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.

A060096 Numerator of coefficients of Euler polynomials (rising powers).

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 1, 0, -3, 1, 0, 1, 0, -2, 1, -1, 0, 5, 0, -5, 1, 0, -3, 0, 5, 0, -3, 1, 17, 0, -21, 0, 35, 0, -7, 1, 0, 17, 0, -28, 0, 14, 0, -4, 1, -31, 0, 153, 0, -63, 0, 21, 0, -9, 1, 0, -155, 0, 255, 0, -126, 0, 30, 0, -5, 1, 691, 0, -1705, 0, 2805, 0, -231, 0, 165, 0, -11, 1, 0, 2073, 0, -3410, 0, 1683, 0, -396
Offset: 0

Views

Author

Wolfdieter Lang, Mar 29 2001

Keywords

Comments

From S. Roman, The Umbral Calculus (see the reference in A048854), p. 101, (4.2.10) (corrected): E(n,x)= sum(sum(binomial(n,m)*((-1/2)^j)*j!*S2(n-m,j),j=0..k)*x^m,m=0..n), with S2(n,m)=A008277(n,m) and S2(n,0)=1 if n=0 else 0 (Stirling2).
From Wolfdieter Lang, Oct 31 2011: (Start)
This is the Sheffer triangle (2/(exp(x)+1),x) (which would be called in the above mentioned S. Roman reference Appell for (exp(t)+1)/2) (see p. 27).
The e.g.f. for the row sums is 2/(1+exp(-x)). The row sums look like A198631(n)/A006519(n+1), n>=0.
The e.g.f. for the alternating row sums is 2/(exp(x)*(exp(x)+1)). These sums look like (-1)^n*A143074(n)/ A006519(n+1).
The e.g.f. for the a-sequence of this Sheffer array is 1. The z-sequence has e.g.f. (1-exp(x))/(2*x). This z-sequence is -1/(2*A000027(n))=-1/(2*(n+1)) (see the link under A006232 for the definition of a- and z-sequences). This leads to the recurrences given below.
The alternating power sums for the first n positive integers are given by sum((-1)^(n-j)*j^k,j=1..n) = (E(k, x=n+1)+(-1)^n*E(k, x=0))/2, k>=1, n>=1,with the row polynomials E(n, x)(see the Abramowitz-Stegun reference, p. 804, 23.1.4, and an addendum in the W. Lang link under A196837).
(End)

Examples

			n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1:  -1    1
2:   0   -1    1
3:   1    0   -3    1
4:   0    1    0   -2    1
5:  -1    0    5    0   -5    1
6:   0   -3    0    5    0   -3    1
7:  17    0  -21    0   35    0   -7    1
8:   0   17    0  -28    0   14    0   -4  1
...
The rational triangle a(n,m)/A060097(n,m) starts
n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1: -1/2   1
2:   0   -1    1
3:  1/4   0  -3/2   1
4:   0    1    0   -2    1
5: -1/2   0   5/2   0  -5/2   1
6:   0   -3    0    5    0   -3    1
7: 17/8   0 -21/2   0  35/4   0  -7/2   1
8:   0   17    0  -28    0   14    0   -4  1
...
		

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. 809.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 20, equations 20:4:1 - 20:4:8 at pages 177-178.

Crossrefs

Cf. A060097.

Programs

  • Maple
    A060096 := proc(n,m) coeff(euler(n,x),x,m) ; numer(%) ;end proc:
    seq(seq(A060096(n,m),m=0..n),n=0..12) ; # R. J. Mathar, Dec 21 2010
  • Mathematica
    Numerator[Flatten[Table[CoefficientList[EulerE[n, x], x], {n, 0, 12}]]] (* Jean-François Alcover, Apr 29 2011 *)

Formula

E(n, x)= sum((a(n, m)/b(n, m))*x^m, m=0..n), denominators b(n, m)= A060097(n, m).
From Wolfdieter Lang, Oct 31 2011: (Start)
E.g.f. for E(n, x) is 2*exp(x*z)/(exp(z)+1).
E.g.f. of column no. m, m>=0, is 2*x^{m+1}/(m!*(exp(x)+1)).
Recurrences for E(n,m):=a(n,m)/A060097(n,m) from the Sheffer a-and z-sequence:
E(n,m)=(n/m)*E(n-1,m-1), n>=1,m>=1.
E(n,0)=-n*sum(E(n-1,j)/(2*(j+1)),j=0..n-1), n>=1, E(0,0)=1.
(see the Sheffer comments above).
(End)
E(n,m) = binomial(n,m)*sum(((-1)^j)*j!*S2(n-m,j)/2^j ,j=0..n-m), 0<=m<=n, with S2 given by A008277. From S. Roman, The umbral calculus, reference under A048854, eq. (4.2.10), p. 101, with a=1, and a misprint corrected: replace 1/k! by binomial(n,k) (also in the two preceding formulas). - Wolfdieter Lang, Nov 03 2011
The first (m=0) column of the rational triangle is conjectured to be E(n,0) = ((-1)^n)*A198631(n) / A006519(n+1). See also the first column shown in A209308 (different signs). - Wolfdieter Lang, Jun 15 2015

Extensions

Table rewritten by Wolfdieter Lang, Oct 31 2011