) < M(y^2) < M(y^2+1)$ then we are done if not then...



If the sequence of writing above ever terminates then we have the lemma; if not then one has that $forall b$ $M(y^{2^b}) > M(y^{2^b}+1)$, i.e



$forall b$ $M(y) > M(y^{2^b}+1)$; this is absurd (I will let you think about why).



I don't think the question above can rely on this trickery.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Twin primes' conjecture implies this: if $p$ and $p+2$ are prime then $M(p-2)<M(p)<M(p+2)$.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    That conjecture is overkill, I "believe" my question is simpler.
    $endgroup$
    – acreativename
    1 hour ago






  • 1




    $begingroup$
    Yes, of course, it was only a comment.
    $endgroup$
    – ajotatxe
    1 hour ago










  • $begingroup$
    no worries all good here
    $endgroup$
    – acreativename
    1 hour ago
















2












$begingroup$


Question; For $a geq 2$, Let $M(a) = $ the largest prime factor of $a$ i.e $M(6) = 3, M(9) = 3, M(69) = 23$ are there infinitely many $n$ so that
$M(n-2) < M(n) < M(n+2)$ ?



One can show that for infinitely many $k$ we have $M(k-1) < M(k) < M(k+1)$ by considering the following lemma;



Lemma; If $M(y) > M(y-1),$ $exists b in mathbb{N} cup {0}$ so that $M(y^{2^b} - 1) < M(y^{2^b}) < M(y^{2^b} +1)$



Proof of Lemma; If one has $M(y-1) < M(y) < M(y+1)$ then we are done, if not then we have $M(y^2-1) < M(y^2)$, If one has $M(y^2-1) < M(y^2) < M(y^2+1)$ then we are done if not then...



If the sequence of writing above ever terminates then we have the lemma; if not then one has that $forall b$ $M(y^{2^b}) > M(y^{2^b}+1)$, i.e



$forall b$ $M(y) > M(y^{2^b}+1)$; this is absurd (I will let you think about why).



I don't think the question above can rely on this trickery.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Twin primes' conjecture implies this: if $p$ and $p+2$ are prime then $M(p-2)<M(p)<M(p+2)$.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    That conjecture is overkill, I "believe" my question is simpler.
    $endgroup$
    – acreativename
    1 hour ago






  • 1




    $begingroup$
    Yes, of course, it was only a comment.
    $endgroup$
    – ajotatxe
    1 hour ago










  • $begingroup$
    no worries all good here
    $endgroup$
    – acreativename
    1 hour ago














2












2








2


1



$begingroup$


Question; For $a geq 2$, Let $M(a) = $ the largest prime factor of $a$ i.e $M(6) = 3, M(9) = 3, M(69) = 23$ are there infinitely many $n$ so that
$M(n-2) < M(n) < M(n+2)$ ?



One can show that for infinitely many $k$ we have $M(k-1) < M(k) < M(k+1)$ by considering the following lemma;



Lemma; If $M(y) > M(y-1),$ $exists b in mathbb{N} cup {0}$ so that $M(y^{2^b} - 1) < M(y^{2^b}) < M(y^{2^b} +1)$



Proof of Lemma; If one has $M(y-1) < M(y) < M(y+1)$ then we are done, if not then we have $M(y^2-1) < M(y^2)$, If one has $M(y^2-1) < M(y^2) < M(y^2+1)$ then we are done if not then...



If the sequence of writing above ever terminates then we have the lemma; if not then one has that $forall b$ $M(y^{2^b}) > M(y^{2^b}+1)$, i.e



$forall b$ $M(y) > M(y^{2^b}+1)$; this is absurd (I will let you think about why).



I don't think the question above can rely on this trickery.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$




Question; For $a geq 2$, Let $M(a) = $ the largest prime factor of $a$ i.e $M(6) = 3, M(9) = 3, M(69) = 23$ are there infinitely many $n$ so that
$M(n-2) < M(n) < M(n+2)$ ?



One can show that for infinitely many $k$ we have $M(k-1) < M(k) < M(k+1)$ by considering the following lemma;



Lemma; If $M(y) > M(y-1),$ $exists b in mathbb{N} cup {0}$ so that $M(y^{2^b} - 1) < M(y^{2^b}) < M(y^{2^b} +1)$



Proof of Lemma; If one has $M(y-1) < M(y) < M(y+1)$ then we are done, if not then we have $M(y^2-1) < M(y^2)$, If one has $M(y^2-1) < M(y^2) < M(y^2+1)$ then we are done if not then...



If the sequence of writing above ever terminates then we have the lemma; if not then one has that $forall b$ $M(y^{2^b}) > M(y^{2^b}+1)$, i.e



$forall b$ $M(y) > M(y^{2^b}+1)$; this is absurd (I will let you think about why).



