An upper bound for integer partitions with unique summands
2
$begingroup$
Let $p_neq (n)$ be the number of all partitions of $n$ such that all summands are distinct (for example $p_neq (6)=4$).
How do we show that $p_neq (n) leq e^{2sqrt n}$?
combinatorics integer-partitions upper-lower-bounds
asked Dec 29 '18 at 3:20
FortoxFortox
6718
$endgroup$
add a comment |
2
$begingroup$
Let $p_neq (n)$ be the number of all partitions of $n$ such that all summands are distinct (for example $p_neq (6)=4$).
How do we show that $p_neq (n) leq e^{2sqrt n}$?
combinatorics integer-partitions upper-lower-bounds
asked Dec 29 '18 at 3:20
FortoxFortox
6718
$endgroup$
2
$begingroup$
My first thought is that the largest summand needs to be greater than $2sqrt n$ because the sum of all the numbers up to that is too small. So we have $p_{neq}(n)=sum_{k=2sqrt n+1}^np_{neq}(n-k)$. I don't know if that works but the coincidence of the exponent is promising.
$endgroup$
– Ross Millikan
Dec 29 '18 at 5:00
1
$begingroup$
Lots of information here Don't see the result you're looking for, though.
$endgroup$
– saulspatz
Dec 29 '18 at 5:57
add a comment |
2
2
2
$begingroup$
Let $p_neq (n)$ be the number of all partitions of $n$ such that all summands are distinct (for example $p_neq (6)=4$).
How do we show that $p_neq (n) leq e^{2sqrt n}$?
combinatorics integer-partitions upper-lower-bounds
asked Dec 29 '18 at 3:20
FortoxFortox
6718
$endgroup$
Let $p_neq (n)$ be the number of all partitions of $n$ such that all summands are distinct (for example $p_neq (6)=4$).
How do we show that $p_neq (n) leq e^{2sqrt n}$?
combinatorics integer-partitions upper-lower-bounds
combinatorics integer-partitions upper-lower-bounds
asked Dec 29 '18 at 3:20
FortoxFortox
6718
asked Dec 29 '18 at 3:20
FortoxFortox
6718
edited Dec 29 '18 at 4:03
Fortox
asked Dec 29 '18 at 3:20
FortoxFortox
6718
asked Dec 29 '18 at 3:20
FortoxFortox
6718
asked Dec 29 '18 at 3:20
FortoxFortox
6718
6718
2
$begingroup$
My first thought is that the largest summand needs to be greater than $2sqrt n$ because the sum of all the numbers up to that is too small. So we have $p_{neq}(n)=sum_{k=2sqrt n+1}^np_{neq}(n-k)$. I don't know if that works but the coincidence of the exponent is promising.
$endgroup$
– Ross Millikan
Dec 29 '18 at 5:00
1
$begingroup$
Lots of information here Don't see the result you're looking for, though.
$endgroup$
– saulspatz
Dec 29 '18 at 5:57
add a comment |
2
$begingroup$
My first thought is that the largest summand needs to be greater than $2sqrt n$ because the sum of all the numbers up to that is too small. So we have $p_{neq}(n)=sum_{k=2sqrt n+1}^np_{neq}(n-k)$. I don't know if that works but the coincidence of the exponent is promising.
$endgroup$
– Ross Millikan
Dec 29 '18 at 5:00
1
$begingroup$
Lots of information here Don't see the result you're looking for, though.
$endgroup$
– saulspatz
Dec 29 '18 at 5:57
2
2
$begingroup$
My first thought is that the largest summand needs to be greater than $2sqrt n$ because the sum of all the numbers up to that is too small. So we have $p_{neq}(n)=sum_{k=2sqrt n+1}^np_{neq}(n-k)$. I don't know if that works but the coincidence of the exponent is promising.
$endgroup$
– Ross Millikan
Dec 29 '18 at 5:00
$begingroup$
My first thought is that the largest summand needs to be greater than $2sqrt n$ because the sum of all the numbers up to that is too small. So we have $p_{neq}(n)=sum_{k=2sqrt n+1}^np_{neq}(n-k)$. I don't know if that works but the coincidence of the exponent is promising.
$endgroup$
– Ross Millikan
Dec 29 '18 at 5:00
1
1
$begingroup$
Lots of information here Don't see the result you're looking for, though.
$endgroup$
– saulspatz
Dec 29 '18 at 5:57
$begingroup$
Lots of information here Don't see the result you're looking for, though.
$endgroup$
– saulspatz
Dec 29 '18 at 5:57
add a comment |
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2
$begingroup$
My first thought is that the largest summand needs to be greater than $2sqrt n$ because the sum of all the numbers up to that is too small. So we have $p_{neq}(n)=sum_{k=2sqrt n+1}^np_{neq}(n-k)$. I don't know if that works but the coincidence of the exponent is promising.
$endgroup$
– Ross Millikan
Dec 29 '18 at 5:00
1
$begingroup$
Lots of information here Don't see the result you're looking for, though.
$endgroup$
– saulspatz
Dec 29 '18 at 5:57