Mixing
The relationship between $sum_{kgeq1}frac{k^{k+a}}{k!(4e)^{k/2}}$ and the number of labeled rooted trees of...
$begingroup$
I was reading Jack D'Aurizio's website matemate.it and learnt that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1tag{1}$$
I worked out some other relatable values:
$$sum_{kgeq1}frac{k^{k+1}}{k!(4e)^{k/2}}=4$$
$$sum_{kgeq1}frac{k^{k+2}}{k!(4e)^{k/2}}=32$$
$$sum_{kgeq1}frac{k^{k+3}}{k!(4e)^{k/2}}=416$$
$$sum_{kgeq1}frac{k^{k+4}}{k!(4e)^{k/2}}=7552$$
$$sum_{kgeq1}frac{k^{k+5}}{k!(4e)^{k/2}}=176128$$
As I typed the number $176128$ into the search bar of OEIS, I recognized that these values coincides with [A005172]number of labeled rooted trees of subsets of an n-set :
Now it seems like
$$sum_{kgeq1}frac{k^{k+n}}{k!(4e)^{k/2}}=a(n+1)tag{2}$$
I read over the linked OEIS webpage and saw that $a(n)$ is related to Stirling formula and Lambert W-function. How should I prove result (2)? and result (1)?
Process:
With the help of Wolfram Alpha, I recognized that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=frac{-Wleft(-frac{1}{2sqrt{e}}right)}{Wleft(-frac{1}{2sqrt{e}}right)+1}$$
where $W$ is Lambert $W$-function. Then, we can use the result that
$$Wleft(-frac{1}{2sqrt{e}}right)=-frac{1}{2}$$
to prove that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1$$
I am trying to work backward. I want to know how to transform the series into Lambert-$W$ function representation.
sequences-and-series lambert-w
asked Jan 18 at 0:35
LarryLarry
2,41331129
$endgroup$
add a comment |
$begingroup$
I was reading Jack D'Aurizio's website matemate.it and learnt that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1tag{1}$$
I worked out some other relatable values:
$$sum_{kgeq1}frac{k^{k+1}}{k!(4e)^{k/2}}=4$$
$$sum_{kgeq1}frac{k^{k+2}}{k!(4e)^{k/2}}=32$$
$$sum_{kgeq1}frac{k^{k+3}}{k!(4e)^{k/2}}=416$$
$$sum_{kgeq1}frac{k^{k+4}}{k!(4e)^{k/2}}=7552$$
$$sum_{kgeq1}frac{k^{k+5}}{k!(4e)^{k/2}}=176128$$
As I typed the number $176128$ into the search bar of OEIS, I recognized that these values coincides with [A005172]number of labeled rooted trees of subsets of an n-set :
Now it seems like
$$sum_{kgeq1}frac{k^{k+n}}{k!(4e)^{k/2}}=a(n+1)tag{2}$$
I read over the linked OEIS webpage and saw that $a(n)$ is related to Stirling formula and Lambert W-function. How should I prove result (2)? and result (1)?
Process:
With the help of Wolfram Alpha, I recognized that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=frac{-Wleft(-frac{1}{2sqrt{e}}right)}{Wleft(-frac{1}{2sqrt{e}}right)+1}$$
where $W$ is Lambert $W$-function. Then, we can use the result that
$$Wleft(-frac{1}{2sqrt{e}}right)=-frac{1}{2}$$
to prove that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1$$
I am trying to work backward. I want to know how to transform the series into Lambert-$W$ function representation.
sequences-and-series lambert-w
asked Jan 18 at 0:35
LarryLarry
2,41331129
$endgroup$
1
$begingroup$
$(1)$ is a consequence of the Lagrange inversion formula (search for Lagrange inversion formula in a nutshell in my notes), so I guess the problem boils down to showing that the recurrence relation fulfilled by $text{LRTS}(n)$ is the same as the recurrence relation fulfilled by your generalized series.
$endgroup$
– Jack D'Aurizio
Jan 18 at 9:27
add a comment |
$begingroup$
I was reading Jack D'Aurizio's website matemate.it and learnt that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1tag{1}$$
I worked out some other relatable values:
$$sum_{kgeq1}frac{k^{k+1}}{k!(4e)^{k/2}}=4$$
$$sum_{kgeq1}frac{k^{k+2}}{k!(4e)^{k/2}}=32$$
$$sum_{kgeq1}frac{k^{k+3}}{k!(4e)^{k/2}}=416$$
$$sum_{kgeq1}frac{k^{k+4}}{k!(4e)^{k/2}}=7552$$
$$sum_{kgeq1}frac{k^{k+5}}{k!(4e)^{k/2}}=176128$$
As I typed the number $176128$ into the search bar of OEIS, I recognized that these values coincides with [A005172]number of labeled rooted trees of subsets of an n-set :
Now it seems like
$$sum_{kgeq1}frac{k^{k+n}}{k!(4e)^{k/2}}=a(n+1)tag{2}$$
I read over the linked OEIS webpage and saw that $a(n)$ is related to Stirling formula and Lambert W-function. How should I prove result (2)? and result (1)?
