%I #33 Mar 15 2022 18:00:41
%S 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,
%T -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,
%U -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
%N Numerator of coefficients of Euler polynomials (rising powers).
%C 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).
%C From _Wolfdieter Lang_, Oct 31 2011: (Start)
%C 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).
%C 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.
%C 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).
%C 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.
%C 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).
%C (End)
%D 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.
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%F E(n, x)= sum((a(n, m)/b(n, m))*x^m, m=0..n), denominators b(n, m)= A060097(n, m).
%F From _Wolfdieter Lang_, Oct 31 2011: (Start)
%F E.g.f. for E(n, x) is 2*exp(x*z)/(exp(z)+1).
%F E.g.f. of column no. m, m>=0, is 2*x^{m+1}/(m!*(exp(x)+1)).
%F Recurrences for E(n,m):=a(n,m)/A060097(n,m) from the Sheffer a-and z-sequence:
%F E(n,m)=(n/m)*E(n-1,m-1), n>=1,m>=1.
%F E(n,0)=-n*sum(E(n-1,j)/(2*(j+1)),j=0..n-1), n>=1, E(0,0)=1.
%F (see the Sheffer comments above).
%F (End)
%F 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
%F 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
%e n\m 0 1 2 3 4 5 6 7 8 ...
%e 0: 1
%e 1: -1 1
%e 2: 0 -1 1
%e 3: 1 0 -3 1
%e 4: 0 1 0 -2 1
%e 5: -1 0 5 0 -5 1
%e 6: 0 -3 0 5 0 -3 1
%e 7: 17 0 -21 0 35 0 -7 1
%e 8: 0 17 0 -28 0 14 0 -4 1
%e ...
%e The rational triangle a(n,m)/A060097(n,m) starts
%e n\m 0 1 2 3 4 5 6 7 8 ...
%e 0: 1
%e 1: -1/2 1
%e 2: 0 -1 1
%e 3: 1/4 0 -3/2 1
%e 4: 0 1 0 -2 1
%e 5: -1/2 0 5/2 0 -5/2 1
%e 6: 0 -3 0 5 0 -3 1
%e 7: 17/8 0 -21/2 0 35/4 0 -7/2 1
%e 8: 0 17 0 -28 0 14 0 -4 1
%e ...
%p A060096 := proc(n,m) coeff(euler(n,x),x,m) ; numer(%) ;end proc:
%p seq(seq(A060096(n,m),m=0..n),n=0..12) ; # _R. J. Mathar_, Dec 21 2010
%t Numerator[Flatten[Table[CoefficientList[EulerE[n, x], x], {n, 0, 12}]]] (* _Jean-François Alcover_, Apr 29 2011 *)
%Y Cf. A060097.
%K sign,easy,tabl,frac
%O 0,9
%A _Wolfdieter Lang_, Mar 29 2001
%E Table rewritten by _Wolfdieter Lang_, Oct 31 2011