I don't think the question above can rely on this trickery.






number-theory elementary-number-theory




number-theory elementary-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









acreativenameacreativename

14718




14718












  • $begingroup$
    Twin primes' conjecture implies this: if $p$ and $p+2$ are prime then $M(p-2)<M(p)<M(p+2)$.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    That conjecture is overkill, I "believe" my question is simpler.
    $endgroup$
    – acreativename
    1 hour ago






  • 1




    $begingroup$
    Yes, of course, it was only a comment.
    $endgroup$
    – ajotatxe
    1 hour ago










  • $begingroup$
    no worries all good here
    $endgroup$
    – acreativename
    1 hour ago


















  • $begingroup$
    Twin primes' conjecture implies this: if $p$ and $p+2$ are prime then $M(p-2)<M(p)<M(p+2)$.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    That conjecture is overkill, I "believe" my question is simpler.
    $endgroup$
    – acreativename
    1 hour ago






  • 1




    $begingroup$
    Yes, of course, it was only a comment.
    $endgroup$
    – ajotatxe
    1 hour ago










  • $begingroup$
    no worries all good here
    $endgroup$
    – acreativename
    1 hour ago
















$begingroup$
Twin primes' conjecture implies this: if $p$ and $p+2$ are prime then $M(p-2)<M(p)<M(p+2)$.
$endgroup$
– ajotatxe
1 hour ago






$begingroup$
Twin primes' conjecture implies this: if $p$ and $p+2$ are prime then $M(p-2)<M(p)<M(p+2)$.
$endgroup$
– ajotatxe
1 hour ago














$begingroup$
That conjecture is overkill, I "believe" my question is simpler.
$endgroup$
– acreativename
1 hour ago




$begingroup$
That conjecture is overkill, I "believe" my question is simpler.
$endgroup$
– acreativename
1 hour ago




1




1




$begingroup$
Yes, of course, it was only a comment.
$endgroup$
– ajotatxe
1 hour ago




$begingroup$
Yes, of course, it was only a comment.
$endgroup$
– ajotatxe
1 hour ago












$begingroup$
no worries all good here
$endgroup$
– acreativename
1 hour ago




$begingroup$
no worries all good here
$endgroup$
– acreativename
1 hour ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

Hint:



For every $age 2$,
$$M(2a)=M(a)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yup that solves the problem thanks.
    $endgroup$
    – acreativename
    1 hour ago










  • $begingroup$
    BTW, this idea proves also $M(n-k)<M(n)<M(n+k)$ when $k$ is a power of two or three.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    It can be extended to be proven for all $k geq 1$
    $endgroup$
    – acreativename
    1 hour ago












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Hint:



For every $age 2$,
$$M(2a)=M(a)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yup that solves the problem thanks.
    $endgroup$
    – acreativename
    1 hour ago










  • $begingroup$
    BTW, this idea proves also $M(n-k)<M(n)<M(n+k)$ when $k$ is a power of two or three.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    It can be extended to be proven for all $k geq 1$
    $endgroup$
    – acreativename
    1 hour ago
















4












$begingroup$

Hint:



For every $age 2$,
$$M(2a)=M(a)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yup that solves the problem thanks.
    $endgroup$
    – acreativename
    1 hour ago










  • $begingroup$
    BTW, this idea proves also $M(n-k)<M(n)<M(n+k)$ when $k$ is a power of two or three.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    It can be extended to be proven for all $k geq 1$
    $endgroup$
    – acreativename
    1 hour ago














4












4








4





$begingroup$

Hint:



For every $age 2$,
$$M(2a)=M(a)$$






share|cite|improve this answer









$endgroup$



Hint:



For every $age 2$,
$$M(2a)=M(a)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









ajotatxeajotatxe

54.6k24090




54.6k24090












  • $begingroup$
    Yup that solves the problem thanks.
    $endgroup$
    – acreativename
    1 hour ago










  • $begingroup$
    BTW, this idea proves also $M(n-k)<M(n)<M(n+k)$ when $k$ is a power of two or three.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    It can be extended to be proven for all $k geq 1$
    $endgroup$
    – acreativename
    1 hour ago


















  • $begingroup$
    Yup that solves the problem thanks.
    $endgroup$
    – acreativename
    1 hour ago










  • $begingroup$
    BTW, this idea proves also $M(n-k)<M(n)<M(n+k)$ when $k$ is a power of two or three.
    $endgroup$
    – ajotatxe
    1 hour ago












  • $begingroup$
    It can be extended to be proven for all $k geq 1$
    $endgroup$
    – acreativename
    1 hour ago
















$begingroup$
Yup that solves the problem thanks.
$endgroup$
– acreativename
1 hour ago




$begingroup$
Yup that solves the problem thanks.
$endgroup$
– acreativename
1 hour ago












$begingroup$
BTW, this idea proves also $M(n-k)<M(n)<M(n+k)$ when $k$ is a power of two or three.
$endgroup$
– ajotatxe
1 hour ago






$begingroup$
BTW, this idea proves also $M(n-k)<M(n)<M(n+k)$ when $k$ is a power of two or three.
$endgroup$
– ajotatxe
1 hour ago














$begingroup$
It can be extended to be proven for all $k geq 1$
$endgroup$
– acreativename
1 hour ago




$begingroup$
It can be extended to be proven for all $k geq 1$
$endgroup$
– acreativename
1 hour ago


















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