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Question relating to a number theoretic function
infinitely many primes p which are not congruent to $-1$ modulo $19$.A question on elementary number theoryThe largest number to break a conjectureUse this sequence to prove that there are infinitely many prime numbers.Any Heuristic Argument that for the distribution of primes $p$, about half of the primes $p$, $u(2$) $=$ $3$ prime chain?Consecutive rising sequence of largest prime factorsA question about algebraic normal numberNumber theoretic olympiad question involving a diophantine equationProve the following number theoretic assertion.A high school Olympiad problem
$begingroup$
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 $M(y) > M(y-1),$ $exists b in mathbbN cup 0$ 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
asked 1 hour ago
acreativenameacreativename
14718
$endgroup$
add a comment |
$begingroup$
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 $M(y) > M(y-1),$ $exists b in mathbbN cup 0$ 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
asked 1 hour ago
acreativenameacreativename
14718
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $M(y) > M(y-1),$ $exists b in mathbbN cup 0$ 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
asked 1 hour ago
acreativenameacreativename
14718
$endgroup$
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 $M(y) > M(y-1),$ $exists b in mathbbN cup 0$ 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
asked 1 hour ago
acreativenameacreativename
14718
asked 1 hour ago
acreativenameacreativename
14718
asked 1 hour ago
acreativenameacreativename
14718
asked 1 hour ago
acreativenameacreativename
14718
asked 1 hour 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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
For every $age 2$,
$$M(2a)=M(a)$$
answered 1 hour ago
ajotatxeajotatxe
54.6k24090
$endgroup$
$begingroup$
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
56 mins ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
For every $age 2$,
$$M(2a)=M(a)$$
answered 1 hour ago
ajotatxeajotatxe
54.6k24090
$endgroup$
$begingroup$
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
56 mins ago
add a comment |
$begingroup$
Hint:
For every $age 2$,
$$M(2a)=M(a)$$
answered 1 hour ago
ajotatxeajotatxe
54.6k24090
$endgroup$
$begingroup$
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
56 mins ago
add a comment |
$begingroup$
Hint:
For every $age 2$,
$$M(2a)=M(a)$$
answered 1 hour ago
ajotatxeajotatxe
54.6k24090
$endgroup$
Hint:
For every $age 2$,
$$M(2a)=M(a)$$
answered 1 hour ago
ajotatxeajotatxe
54.6k24090
answered 1 hour ago
ajotatxeajotatxe
54.6k24090
answered 1 hour ago
ajotatxeajotatxe
54.6k24090
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
56 mins ago
add a comment |
$begingroup$
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
56 mins 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
56 mins ago
$begingroup$
It can be extended to be proven for all $k geq 1$
$endgroup$
– acreativename
56 mins ago
add a comment |
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$begingroup$
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