OFFSET
0,5
COMMENTS
E.g.f. A(x) = y satisfies (y + y' + y'') * y - y'^2 = 0. - Michael Somos, Mar 11 2004
The 10-adic sum: B(n) = Sum_{k>=0} k^n*k! simplifies to: B(n) = A014182(n)*B(0) + A014619(n) for n>=0, where B(0) is the 10-adic sum of factorials (A025016); a result independent of base. - Paul D. Hanna, Aug 12 2006
Equals row sums of triangle A143987 and (shifted) = right border of A143987. [Gary W. Adamson, Sep 07 2008]
From Gary W. Adamson, Dec 31 2008: (Start)
Equals the eigensequence of the inverse of Pascal's triangle, A007318.
Binomial transform shifts to the right: (1, 1, 0, -1, 1, 2, -9, ...).
Double binomial transform = A109747. (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
Mohammad Ghorbani, Mehdi Hassani, and Hossein Moshtagh, Moments and asymptotic expansion of derangement polynomials in terms of Touchard polynomials, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 4, 832-842. See pp. 834, 839.
FORMULA
E.g.f.: exp(1-x-exp(-x)).
a(n) = Sum_{k=0..n} (-1)^(n-k)*Stirling2(n+1,k+1). - Paul D. Hanna, Aug 12 2006
A000587(n+1) = -a(n). - Michael Somos, May 12 2012
G.f.: 1/x/(U(0)-x) -1/x where U(k)= 1 - x + x*(k+1)/(1 - x/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Oct 12 2012
G.f.: 1/(U(0) - x) where U(k) = 1 + x*(k+1)/(1 - x/U(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2012
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+k*x+x)/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: G(0)/(1+x)-1 where G(k) = 1 + 1/(1 + k*x - x*(1+k*x)*(1+k*x+x)/(x*(1+k*x+x) + (1+k*x+2*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: S-1 where S = Sum_{k>=0} (2 + x*k)*x^k/Product_{i=0..k} (1+x+x*i). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: G(0)*x^2/(1+x)/(1+2*x) + 2/(1+x) - 1 where G(k) = 1 + 2/(x + k*x - x^3*(k+1)*(k+2)/(x^2*(k+2) + 2*(1+k*x+3*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
G.f.: 1/(x*Q(0)) -1/x, where Q(k) = 1 - x/(1 + (k+1)*x/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Sep 27 2013
G.f.: G(0)/(1-x)/x - 1/x, where G(k) = 1 - x^2*(k+1)/(x^2*(k+1) + (x*k + 1 - x)*(x*k + 1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Feb 06 2014
G.f.: (1 - Sum_{k>0} k * x^k / ((1 + x) * (1 + 2*x) + ... (1 + k*x))) / (1 - x). - Michael Somos, Nov 07 2014
a(n) = exp(1) * (-1)^n * Sum_{k>=0} (-1)^k * (k + 1)^n / k!. - Ilya Gutkovskiy, Dec 20 2019
EXAMPLE
G.f. = 1 - x^2 + x^3 + 2*x^4 - 9*x^5 + 9*x^6 + 50*x^7 - 267*x^8 + 413*x^9 + ...
MATHEMATICA
With[{nn=30}, CoefficientList[Series[Exp[1-x-Exp[-x]], {x, 0, nn}], x] Range[0, nn]!] (* Harvey P. Dale, Jan 15 2012 *)
a[ n_] := SeriesCoefficient[ (1 - Sum[ k / Pochhammer[ 1/x + 1, k], {k, n}]) / (1 - x), {x, 0, n} ]; (* Michael Somos, Nov 07 2014 *)
PROG
(PARI) {a(n)=sum(j=0, n, (-1)^(n-j)*Stirling2(n+1, j+1))}
{Stirling2(n, k)=(1/k!)*sum(i=0, k, (-1)^(k-i)*binomial(k, i)*i^n)} \\ Paul D. Hanna, Aug 12 2006
(PARI) {a(n) = if( n<0, 0, n! * polcoeff( exp( 1 - x - exp( -x + x * O(x^n))), n))} /* Michael Somos, Mar 11 2004 */
(Sage)
def A014182_list(len): # len>=1
T = [0]*(len+1); T[1] = 1; R = [1]
for n in (1..len-1):
a, b, c = 1, 0, 0
for k in range(n, -1, -1):
r = a - k*b - (k+1)*c
if k < n : T[k+2] = u;
a, b, c = T[k-1], a, b
u = r
T[1] = u; R.append(u)
return R
A014182_list(27) # Peter Luschny, Nov 01 2012
CROSSREFS
KEYWORD
sign,easy,nice
AUTHOR
STATUS
approved