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.

Showing 1-3 of 3 results.

A225117 Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 3}(x) in descending order.

Original entry on oeis.org

1, 2, 1, 4, 13, 1, 8, 93, 60, 1, 16, 545, 1131, 251, 1, 32, 2933, 14498, 10678, 1018, 1, 64, 15177, 154113, 262438, 88998, 4089, 1, 128, 77101, 1475736, 4890287, 3870352, 692499, 16376, 1, 256, 388321, 13270807, 77404933, 117758659, 50476003, 5175013, 65527, 1
Offset: 0

Views

Author

Peter Luschny, May 02 2013

Keywords

Comments

The row sums equal the triple factorial numbers A032031 and the alternating row sums, i.e., Sum_{k=0..n}(-1)^k*T(n,k), are up to a sign A000810. - Johannes W. Meijer, May 04 2013

Examples

			[0]  1
[1]  2*x   +   1
[2]  4*x^2 +  13*x   +    1
[3]  8*x^3 +  93*x^2 +   60*x   +   1
[4] 16*x^4 + 545*x^3 + 1131*x^2 + 251*x + 1
...
The triangle T(n, k) begins:
n \ k 0      1        2        3       4      5     6 7 ...
0:    1
1:    2      1
2:    4     13        1
3:    8     93       60        1
5:   16    545     1131      251       1
6:   32   2933    14498    10678    1018      1
7:   64  15177   154113   262438   88998   4089     1
8:  128  77101  1475736  4890287 3870352 692499 16376 1
...  - _Wolfdieter Lang_, Apr 08 2017
Three term recurrence: T(2,1) = (3*(2-1)+1)*2 + (3*1+2)*1 = 13. - _Wolfdieter Lang_, Apr 10 2017
		

Crossrefs

Coefficients of A_{n,1}(x) = A008292, coefficients of A_{n,2}(x) = A060187, coefficients of A_{n,4}(x) = A225118.

Programs

  • Maple
    gf := proc(n, k) local f; f := (x,t) -> x*exp(t*x/k)/(1-x*exp(t*x));
    series(f(x,t), t, n+2); ((1-x)/x)^(n+1)*k^n*n!*coeff(%, t, n):
    collect(simplify(%), x) end:
    seq(print(seq(coeff(gf(n, 3), x, n-k), k=0..n)), n=0..6);
    # Recurrence
    P := proc(n,x) option remember; if n = 0 then 1 else
      (n*x+(1/3)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
      expand(%) fi end:
    A225117 := (n,k) -> 3^n*coeff(P(n,x),x,n-k):
    seq(print(seq(A225117(n,k), k=0..n)), n=0..5);  # Peter Luschny, Mar 08 2014
  • Mathematica
    gf[n_, k_] := Module[{f, s}, f[x_, t_] := x*Exp[t*x/k]/(1-x*Exp[t*x]); s = Series[f[x, t], {t, 0, n+2}]; ((1-x)/x)^(n+1)*k^n*n!*SeriesCoefficient[s, {t, 0, n}]]; Table[Table[SeriesCoefficient[gf[n, 3], {x, 0, n-k}], {k, 0, n}], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
  • PARI
    T(n, k) = sum(j=0, n - k, (-1)^(n - k - j)*binomial(n + 1, n - k - j)*(1 + 3*j)^n);
    for(n=0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 10 2017
    
  • Python
    from sympy import binomial
    def T(n,k): return sum((-1)**(n - k - j)* binomial(n + 1, n - k - j)*(1 + 3*j)**n for j in range(n - k + 1))
    for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 10 2017
  • Sage
    @CachedFunction
    def EB(n, k, x):  # Modified cardinal B-splines
        if n == 1: return 0 if (x < 0) or (x >= 1) else 1
        return k*x*EB(n-1, k, x) + k*(n-x)*EB(n-1, k, x-1)
    def EulerianPolynomial(n, k): # Generalized Eulerian polynomials
        R. = ZZ[]
        if x == 0: return 1
        return add(EB(n+1, k, m+1/k)*x^m for m in (0..n))
    [EulerianPolynomial(n, 3).coefficients()[::-1] for n in (0..5)]
    

