Predicting the Next Term in Recursive Derivatives Series

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Can anybody predict the next line in the sequence?

-1
1 -1
-3 7 -4
15 -63 82 -34
-105 729 -1638 1510 -496
945 -10545 36684 -56568 40540 -11056


Good Luck!
 
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Just for my curiosity, where is this question from?
 
first element in row : : : last element in row
-10,395 : : : : : : : : sum of previous elements in row * -1

I'm likewise curious to as to what this sequence is to. It might help to figure out the sequence.

It would look roughly something like:
-10,395 161,051 -1,054,500 1,907,568 –1,207,882 1,641,870 -278,300

That is, each element should roughly be around the value given (within 25% error maybe?), except that the first value is correct.
 
Hi All...

The sequence comes from the series solution to a differential equation,,,,

and the next line is...

-10395 186075 -946392 2174820 -2552292 1497688 -349504
 
What's the formula for producing some of these elements (like row 5 for example)?
 
I've just posted this to the differential equations section...it should explain everything

I've been working on finding a closed form solution or an approximate solution for a variant of the Poisson-Boltzmann equation.

y''[x]=A*(1-e^(-y[x])

I'm unable to solve it using methods such as separation of variables, but I have been able to generate a power series using recursive derivatives. I've pasted the mathematica code below...I've been unable to find the pattern to generate the series...can anybody help?

In[2]:= (*Define Recursive Derivative Function *)

In[3]:= Derivative[N_][Y][0]:=D[Y''[x],{x,N-2}] /. x->0

In[4]:= (*Define the Problem*)

In[5]:= Y''[x]=A*(1-E^(-Y[x]))

Out[5]= A (1-E^-Y[x])

In[6]:= Y''[0]=Y''[x] /. x->0

Out[6]= A (1-E^-Y[0])

In[7]:= Y'[0]=0

Out[7]= 0

In[8]:= (*Check the Series*)

In[9]:= Y[0]+Integrate[Integrate[Series[Y''[x],{x,0,10}],x],x]

Out[9]= Y[0]+1/2 (A-A E^-Y[0]) x^2+1/24 A^2 E^(-2 Y[0]) (-1+E^Y[0]) x^4-1/720 (A^3 E^(-3 Y[0]) (4-7 E^Y[0]+3 E^(2 Y[0]))) x^6+(A^4 E^(-4 Y[0]) (-34+82 E^Y[0]-63 E^(2 Y[0])+15 E^(3 Y[0])) x^8)/40320-((A^5 E^(-5 Y[0]) (496-1510 E^Y[0]+1638 E^(2 Y[0])-729 E^(3 Y[0])+105 E^(4 Y[0]))) x^10)/3628800+(A^6 E^(-6 Y[0]) (-11056+40540 E^Y[0]-56568 E^(2 Y[0])+36684 E^(3 Y[0])-10545 E^(4 Y[0])+945 E^(5 Y[0])) x^12)/479001600+O[x]^13
 

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