A081733 Triangle read by rows, T(n,k) = 2^(n-k)*[x^k] Euler_polynomial(n, x), for n >= 0, k >= 0.
1, -1, 1, 0, -2, 1, 2, 0, -3, 1, 0, 8, 0, -4, 1, -16, 0, 20, 0, -5, 1, 0, -96, 0, 40, 0, -6, 1, 272, 0, -336, 0, 70, 0, -7, 1, 0, 2176, 0, -896, 0, 112, 0, -8, 1, -7936, 0, 9792, 0, -2016, 0, 168, 0, -9, 1, 0, -79360, 0, 32640, 0, -4032, 0, 240, 0, -10, 1, 353792, 0, -436480, 0, 89760, 0, -7392, 0, 330, 0, -11, 1, 0, 4245504, 0
Offset: 0
Examples
The coefficient lists of the first 5 Euler polynomials are {1}, {-1/2, 1}, {0, -1, 1}, {1/4, 0, -3/2, 1}, {0, 1, 0, -2, 1}. Multiply by 2^(n-k) to get
1,
-1, 1,
0, -2, 1,
2, 0, -3, 1,
0, 8, 0, -4, 1.
Programs
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Maple
T := (n,k) -> 2^(n-k)*coeff(euler(n,x),x,k): T := (n,k) -> 2^(n-k)*binomial(n,k)*euler(n-k,1): # Peter Luschny, Jan 25 2009
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Mathematica
Table[2^n (1/2)^(Range[0, n]) CoefficientList[EulerE[n, x], x], {n, 0, 16}] -
Sage
def A081733(n, k) : return (-2)^(n-k)*binomial(n,k)*euler_polynomial(n-k,1) # Peter Luschny, Jul 18 2012
Formula
T(n,k) = C(n,k)*2^(n-k)*E_{n-k}(0) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
Let skp{n}(x) denote the Swiss-Knife polynomials A153641. Then T(n,k) = [x^(n-k)]((skp{n}(x-1) - skp{n}(x+1))/2 + x^n). - Peter Luschny, Jul 22 2012
E.g.f.: exp(z*x)*(1-tanh(x)). - Peter Luschny, Aug 01 2012
E.g.f.: [2/(e^(2t)+1)] e^(tx) = e^(P.(x)t), so this is an Appell sequence with lowering operator D = d/dx and raising operator R = x - 2/[e^(-2D)+1], i.e., D P_n(x) = n P_{n-1}(x) and R p_n(x) = P_{n+1}(x). Also, (P.(x)+y)^n = P_n(x+y), umbrally. R = x - 1 - D + 2 D^3/3! + ... contains the e.g.f.(D) mod signs of A009006 and A155585 and signed, aerated A000182, the zag numbers, and P_n(0) are the coefficients (mod signs/shift) of these sequences. The polynomials PI_n(x) of A119468 are the umbral compositional inverses of this sequence, i.e., P_n(PI.(x)) = x^n = PI_n(P.(x)) under umbral composition. Note that 2/[e^(2t)+1] = 2 Sum_{n >= 0} Eta(-n) (-2t)^n/n!], where Eta(s) is the Dirichlet eta function, and b_n = 2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers, so P_n(x) = (b. + x)^n, as an Appell polynomial. - Tom Copeland, Sep 27 2015
Extensions
Corrected T(0,0) = Euler(0) = 1 (was 0), Peter Luschny, Sep 30 2010
New name from Peter Luschny, Jul 18 2012
Comments