$ t(n) = t( x_1 x_2 x_3 …) = t(x_1) + t(x_2) + t(x_3) + … + t( x_1 + x_2 + x_3 + … ) $
$begingroup$
Let $ n > 1 $ be an integer.
Consider The prime factorization
$$ n = x_1 x_2 x_3 ... $$
Now define
$$ t(n) = t( x_1 x_2 x_3 ...) = t(x_1) + t(x_2) + t(x_3) + ... + t( x_1 + x_2 + x_3 + ... ) $$
Clearly this function is completely determined by its values at primes. This gives us multiple solutions.
Im fascinated by this function.
Has this been studied before ?
—-
Some more specific questions :
Can we find solutions such that
$$ t(n) leq t(n+1) leq t(n+2) $$
And if so , which one grows the slowest ?
And how fast does that slowest solution grow asymptotically ?
How about The fastest ?
Are they easy questions or do they depend on a lot of theory or open conjectures ?
A reference or a few plots would be nice too.
Im not sure how to handle this.
—-
Update :
There seems to be a problem with defining $t(2n)$ so for now i focus on odd $n$.
From $t(3^{3^k}) $ we conclude that f grows like $ O( ln(n) + ln(ln(n)) + ln(ln(ln(n))) + ... ) $
number-theory inequality asymptotics functional-equations closed-form
asked Jan 20 at 23:39
mickmick
5,15922164
$endgroup$
|
show 6 more comments
$begingroup$
Let $ n > 1 $ be an integer.
Consider The prime factorization
$$ n = x_1 x_2 x_3 ... $$
Now define
$$ t(n) = t( x_1 x_2 x_3 ...) = t(x_1) + t(x_2) + t(x_3) + ... + t( x_1 + x_2 + x_3 + ... ) $$
Clearly this function is completely determined by its values at primes. This gives us multiple solutions.
Im fascinated by this function.
Has this been studied before ?
—-
Some more specific questions :
Can we find solutions such that
$$ t(n) leq t(n+1) leq t(n+2) $$
And if so , which one grows the slowest ?
And how fast does that slowest solution grow asymptotically ?
How about The fastest ?
Are they easy questions or do they depend on a lot of theory or open conjectures ?
A reference or a few plots would be nice too.
Im not sure how to handle this.
—-
Update :
There seems to be a problem with defining $t(2n)$ so for now i focus on odd $n$.
From $t(3^{3^k}) $ we conclude that f grows like $ O( ln(n) + ln(ln(n)) + ln(ln(ln(n))) + ... ) $
number-theory inequality asymptotics functional-equations closed-form
asked Jan 20 at 23:39
mickmick
5,15922164
$endgroup$
$begingroup$
The $x_i$ are not NEC distinct ! In case that might confuse.
$endgroup$
– mick
Jan 20 at 23:41
$begingroup$
I am not sure your function is well-defined. What is $t(p^k)$, just for my understanding?
$endgroup$
– Jack D'Aurizio
Jan 20 at 23:44
$begingroup$
$t(p^k) = t(p p p ... ) $ that is what i meant by not necc distinct !
$endgroup$
– mick
Jan 20 at 23:47
$begingroup$
For instance $t(8) = t( 2 *2 *2 ) = t(2) + t(2) + t(2) + t(2+2+2) = 3 t(2) + t(6) $ or $ t(p^k) = k t(p) + t(p k) $ for $ k > 1$.
$endgroup$
– mick
Jan 20 at 23:51
1
$begingroup$
@Jack No, that's quite clear: $t(p^3)=3t(p)+t(3p)=4t(p)+t(3)$. I have another question, though: how do you define $t(4)$?
$endgroup$
– Ivan Neretin
Jan 21 at 8:10
|
show 6 more comments
$begingroup$
Let $ n > 1 $ be an integer.
