cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A006569 Numerators of generalized Bernoulli numbers.

Original entry on oeis.org

1, -1, 1, 1, -1, -5, -1, 7, 13, -307, -479, 1837, 100921, 15587, -23737, -5729723, 14731223, 9129833, 2722952839, -4700745901, -1556262845, 190717213397, 24684889339847, -50242799489, -148437433077277, -8592042383621, 221844330989749, 176585172615885307, -9245931549625447
Offset: 0

Views

Author

Keywords

References

  • F. T. Howard, A sequence of numbers related to the exponential function, Duke Math. J., 34 (1967), 599-615.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    eq:=n->bernoulli(n+1)=a[n+1]-sum(binomial(n+1,r)*bernoulli(r)*a[n+2-r],r=1..n+1): a[0]:=1:for n from 0 to 28 do a[n+1]:=solve(eq(n),a[n+1]) od: seq(numer(a[n]),n=0..29); # Emeric Deutsch, Jan 23 2005
  • Mathematica
    rows = 29; M = Table[If[n-1 <= k <= n, 0, Binomial[n, k]], {n, 2, rows+1}, {k, 0, rows-1}] // Inverse;
    M[[All, 1]] // Numerator (* Jean-François Alcover, Jul 14 2018 *)
  • Sage
    def A006569_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+2)
            C[0] = -sum(C[k] for k in (1..n))
            R.append((C[0]*f).numerator())
        return R
    print(A006569_list(29)) # Peter Luschny, Feb 20 2016

Formula

Recurrence relation: Bernoulli(n+1) = a(n+1) - Sum_{r=1..n+1} binomial(n+1, r)*Bernoulli(r)*a(n+2-r); a(0)=1 (p. 603 of the Howard reference). - Emeric Deutsch, Jan 23 2005
E.g.f. for fractions: x^2/2 / (e^x-1-x). - Franklin T. Adams-Watters, Nov 04 2009

Extensions

More terms from Emeric Deutsch, Jan 23 2005

A209518 Triangle by rows, reversal of A104712.

Original entry on oeis.org

1, 1, 3, 1, 4, 6, 1, 5, 10, 10, 1, 6, 15, 20, 15, 1, 7, 21, 35, 35, 21, 1, 8, 28, 56, 70, 56, 28, 1, 9, 36, 84, 126, 126, 84, 36, 1, 10, 45, 120, 210, 252, 210, 120, 45, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55
Offset: 0

Views

Author

Gary W. Adamson, Mar 09 2012

Keywords

Comments

The offset is chosen as "0" to match the generalized or compositional Bernoulli numbers.
Following [Blandin and Diaz], we can generalize a subset of Bernoulli numbers to comply with the origin of the triangle (the Pascal matrix A007318 beheaded once: (A074909), twice: (this triangle), and so on...); and a corresponding Bernoulli sequence that equals the inverse of the triangle, extracting the left border. This procedure done with A074909 results in The Bernoulli numbers (A027641/A026642) starting (1, -1/2, 1/6,...). Done with this triangle we obtain A006568/A006569: (1, -1/3, 1/18, 1/90,...).
A generalized algebraic property of the subset of such triangles and compositional Bernoulli numbers is that the triangle M * [corresponding Bernoulli sequence considered as a vector, V] = [1, 0, 0, 0,...].
The infinite set of generalized Bernoulli number sequences thus generated from variants of Pascal's triangle begins: [(1, -1/2, 1/6,...); (1, -1/3, 1/18,...); (1, -1/4, 1/40,...); (1, -1/5, 1/75,...); where the third term denominators = A002411 (1, 6, 18, 40, 75,...) after the "1".
Row sums of the triangle = A000295 starting (1, 4, 11, 26, 57,...).

Examples

			First few rows of the triangle =
1;
1, 3;
1, 4, 6;
1, 5, 10, 10;
1, 6, 15, 20, 15;
1, 7, 21, 35, 35, 21;
1, 8, 28, 56, 70, 56, 28;
1, 9, 36, 84, 126, 126, 84, 36;
1, 10, 45, 120, 210, 252, 210, 120, 45;
1, 11, 55, 165, 330, 462, 462, 30, 165, 55;
...
		

Crossrefs

Programs

  • Mathematica
    Table[Binomial[n+2, k+2], {n, 0, 9}, {k, n , 0, -1}] // Flatten (* Jean-François Alcover, Aug 08 2018 *)

Formula

Doubly beheaded variant of Pascal's triangle in which two rightmost diagonals are deleted.
T(n,k)=T(n-1,k)+3*T(n-1,k-1)-2*T(n-2,k-1)-3*T(n-2,k-2)+T(n-3,k-2)+T(n-3,k-3), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
Showing 1-2 of 2 results.