A006252 Expansion of e.g.f. 1/(1 - log(1+x)).

1, 1, 1, 2, 4, 14, 38, 216, 600, 6240, 9552, 319296, -519312, 28108560, -176474352, 3998454144, -43985078784, 837126163584, -12437000028288, 237195036797184, -4235955315745536, 85886259443020800, -1746536474655406080, 38320721602434017280, -864056965711935974400

Offset: 0

Author

N. J. A. Sloane

Keywords

• sign

Comments

From Michael Somos, Mar 04 2004: (Start)

Stirling transform of a(n+1)=[1,2,4,14,38,...] is A000255(n)=[1,3,11,53,309,...].

Stirling transform of 2*a(n)=[2,2,4,8,28,...] is A052849(n)=[2,4,12,48,240,...].

Stirling transform of a(n)=[1,1,2,4,14,38,216,...] is A000142(n)=[1,2,6,24,120,...].

Stirling transform of a(n-1)=[1,1,1,2,4,14,38,...] is A000522(n-1)=[1,2,5,16,65,...].

Stirling transform of a(n-1)=[0,1,1,2,4,14,38,...] is A007526(n-1)=[0,1,4,15,64,...].

(End)

For n > 0: a(n) = sum of n-th row in triangle A048594. - Reinhard Zumkeller, Mar 02 2014

Coefficients in a factorial series representation of the exponential integral: exp(z)*E_1(z) = Sum_{n >= 0} (-1)^n*a(n)/(z)n, where (z)_n denotes the rising factorial z*(z + 1)*...*(z + n) and E_1(z) = Integrate{t = z..inf} exp(-t)/t dt. See Weninger, equation 6.4. - Peter Bala, Feb 12 2019

References

• G. Pólya, Induction and Analogy in Mathematics. Princeton Univ. Press, 1954, p. 9.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..400
• Beáta Bényi and Daniel Yaqubi, Mixed coloured permutations, arXiv:1903.07450 [math.CO], 2019.
• Takao Komatsu and Amalia Pizarro-Madariaga, Harmonic numbers associated with inversion numbers in terms of determinants, Turkish Journal of Mathematics (2019) Vol. 43, 340-354.
• E. J. Weniger, Summation of divergent power series by means of factorial series arXiv:1005.0466v1 [math.NA], 2010.

Crossrefs

Column k=1 of A320080.

Cf. A007840.

Programs

Haskell

a006252 0 = 1
a006252 n = sum $ a048594_row n  -- Reinhard Zumkeller, Mar 02 2014

Mathematica

With[{nn=30},CoefficientList[Series[1/(1-Log[1+x]),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Aug 12 2016 *)

PARI

a(n)=if(n<0,0,n!*polcoeff(1/(1-log(1+x+x*O(x^n))),n))

PARI

{a(n)=local(CF=1+x*O(x^n)); for(k=0, n-1, CF=1/((n-k+1)-(n-k)*x+(n-k+1)^2*x*CF)); n!*polcoeff(1+x/(1-x+x*CF), n, x)} /* Paul D. Hanna, Dec 31 2011 */

PARI

a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=1, i, (-1)^(j-1)*(j-1)!*binomial(i, j)*v[i-j+1])); v; \\ Seiichi Manyama, May 22 2022

Sage

def A006252_list(len):
    f, R, C = 1, [1], [1]+[0]*len
    for n in (1..len):
        f *= n
        for k in range(n, 0, -1):
            C[k] = -C[k-1]*((k-1)/(k) if k>1 else 1)
        C[0] = -sum(C[k] for k in (1..n))
        R.append(C[0]*f)
    return R
print(A006252_list(24)) # Peter Luschny, Feb 21 2016

Formula

a(n) = Sum_{k=0..n} k!*stirling1(n, k). - Vladeta Jovovic, Sep 08 2002

a(n) = D^n(1/(1-x)) evaluated at x = 0, where D is the operator exp(-x)*d/dx. Row sums of A048594. Cf. A007840. - Peter Bala, Nov 25 2011

E.g.f.: 1/(1-log(1+x)) = 1 + x/(1-x + x/(2-x + 4*x/(3-2*x + 9*x/(4-3*x + 16*x/(5-4*x + 25*x/(6-5*x +...)))))), a continued fraction. - Paul D. Hanna, Dec 31 2011

a(n)/n! ~ -(-1)^n / (n * (log(n))^2) * (1 - 2*(1 + gamma)/log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jul 01 2018

a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k-1) * (k-1)! * binomial(n,k) * a(n-k). - Seiichi Manyama, May 22 2022