Consider The prime factorization
$$ n = x_1 x_2 x_3 ... $$
Now define
$$ t(n) = t( x_1 x_2 x_3 ...) = t(x_1) + t(x_2) + t(x_3) + ... + t( x_1 + x_2 + x_3 + ... ) $$
Clearly this function is completely determined by its values at primes. This gives us multiple solutions.
Im fascinated by this function.
Has this been studied before ?
—-
Some more specific questions :
Can we find solutions such that
$$ t(n) leq t(n+1) leq t(n+2) $$
And if so , which one grows the slowest ?
And how fast does that slowest solution grow asymptotically ?
How about The fastest ?
Are they easy questions or do they depend on a lot of theory or open conjectures ?
A reference or a few plots would be nice too.
Im not sure how to handle this.
—-
Update :
There seems to be a problem with defining $t(2n)$ so for now i focus on odd $n$.
From $t(3^{3^k}) $ we conclude that f grows like $ O( ln(n) + ln(ln(n)) + ln(ln(ln(n))) + ... ) $
number-theory inequality asymptotics functional-equations closed-form
asked Jan 20 at 23:39
mickmick
5,15922164
$endgroup$
Let $ n > 1 $ be an integer.
Consider The prime factorization
$$ n = x_1 x_2 x_3 ... $$
Now define
$$ t(n) = t( x_1 x_2 x_3 ...) = t(x_1) + t(x_2) + t(x_3) + ... + t( x_1 + x_2 + x_3 + ... ) $$
Clearly this function is completely determined by its values at primes. This gives us multiple solutions.
Im fascinated by this function.
Has this been studied before ?
—-
Some more specific questions :
Can we find solutions such that
$$ t(n) leq t(n+1) leq t(n+2) $$
And if so , which one grows the slowest ?
And how fast does that slowest solution grow asymptotically ?
How about The fastest ?
Are they easy questions or do they depend on a lot of theory or open conjectures ?
A reference or a few plots would be nice too.
Im not sure how to handle this.
—-
Update :
There seems to be a problem with defining $t(2n)$ so for now i focus on odd $n$.
From $t(3^{3^k}) $ we conclude that f grows like $ O( ln(n) + ln(ln(n)) + ln(ln(ln(n))) + ... ) $
number-theory inequality asymptotics functional-equations closed-form
number-theory inequality asymptotics functional-equations closed-form
asked Jan 20 at 23:39
mickmick
5,15922164
asked Jan 20 at 23:39
mickmick
5,15922164
edited Jan 21 at 19:20
mick
asked Jan 20 at 23:39
mickmick
5,15922164
asked Jan 20 at 23:39
mickmick
5,15922164
asked Jan 20 at 23:39
mickmick
5,15922164
5,15922164
$begingroup$
The $x_i$ are not NEC distinct ! In case that might confuse.
$endgroup$
– mick
Jan 20 at 23:41
$begingroup$
I am not sure your function is well-defined. What is $t(p^k)$, just for my understanding?
$endgroup$
– Jack D'Aurizio
Jan 20 at 23:44
$begingroup$
$t(p^k) = t(p p p ... ) $ that is what i meant by not necc distinct !
$endgroup$
– mick
Jan 20 at 23:47
$begingroup$
For instance $t(8) = t( 2 *2 *2 ) = t(2) + t(2) + t(2) + t(2+2+2) = 3 t(2) + t(6) $ or $ t(p^k) = k t(p) + t(p k) $ for $ k > 1$.
$endgroup$
– mick
Jan 20 at 23:51
1
$begingroup$
@Jack No, that's quite clear: $t(p^3)=3t(p)+t(3p)=4t(p)+t(3)$. I have another question, though: how do you define $t(4)$?
$endgroup$
– Ivan Neretin
Jan 21 at 8:10
|
show 6 more comments
$begingroup$
The $x_i$ are not NEC distinct ! In case that might confuse.
$endgroup$
– mick
Jan 20 at 23:41
$begingroup$
I am not sure your function is well-defined. What is $t(p^k)$, just for my understanding?
