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scipy.special.

# bernoulli

scipy.special.bernoulli(n)

  Bernoulli numbers B0..Bn (inclusive).

  Parameters:
    nint

      Indicated the number of terms in the Bernoulli series to generate.

  Returns:
    ndarray

      The Bernoulli numbers `[B(0), B(1), ..., B(n)]`.

  References

  [1]

  Zhang, Shanjie and Jin, Jianming. “Computation of Special Functions”, John Wiley and Sons, 1996. https://people.sc.fsu.edu/~jburkardt/f77_src/special_functions/special_functions.html

  [2]

  “Bernoulli number”, Wikipedia, https://en.wikipedia.org/wiki/Bernoulli_number

  Examples

      >>> import numpy as np
      >>> from scipy.special import bernoulli, zeta
      >>> bernoulli(4)
      array([ 1.        , -0.5       ,  0.16666667,  0.        , -0.03333333])

  The Wikipedia article ([2]) points out the relationship between the Bernoulli numbers and the zeta function, `B_n^+ = -n * zeta(1 - n)` for `n > 0`:

      >>> n = np.arange(1, 5)
      >>> -n * zeta(1 - n)
      array([ 0.5       ,  0.16666667, -0.        , -0.03333333])

  Note that, in the notation used in the wikipedia article, `bernoulli` computes `B_n^-` (i.e. it used the convention that `B_1` is -1/2). The relation given above is for `B_n^+`, so the sign of 0.5 does not match the output of `bernoulli(4)`.