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.

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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.

Original entry on oeis.org

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

Views

Author

Wouter Meeussen, Apr 06 2003

Keywords

Comments

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

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.

Crossrefs

Cf. A002425, A058940, A006519.
Cf. A009006, A155585, A000182, A119468.

Programs

  • 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
  • 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
Matrix inverse is A119468 and central column is A214447. - Peter Luschny, Jul 18 2012
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

A363393 Triangle read by rows. T(n, k) = [x^k] (2 - Sum_{k=0..n} binomial(n, k)*Euler(k, 1)*(-2*x)^k).

Original entry on oeis.org

1, 1, 1, 1, 2, 0, 1, 3, 0, -2, 1, 4, 0, -8, 0, 1, 5, 0, -20, 0, 16, 1, 6, 0, -40, 0, 96, 0, 1, 7, 0, -70, 0, 336, 0, -272, 1, 8, 0, -112, 0, 896, 0, -2176, 0, 1, 9, 0, -168, 0, 2016, 0, -9792, 0, 7936, 1, 10, 0, -240, 0, 4032, 0, -32640, 0, 79360, 0
Offset: 0

Views

Author

Peter Luschny, Jun 04 2023

Keywords

Comments

The Swiss-Knife polynomials (A081658 and A153641) generate the dual triangle ('dual' in the sense of Euler-tangent versus Euler-secant numbers).

Examples

			The triangle T(n, k) starts:
[0] 1;
[1] 1, 1;
[2] 1, 2, 0;
[3] 1, 3, 0,   -2;
[4] 1, 4, 0,   -8, 0;
[5] 1, 5, 0,  -20, 0,   16;
[6] 1, 6, 0,  -40, 0,   96, 0;
[7] 1, 7, 0,  -70, 0,  336, 0,  -272;
[8] 1, 8, 0, -112, 0,  896, 0, -2176, 0;
[9] 1, 9, 0, -168, 0, 2016, 0, -9792, 0, 7936;

Links

Crossrefs

Cf. A122045 (alternating row sums), A119880 (row sums), A214447 (central column), A155585 (main diagonal), A109573 (subdiagonal), A162660 (variant), A000364.
Cf. A081658, A153641, A220901, A318254, A346838.

Programs

  • Maple
    P := n -> add(binomial(n + 1, j)*bernoulli(j, 1)*(4^j - 2^j)*x^(j-1), j = 0..n+1) / (n + 1):  T := (n, k) -> coeff(P(n), x, k):
seq(seq(T(n, k), k = 0..n), n = 0..9);
# Second program, based on the generating functions of the columns:
ogf := n -> -(-2)^n * euler(n, 1) / (x - 1)^(n + 1):
ser := n -> series(ogf(n), x, 16):
T := (n, k) -> coeff(ser(k), x, n - k):
for n from 0 to 9 do seq(T(n, k), k = 0..n) od;
# Alternative, based on the bivariate generating function:
egf :=  exp(x*y) * (1 + tanh(y)): ord := 20:
sery := series(egf, y, ord): polx := n -> coeff(sery, y, n):
coefx := n -> seq(n! * coeff(polx(n), x, n - k), k = 0..n):
for n from 0 to 9 do coefx(n) od;
  • Python
    from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k == 0: return 1
    if k % 2 == 0:  return 0
    if k == n: return 1 - sum(T(n, j) for j in range(1, n, 2))
    return (T(n - 1, k) * n) // (n - k)
for n in range(10): print([T(n, k) for k in range(n + 1)])
  • SageMath
    def B(n: int):
    return bernoulli_polynomial(1, n)
def P(n: int):
    return sum(binomial(n + 1, j) * B(j) * (4^j - 2^j) * x^(j - 1)
           for j in range(n + 2)) / (n + 1)
for n in range(10): print(P(n).list())

Formula

For a recursion see the Python program.
T(n, k) = [x^k] P(n, x) where P(n, x) = (1 / (n + 1)) * Sum_{j=0..n+1} binomial(n + 1, j) * Bernoulli(j, 1) * (4^j - 2^j) * x^(j - 1).
Integral_{x=-n..n} P(n, x)/2 dx = n.
T(n, k) = [x^(n - k)] -(-2)^k * Euler(k, 1) / (x - 1)^(k + 1).
T(n, k) = n! * [x^(n - k)][y^n] exp(x*y) * (1 + tanh(y)).

Extensions

Simpler name by Peter Luschny, Nov 17 2024
Showing 1-2 of 2 results.

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