$endgroup$
– Jack D'Aurizio
Jan 20 at 23:44
$begingroup$
$t(p^k) = t(p p p ... ) $ that is what i meant by not necc distinct !
$endgroup$
– mick
Jan 20 at 23:47
$begingroup$
For instance $t(8) = t( 2 *2 *2 ) = t(2) + t(2) + t(2) + t(2+2+2) = 3 t(2) + t(6) $ or $ t(p^k) = k t(p) + t(p k) $ for $ k > 1$.
$endgroup$
– mick
Jan 20 at 23:51
1
$begingroup$
@Jack No, that's quite clear: $t(p^3)=3t(p)+t(3p)=4t(p)+t(3)$. I have another question, though: how do you define $t(4)$?
$endgroup$
– Ivan Neretin
Jan 21 at 8:10
$begingroup$
The $x_i$ are not NEC distinct ! In case that might confuse.
$endgroup$
– mick
Jan 20 at 23:41
$begingroup$
The $x_i$ are not NEC distinct ! In case that might confuse.
$endgroup$
– mick
Jan 20 at 23:41
$begingroup$
I am not sure your function is well-defined. What is $t(p^k)$, just for my understanding?
$endgroup$
– Jack D'Aurizio
Jan 20 at 23:44
$begingroup$
I am not sure your function is well-defined. What is $t(p^k)$, just for my understanding?
$endgroup$
– Jack D'Aurizio
Jan 20 at 23:44
$begingroup$
$t(p^k) = t(p p p ... ) $ that is what i meant by not necc distinct !
$endgroup$
– mick
Jan 20 at 23:47
$begingroup$
$t(p^k) = t(p p p ... ) $ that is what i meant by not necc distinct !
$endgroup$
– mick
Jan 20 at 23:47
$begingroup$
For instance $t(8) = t( 2 *2 *2 ) = t(2) + t(2) + t(2) + t(2+2+2) = 3 t(2) + t(6) $ or $ t(p^k) = k t(p) + t(p k) $ for $ k > 1$.
$endgroup$
– mick
Jan 20 at 23:51
$begingroup$
For instance $t(8) = t( 2 *2 *2 ) = t(2) + t(2) + t(2) + t(2+2+2) = 3 t(2) + t(6) $ or $ t(p^k) = k t(p) + t(p k) $ for $ k > 1$.
$endgroup$
– mick
Jan 20 at 23:51
1
1
$begingroup$
@Jack No, that's quite clear: $t(p^3)=3t(p)+t(3p)=4t(p)+t(3)$. I have another question, though: how do you define $t(4)$?
$endgroup$
– Ivan Neretin
Jan 21 at 8:10
$begingroup$
@Jack No, that's quite clear: $t(p^3)=3t(p)+t(3p)=4t(p)+t(3)$. I have another question, though: how do you define $t(4)$?
$endgroup$
– Ivan Neretin
Jan 21 at 8:10
|
show 6 more comments
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$begingroup$
The $x_i$ are not NEC distinct ! In case that might confuse.
$endgroup$
– mick
Jan 20 at 23:41
$begingroup$
I am not sure your function is well-defined. What is $t(p^k)$, just for my understanding?
$endgroup$
– Jack D'Aurizio
Jan 20 at 23:44
$begingroup$
$t(p^k) = t(p p p ... ) $ that is what i meant by not necc distinct !
$endgroup$
– mick
Jan 20 at 23:47
$begingroup$
For instance $t(8) = t( 2 *2 *2 ) = t(2) + t(2) + t(2) + t(2+2+2) = 3 t(2) + t(6) $ or $ t(p^k) = k t(p) + t(p k) $ for $ k > 1$.
$endgroup$
– mick
Jan 20 at 23:51
1
$begingroup$
@Jack No, that's quite clear: $t(p^3)=3t(p)+t(3p)=4t(p)+t(3)$. I have another question, though: how do you define $t(4)$?
$endgroup$
– Ivan Neretin
Jan 21 at 8:10