Finding a more explicit way to express a coefficient in this summation

Question

I am sorry about the vague title, i really don't know how else to ask the question. So i have came up with the following:

$$ 1 = \sum_{k=0}^{n} \sum_{v=0}^{n} \frac{x^{n-k+1} T_{n+1,k+1} (-1)^{n-v+1}}{(x-v)v!(n-v)!} $$

Where as $T_{n+1,k+1}$ is the coefficient that i am trying to find a more explicit definition of. I have come to find that $T_{n,k}$ holds some values such as:

$$ T_{n,1} = 1 $$

$$ T_{n,2} = \frac{-n(n-1)}{2} $$

$$ T_{n,n} = (-1)^{n-1}(n-1)! $$

And if $2<c<n$

$$ T_{n,c} = T_{n-1,c} - (n-1)T_{n-1,c-1} $$

These properties are good and all but i don't like how there is a recurrence relation. If there is a recursive definition should i just be satisfied and leave it at that? Is there no way for me to find out a more explicit way to express this? Thank you in advance.

Answer 1

From you recursive definition I would say that you found the signed Stirling numbers of the first kind with only the order changed : $$T_{n,c}=s(n,\,n+1-c)= (-1)^{c-1}\left[\matrix {n\\n+1-c}\right]$$ (concerning your first equality I obtain $-1$ at the left)

Further references : Gould "Combinatorial Numbers and Associated Identities" vol $7$.
the recommended Book "Concrete mathematics" by Graham&Knuth&Patashnik.

pari/gp scripts used :

t(n,c)=if(c<=2,if(c<=1,1,-n*(n-1)/2),if(c>=n,(-1)^(n-1)*(n-1)!,t(n-1,c)-(n-1)*t(n-1,c-1)))
ss(n)=sum(k=0,n,sum(v=0,n,(x^(n-k+1)*stirling(n+1,n+2-(k+1)) *(-1)^(n-v+1))/((x-v)v!(n-v)!)))
\\we may replace t(n,c) by stirling(n,n+1-c) in the following line for(n=1,6,print(vector(n,c,t(n,c))))

[1]
[1, -1]
[1, -3, 2]
[1, -6, 11, -6]
[1, -10, 35, -50, 24]
[1, -15, 85, -225, 274, -120]