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 A081733 Triangle read by rows, T(n,k) = 2^(n-k)*[x^k] Euler_polynomial(n, x), for n >= 0, k >= 0. 5

%I

%S 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,

%T 272,0,-336,0,70,0,-7,1,0,2176,0,-896,0,112,0,-8,1,-7936,0,9792,0,

%U -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

%N Triangle read by rows, T(n,k) = 2^(n-k)*[x^k] Euler_polynomial(n, x), for n >= 0, k >= 0.

%C Sum of row n equals Euler(n) (in the sense of the non-official version A122045; R. P. Stanley calls A000111 Euler numbers.)

%F 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

%F Matrix inverse is A119468 and central column is A214447. - _Peter Luschny_, Jul 18 2012

%F 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

%F E.g.f.: exp(z*x)*(1-tanh(x)). - _Peter Luschny_, Aug 01 2012

%F 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

%e 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

%e 1,

%e -1, 1,

%e 0, -2, 1,

%e 2, 0, -3, 1,

%e 0, 8, 0, -4, 1.

%p T := (n,k) -> 2^(n-k)*coeff(euler(n,x),x,k):

%p T := (n,k) -> 2^(n-k)*binomial(n,k)*euler(n-k,1): # _Peter Luschny_, Jan 25 2009

%t Table[2^n (1/2)^(Range[0, n]) CoefficientList[EulerE[n, x], x], {n, 0, 16}]

%o (Sage)

%o def A081733(n, k) : return (-2)^(n-k)*binomial(n,k)*euler_polynomial(n-k,1)

%o # _Peter Luschny_, Jul 18 2012

%Y Cf. A002425, A058940, A006519.

%Y Cf. A009006, A155585, A000182, A119468.

%K sign,tabl

%O 0,5

%A _Wouter Meeussen_, Apr 06 2003

%E Corrected T(0,0) = Euler(0) = 1 (was 0), _Peter Luschny_, Sep 30 2010

%E New name from _Peter Luschny_, Jul 18 2012

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Last modified February 20 09:15 EST 2019. Contains 320325 sequences. (Running on oeis4.)