A006252 Expansion of e.g.f. 1/(1 - log(1+x)). (Formerly M1275)

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

0,4

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.

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

Mathematica

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

Prog

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

Crossrefs

Column k=1 of A320080.

Cf. A007840.

Sequence in context: A035010 A349450 A055540 * A079995 A279322 A152011

Adjacent sequences: A006249 A006250 A006251 * A006253 A006254 A006255

Keyword

sign

Author

N. J. A. Sloane

Status

approved