Process:
With the help of Wolfram Alpha, I recognized that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=frac{-Wleft(-frac{1}{2sqrt{e}}right)}{Wleft(-frac{1}{2sqrt{e}}right)+1}$$
where $W$ is Lambert $W$-function. Then, we can use the result that
$$Wleft(-frac{1}{2sqrt{e}}right)=-frac{1}{2}$$
to prove that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1$$
I am trying to work backward. I want to know how to transform the series into Lambert-$W$ function representation.
sequences-and-series lambert-w
asked Jan 18 at 0:35
LarryLarry
2,41331129
$endgroup$
I was reading Jack D'Aurizio's website matemate.it and learnt that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1tag{1}$$
I worked out some other relatable values:
$$sum_{kgeq1}frac{k^{k+1}}{k!(4e)^{k/2}}=4$$
$$sum_{kgeq1}frac{k^{k+2}}{k!(4e)^{k/2}}=32$$
$$sum_{kgeq1}frac{k^{k+3}}{k!(4e)^{k/2}}=416$$
$$sum_{kgeq1}frac{k^{k+4}}{k!(4e)^{k/2}}=7552$$
$$sum_{kgeq1}frac{k^{k+5}}{k!(4e)^{k/2}}=176128$$
As I typed the number $176128$ into the search bar of OEIS, I recognized that these values coincides with [A005172]number of labeled rooted trees of subsets of an n-set :
Now it seems like
$$sum_{kgeq1}frac{k^{k+n}}{k!(4e)^{k/2}}=a(n+1)tag{2}$$
I read over the linked OEIS webpage and saw that $a(n)$ is related to Stirling formula and Lambert W-function. How should I prove result (2)? and result (1)?
Process:
With the help of Wolfram Alpha, I recognized that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=frac{-Wleft(-frac{1}{2sqrt{e}}right)}{Wleft(-frac{1}{2sqrt{e}}right)+1}$$
where $W$ is Lambert $W$-function. Then, we can use the result that
$$Wleft(-frac{1}{2sqrt{e}}right)=-frac{1}{2}$$
to prove that
$$sum_{kgeq1}frac{k^{k}}{k!(4e)^{k/2}}=1$$
I am trying to work backward. I want to know how to transform the series into Lambert-$W$ function representation.
sequences-and-series lambert-w
sequences-and-series lambert-w
asked Jan 18 at 0:35
LarryLarry
2,41331129
asked Jan 18 at 0:35
LarryLarry
2,41331129
edited Jan 18 at 0:42
Larry
asked Jan 18 at 0:35
LarryLarry
2,41331129
asked Jan 18 at 0:35
LarryLarry
2,41331129
asked Jan 18 at 0:35
LarryLarry
2,41331129
2,41331129
1
$begingroup$
$(1)$ is a consequence of the Lagrange inversion formula (search for Lagrange inversion formula in a nutshell in my notes), so I guess the problem boils down to showing that the recurrence relation fulfilled by $text{LRTS}(n)$ is the same as the recurrence relation fulfilled by your generalized series.
$endgroup$
– Jack D'Aurizio
Jan 18 at 9:27
add a comment |
1
$begingroup$
$(1)$ is a consequence of the Lagrange inversion formula (search for Lagrange inversion formula in a nutshell in my notes), so I guess the problem boils down to showing that the recurrence relation fulfilled by $text{LRTS}(n)$ is the same as the recurrence relation fulfilled by your generalized series.
$endgroup$
– Jack D'Aurizio
Jan 18 at 9:27
1
1
$begingroup$
$(1)$ is a consequence of the Lagrange inversion formula (search for Lagrange inversion formula in a nutshell in my notes), so I guess the problem boils down to showing that the recurrence relation fulfilled by $text{LRTS}(n)$ is the same as the recurrence relation fulfilled by your generalized series.
$endgroup$
– Jack D'Aurizio
Jan 18 at 9:27
$begingroup$
$(1)$ is a consequence of the Lagrange inversion formula (search for Lagrange inversion formula in a nutshell in my notes), so I guess the problem boils down to showing that the recurrence relation fulfilled by $text{LRTS}(n)$ is the same as the recurrence relation fulfilled by your generalized series.
$endgroup$
– Jack D'Aurizio
Jan 18 at 9:27
add a comment |
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$begingroup$
$(1)$ is a consequence of the Lagrange inversion formula (search for Lagrange inversion formula in a nutshell in my notes), so I guess the problem boils down to showing that the recurrence relation fulfilled by $text{LRTS}(n)$ is the same as the recurrence relation fulfilled by your generalized series.
$endgroup$
– Jack D'Aurizio
Jan 18 at 9:27