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A027641 Numerator of Bernoulli number B_n.

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0, 2929993913841559, 0, -261082718496449122051
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

a(n)/A027642(n) (Bernoulli numbers) provide the a-sequence for the Sheffer matrix A094816 (coefficients of orthogonal Poisson-Charlier polynomials). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The corresponding z-sequence is given by the rationals A130189(n)/A130190(n).
Harvey (2008) describes a new algorithm for computing Bernoulli numbers. His method is to compute B(k) modulo p for many small primes p and then reconstruct B(k) via the Chinese Remainder Theorem. The time complexity is O(k^2 log(k)^(2+eps)). The algorithm is especially well-suited to parallelization. - Jonathan Vos Post, Jul 09 2008
Regard the Bernoulli numbers as forming a vector = B_n, and the variant starting (1, 1/2, 1/6, 0, -1/30, ...), (i.e., the first 1/2 has sign +) as forming a vector Bv_n. The relationship between the Pascal triangle matrix, B_n, and Bv_n is as follows: The binomial transform of B_n = Bv_n. B_n is unchanged when multiplied by the Pascal matrix with rows signed (+-+-, ...), i.e., (1; -1,-1; 1,2,1; ...). Bv_n is unchanged when multiplied by the Pascal matrix with columns signed (+-+-, ...), i.e., (1; 1,-1; 1,-2,1; 1,-3,3,-1; ...). - Gary W. Adamson, Jun 29 2012
The sequence of the Bernoulli numbers B_n = a(n)/A027642(n) is the inverse binomial transform of the sequence {A164555(n)/A027642(n)}, illustrated by the fact that they appear as top row and left column in A190339. - Paul Curtz, May 13 2016
Named by de Moivre (1773; "the numbers of Mr. James Bernoulli") after the Swiss mathematician Jacob Bernoulli (1655-1705). - Amiram Eldar, Oct 02 2023

Examples

			B_n sequence begins 1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, -691/2730, 0, 7/6, 0, -3617/510, ...

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • Harold T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 230.
  • Harold M. Edwards, Riemann's Zeta Function, Academic Press, NY, 1974; see p. 11.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.1.
  • Herman H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.6.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 137.
  • Hans Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1.

Links

Crossrefs

This is the main entry for the Bernoulli numbers and has all the references, links and formulas. Sequences A027642 (the denominators of B_n) and A000367/A002445 = B_{2n} are also important!
A refinement is A194587.
Cf. A000146, A002882, A003245, A127187, A127188, A074909, A164555, A176327, A190339, A131689.
Cf. A094816, A130189, A130190, A155585.

Programs

  • Magma
    [Numerator(Bernoulli(n)): n in [0..40]]; // Vincenzo Librandi, Mar 17 2014
  • Maple
    B := n -> add((-1)^m*m!*Stirling2(n, m)/(m+1), m=0..n);
B := n -> bernoulli(n);
seq(numer(bernoulli(n)), n=0..40); # Zerinvary Lajos, Apr 08 2009
  • Mathematica
    Table[ Numerator[ BernoulliB[ n]], {n, 0, 40}] (* Robert G. Wilson v, Oct 11 2004 *)
Numerator[ Range[0, 40]! CoefficientList[ Series[x/(E^x - 1), {x, 0, 40}], x]]
Numerator[CoefficientList[Series[PolyGamma[1, 1/x]/x - x, {x, 0, 40}, Assumptions -> x > 0], x]] (* Vladimir Reshetnikov, Apr 24 2013 *)
  • Maxima
    B(n):=(-1)^((n))*sum((stirling1(n,k)*stirling2(n+k,n))/binomial(n+k,k),k,0,n);
makelist(num(B(n)),n,0,20); /* Vladimir Kruchinin, Mar 16 2013 */
  • PARI
    a(n)=numerator(bernfrac(n))
  • Python
    from sympy import bernoulli
from fractions import Fraction
[bernoulli(i).as_numer_denom()[0] for i in range(51)]  # Indranil Ghosh, Mar 18 2017
  • Python
    from sympy import bernoulli
def A027641(n): return bernoulli(n).p
print([A027641(n) for n in range(80)])  # M. F. Hasler, Jun 11 2019
  • SageMath
    [bernoulli(n).numerator() for n in range(41)]  # Peter Luschny, Feb 19 2016
  • SageMath
    # Alternatively:
def A027641_list(len):
    f, R, C = 1, [1], [1]+[0]*(len-1)
    for n in (1..len-1):
        f *= n
        for k in range(n, 0, -1):
            C[k] = C[k-1] / (k+1)
        C[0] = -sum(C[k] for k in (1..n))
        R.append((C[0]*f).numerator())
    return R
A027641_list(41)  # Peter Luschny, Feb 20 2016