Formula

Generating function of the polynomials is gf(n, k) = k^n*n!*(1/x-1)^(n+1)[t^n](x*e^(t*x/k)*(1-x*e(t*x))^(-1)) for k = 3; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
From Wolfdieter Lang, Apr 10 2017: (Start)
T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*binomial(n+1, n-k-j)*(1+3*j)^n, 0 <= k <= n.
T(n, k) = Sum_{m=0..n-k} (-1)^(n-k-m)*binomial(n-m, k)*A284861(n, m), 0 <= k <= n.
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k are R(n, x) = (x-1)^n*Sum_{m=0} A284861(n, m)*(1/(x-1))^m, n >= 0, i.e. the row polynomials of A284861 in the variable 1/(x-1) multiplied by (x-1)^n.
The row polynomials with falling powers are P(n, x) = (1-x)^n*Sum_{m=0..n} A284861(n, m)*(x/(1-x))^m, n >= 0.
The e.g.f. of the row polynomials in falling powers of x (A_{n, 3}(x) of the name) is exp((1-x)*z)/(1 - (x/(1 - x)) * (exp(3*(1-x)*z) - 1)) = (1-x)*exp((1-x)*z)/(1 - x*exp(3*(1-x)*z)).
The e.g.f. of the row polynomials R(n, x) (rising powers of x) is then (1-x)*exp(2*(1-x)*z)/(1 - x*exp(3*(1-x)*z)).
Three term recurrence: T(n, k) = 0 if n < k , T(n, -1) = 0, T(0,0) = 1, T(n, k) = (3*(n-k)+1)*T(n-1, k-1) + (3*k+2)*T(n-1, k) for n >= 1, k=0..n. (End)

A225147 a(n) = Im((1-I)^(1-n)*A_{n, 3}(I)) where A_{n, k}(x) are the generalized Eulerian polynomials.

Original entry on oeis.org

-1, 2, 5, -46, -205, 3362, 22265, -515086, -4544185, 135274562, 1491632525, -54276473326, -718181418565, 30884386347362, 476768795646785, -23657073914466766, -417370516232719345, 23471059057478981762, 465849831125196593045, -29279357851856595135406
Offset: 0

Views

Author

Peter Luschny, Apr 30 2013

Keywords

Crossrefs

Cf. A000810 (real part (up to sign)), A212435 (k=2), A122045 (k=1), A002439.
Bisections are A002438 and A000191.

Programs

  • Maple
    B := proc(n, u, k) option remember;
    if n = 1 then if (u < 0) or (u >= 1) then 0 else 1 fi
    else k*u*B(n-1, u, k) + k*(n-u)*B(n-1, u-1, k) fi end:
    EulerianPolynomial := proc(n, k, x) local m; if x = 0 then RETURN(1) fi;
    add(B(n+1, m+1/k, k)*u^m, m = 0..n); subs(u=x, %) end:
    seq(Im((1-I)^(1-n)*EulerianPolynomial(n, 3, I)), n=0..19);
  • Mathematica
    CoefficientList[Series[-E^(-2*x)*Sech[3*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Sep 29 2014 after Sergei N. Gladkovskii *)
    Table[-6^n EulerE[n,1/6], {n,0,19}] (* Peter Luschny, Nov 16 2016 after Peter Bala *)
  • Sage
    from mpmath import mp, polylog, im
    mp.dps = 32; mp.pretty = True
    def A225147(n): return im(-2*I*(1+add(binomial(n,j)*polylog(-j,I)*3^j for j in (0..n))))
    [int(A225147(n)) for n in (0..19)]

