A167894 Expansion of g.f.: 1/(Sum_{k >= 0} k!*x^k).

1, -1, -1, -3, -13, -71, -461, -3447, -29093, -273343, -2829325, -31998903, -392743957, -5201061455, -73943424413, -1123596277863, -18176728317413, -311951144828863, -5661698774848621, -108355864447215063

Offset: 0

Author

Philippe Deléham, Nov 15 2009

Keywords

• sign

Comments

Essentially the same as A003319, which is the main entry for these numbers. - N. J. A. Sloane, Jun 11 2013

References

• M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 40.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A003319, A158882.

Programs

Magma

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 1/(&+[Factorial(k)*x^k: k in [0..m+1]]) )); // G. C. Greubel, Feb 07 2019

Mathematica

CoefficientList[Series[1/(Sum[k!*x^k, {k, 0, 25}]), {x, 0, 20}], x] (* G. C. Greubel, Jun 30 2016 *)

Maxima

a(n) := if n=0 then 1 else -sum( a(i)*(n-i)!,i,0,n-1); /* Vladimir Kruchinin, Oct 10 2024 */

PARI

m=20; my(x='x+O('x^m)); Vec(1/sum(k=0,m+1, k!*x^k)) \\ G. C. Greubel, Feb 07 2019

Sage

def A167894_list(len):
    R, C = [1], [1]+[0]*(len-1)
    for n in (1..len-1):
        for k in range(n, 0, -1):
            C[k] = C[k-1] * k
        C[0] = -sum(C[k] for k in (1..n))
        R.append(C[0])
    return R
print(A167894_list(20)) # Peter Luschny, Feb 19 2016

Sage

m=20; (1/sum(factorial(k)*x^k for k in range(m+1))).series(x, m).coefficients(x, sparse=False) # G. C. Greubel, Feb 07 2019

Formula

a(n) = - Sum_{i=0..n-1} a(i)*(n-i)! for n > 0 with a(0) = 1. - Vladimir Kruchinin, Oct 10 2024

From Sergei N. Gladkovskii, Jun 24 2012, Oct 15 2012, Nov 18 2012, Dec 26 2012, Apr 25 2013, May 29 2013, Aug 08 2013, Nov 19 2013: (Start) Continued fractions:

G.f.: 1 - x/Q(0), where Q(k) = 1 - (k+1)*x/(1 - (k+2)*x/Q(k+1)).

G.f.: U(0) where U(k) = 1 - x*(k+1)/(1 - x*(k+1)/U(k+1)).

G.f.: 1/G(0) where G(k) = 1 + x*(2*k+1)/(1 - 2*x*(k+1)/(2*x*(k+1) + 1/G(k+1))).

G.f.: A(x) = 1 - x/G(0) where G(k) = 1 + (k+1)*x - x*(k+2)/G(k+1).

G.f.: x*Q(0), where Q(k) = 1/x - 1 - 2*k - (k+1)^2/Q(k+1).

G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 1/(1 - 1/(2*x*(k+1)) + 1/G(k+1))).

G.f.: 2/Q(0), where Q(k) = 1 + 1/(1 - x*(k+1)/(x*(k+1) + 1/Q(k+1) )).

G.f.: conjecture: Q(0), where Q(k) = 1 + k*x - (k+1)*x/Q(k+1). (End)

a(n) ~ -n! * (1 - 2/n - 1/n^2 - 5/n^3 - 32/n^4 - 253/n^5 - 2381/n^6 - 25912/n^7 - 319339/n^8 - 4388949/n^9 - 66495386/n^10 - ...). - Vaclav Kotesovec, Dec 08 2020