Formula

E.g.f: x/(exp(x) - 1); take numerators.
Recurrence: B^n = (1+B)^n, n >= 2 (interpreting B^j as B_j).
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k>=1} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/(2*Pi)^(2*n).
Sum_{i=1..n-1} i^k = ((n+B)^(k+1)-B^(k+1))/(k+1) (interpreting B^j as B_j).
B_{n-1} = - Sum_{r=1..n} (-1)^r binomial(n, r) r^(-1) Sum_{k=1..r} k^(n-1). More concisely, B_n = 1 - (1-C)^(n+1), where C^r is replaced by the arithmetic mean of the first r n-th powers of natural numbers in the expansion of the right-hand side. [Bergmann]
Sum_{i>=1} 1/i^(2k) = zeta(2k) = (2*Pi)^(2k)*|B_{2k}|/(2*(2k)!).
B_{2n} = (-1)^(m-1)/2^(2m+1) * Integral{-inf..inf, [d^(m-1)/dx^(m-1) sech(x)^2 ]^2 dx} (see Grosset/Veselov).
Let B(s,z) = -2^(1-s)(i/Pi)^s s! PolyLog(s,exp(-2*i*Pi/z)). Then B(2n,1) = B_{2n} for n >= 1. Similarly the numbers B(2n+1,1), which might be called Co-Bernoulli numbers, can be considered, and it is remarkable that Leonhard Euler in 1755 already calculated B(3,1) and B(5,1) (Opera Omnia, Ser. 1, Vol. 10, p. 351). (Cf. the Luschny reference for a discussion.) - Peter Luschny, May 02 2009
The B_n sequence is the left column of the inverse of triangle A074909, the "beheaded" Pascal's triangle. - Gary W. Adamson, Mar 05 2012
From Sergei N. Gladkovskii, Dec 04 2012: (Start)
E.g.f. E(x)= 2 - x/(tan(x) + sec(x) - 1)= Sum_{n>=0} a(n)*x^n/n!, a(n)=|B(n)|, where B(n) is Bernoulli number B_n.
E(x)= 2 + x - B(0), where B(k)= 4*k+1 + x/(2 + x/(4*k+3 - x/(2 - x/B(k+1)))); (continued fraction, 4-step). (End)
E.g.f.: x/(exp(x)-1)= U(0); U(k)= 2*k+1 - x(2*k+1)/(x + (2*k+2)/(1 + x/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 05 2012
E.g.f.: 2*(x-1)/(x*Q(0)-2) where Q(k) = 1 + 2*x*(k+1)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)^2/(x*(2*k+3) + 4*(k+1)*(k+2)/Q(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 26 2013
a(n) = numerator(B(n)), B(n) = (-1)^n*Sum_{k=0..n} Stirling1(n,k) * Stirling2(n+k,n) / binomial(n+k,k). - Vladimir Kruchinin, Mar 16 2013
E.g.f.: x/(exp(x)-1) = E(0) where E(k) = 2*k+1 - x/(2 + x/E(k+1)); (continued fraction). - Sergei N. Gladkovskii, Mar 16 2013
G.f. for Bernoulli(n) = a(n)/A027642(n): psi_1(1/x)/x - x, where psi_n(z) is the polygamma function, psi_n(z) = (d/dz)^(n+1) log(Gamma(z)). - Vladimir Reshetnikov, Apr 24 2013
E.g.f.: 2*E(0) - 2*x, where E(k)= x + (k+1)/(1 + 1/(1 - x/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 10 2013
B_n = Sum_{m=0..n} (-1)^m *A131689(n, m)/(m + 1), n >= 0. See one of the Maple programs. - Wolfdieter Lang, May 05 2017
a(n) = numerator((-1)^n*A155585(n-1)*n/(4^n-2^n)), for n>=1. - Mats Granvik, Nov 26 2017
From Artur Jasinski, Dec 30 2020: (Start)
a(n) = numerator(-2*cos(Pi*n/2)*Gamma(n+1)*zeta(n)/(2*Pi)^n), for n=0 and n>1.
a(n) = numerator(-n*zeta(1-n)), for n=0 and n>1. (End)
a(n) = numerator(Sum_{k=0..n-1} (-1)^(k-1)*k!*Stirling2(n-1,k) / ((k+1)*(k+2))), for n>0 (see Jha link). - Bill McEachen, Jul 17 2025

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