Formula

a(n) = Im(-2*i*(1+Sum_{j=0..n}(binomial(n,j)*Li{-j}(i)*3^j))).
For a recurrence see the Maple program.
G.f.: conjecture -T(0)/(1+2*x), where T(k) = 1 - 9*x^2*(k+1)^2/(9*x^2*(k+1)^2 + (1+2*x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013
a(n) = -(-3)^n*skp(n, 2/3), where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 19 2014
G.f.: A225147 = -1/T(0), where T(k) = 1 + 2*x + (k+1)^2*(3*x)^2/ T(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 29 2014
E.g.f.: -exp(-2*x)*sech(3*x). - Sergei N. Gladkovskii, Sep 29 2014
a(n) ~ n! * (sqrt(3)*sin(Pi*n/2) - cos(Pi*n/2)) * 2^(n+1) * 3^n / Pi^(n+1). - Vaclav Kotesovec, Sep 29 2014
From Peter Bala, Nov 13 2016: (Start)
a(n) = - 6^n*E(n,1/6), where E(n,x) denotes the Euler polynomial of order n.
a(2*n) = (-1)^(n+1)*A002438(n); a(2*n+1) = (1/2)*(-1)^n*A002439(n). (End)

A378066 Array read by ascending antidiagonals: A(n, k) = (-2*n)^k * Euler(k, (n - 1)/(2*n)) for n >= 1 and A(0, k) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, -3, -2, 1, 1, 1, -8, -11, 0, 1, 1, 1, -15, -26, 57, 16, 1, 1, 1, -24, -47, 352, 361, 0, 1, 1, 1, -35, -74, 1185, 1936, -2763, -272, 1, 1, 1, -48, -107, 2976, 6241, -38528, -24611, 0, 1
Offset: 0

Views

Author

Peter Luschny, Nov 15 2024

Keywords

Comments

This is the counterpart of A377666, where A(1, n) are the secant numbers A122045(n). Here A(1, n) are the tangent numbers A155585(n).

Examples

			Array starts:
  [0]  1, 1,   1,    1,     1,     1,        1, ...  A000012
  [1]  1, 1,   0,   -2,     0,    16,        0, ...  A155585
  [2]  1, 1,  -3,  -11,    57,   361,    -2763, ...  A188458
  [3]  1, 1,  -8,  -26,   352,  1936,   -38528, ...  A000810
  [4]  1, 1, -15,  -47,  1185,  6241,  -230895, ...  A000813
  [5]  1, 1, -24,  -74,  2976, 15376,  -906624, ...  A378065
  [6]  1, 1, -35, -107,  6265, 32041, -2749355, ...
  [7]  1, 1, -48, -146, 11712, 59536, -6997248, ...
		

Crossrefs

Columns: A005563 (k=2), A080663 (k=3), A378064 (k=4).
Cf. A378063 (main diagonal), A377666 (secant), A081658 (column generating polynomials).

Programs

  • Maple
    A := (n, k) -> ifelse(n = 0, 1, (-2*n)^k * euler(k, (n - 1) / (2*n))):
    for n from 0 to 7 do seq(A(n, k), k = 0..9) od; # row by row
    # Alternative:
    A := proc(n, k) local j; add(binomial(k, j)*euler(j, 1/2)*(-2*n)^j, j = 0..k) end: seq(seq(A(n - k, k), k = 0..n), n = 0..10);
    # Using generating functions:
    egf := n -> exp(x)/cosh(n*x): ser := n -> series(egf(n), x, 14):
    row := n -> local k; seq(k!*coeff(ser(n), x, k), k = 0..7):
    seq(lprint(row(n)), n = 0..7);

Formula

A(n, k) = k! * [x^k] exp(x)/cosh(n*x).
A(n, k) = Sum_{j = 0..k} binomial(k, j) * Euler(j, 1/2) *(-2*n)^j.
Showing 1-3 of 3 results.