A025016: 10-adic Sum of Factorials

[Paul D. Hanna]

Franklin, and Seqfans,
     Thanks for the comment; it led to a formula for A014182(n). 
> Generally, n^k * n! can be written as a linear function of n!, ..., 
> (n+k-1)!; I suspect this will lead to your general relation, but I 
> don't have time to dig into it now. 
> Franklin T. Adams-Watters 
Yes; it is easy to see: 
n*n! = (n+1)! - n!
n^2*n! = (n+2)! - 3*(n+1)! + n!
n^3*n! = (n+3)! - 6*(n+2)! + 7*(n+1)! - n!
n^4*n! = (n+4)! - 10*(n+3)! + 25*(n+2)! - 15*(n+1)! + n!
etc. 
Coefficients are signed Stirling numbers of the 2nd kind, 
where S2(n,k) = A008277(n,k), n>=k>=1. 
So  
   n^k*n! = Sum_{j=0..k} (-1)^(k-j)*S2(k+1,j+1)*(n+j)! 
 
Multiplying by y^n and summing over n, 
Sum_{n>=0} n^k*n!*y^n = 
   Sum_{j=0..k} (-1)^(k-j)*S2(k+1,j+1) * [Sum_{n>=0} (n+j)!*y^n]
 
Since the sum is 10-adic, we ignore some problems with divergence, 
so that we can set y=1 to get: 

B(k) = Sum_{j=0..k} (-1)^(k-j)*S2(k+1,j+1)*[B(0) - Sum_{i=0..j-1} i!] 
  
This agrees with my earlier formula 
   B(n) = A014182(n)*B(0) + A014619(n)
when we substitute:
  
A014182(n) = Sum_{k=0..n} (-1)^(n-k)*S2(n+1,k+1). 
A014619(n) = Sum_{k=0..n} (-1)^(n-k+1)*S2(n+1,k+1)*[Sum_{i=0..k-1} i!]. 
    
The first formula is not given in OEIS for A014182 (will submit), 
and the formula for A014619 is given in OEIS by Vladeta Jovovic. 
  
An easy yet fun exercise. 
   Paul 
 
/* PARI code to validate formulas: */
{S2(n,k)=(1/k!)*sum(i=0,k,(-1)^(k-i)*binomial(k,i)*i^n)}
{A014182(n)=sum(j=0,n,(-1)^(n-j)*S2(n+1,j+1))}
{A014619(n)=sum(j=0,n,(-1)^(n-j+1)*S2(n+1,j+1)*sum(i=0,j-1,i!))}
for(n=0,30,print1(A014182(n),","))
for(n=0,30,print1(A014619(n),",")) 
/* Compare to G.F. given in A014182: */
{G014182(n)=if(n<0,0,n!*polcoeff(exp(1-x-exp(-x+x*O(x^n))),n))}
for(n=0,120,print1(A014182(n)-G014182(n),","))  /* = all zeros */ 
  
> -----Original Message----- 
> From: Paul D. Hanna <pauldhanna at juno.com> 
>  
> Seqfans, 
>      Consider David Wilson's nice 10-adic constant A025016: 
> x = Sum_{n>=0} n! (10-adic) 
>
=...92247479478684738621107994804323593105039052556442336528920420940314 
> ...
> Investigating the Bell number analogue: 
>    B(k) = Sum_{n>=0} n^k*n!  (10-adic) 
> ...   
>    B(n) = A014182(n)*B(0) + A014619(n) 
> ...