A076743 -id:A076743 - cp's OEIS frontend

A076256 Coefficients of the polynomials in the numerator of 1/(1+x^2) and its successive derivatives, starting with the constant term.

1, 0, -2, -2, 0, 6, 0, 24, 0, -24, 24, 0, -240, 0, 120, 0, -720, 0, 2400, 0, -720, -720, 0, 15120, 0, -25200, 0, 5040, 0, 40320, 0, -282240, 0, 282240, 0, -40320, 40320, 0, -1451520, 0, 5080320, 0, -3386880, 0, 362880, 0, -3628800, 0, 43545600, 0, -91445760, 0, 43545600, 0, -3628800

Offset: 0

Mohammad K. Azarian, Nov 05 2002

Let T(n, k) be the coefficient of x^k in the numerator of the n-th derivative of 1/(1+x^2).

The denominators are (1+x^2)^(n+1), whose coefficients are the binomial coefficients, A007318.

			Triangle begins:
     1;
     0,    -2;
    -2,     0,     6;
     0,    24,     0,     -24;
    24,     0,  -240,       0,    120;
     0,  -720,     0,    2400,      0,   -720;
  -720,     0, 15120,       0, -25200,      0, 5040;
     0, 40320,     0, -282240,      0, 282240,    0, -40320;
   ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Cf. A000165, A007318, A076257, A076741, A076743.

a[n_, k_] := Coefficient[Expand[Together[(1+x^2)^(n+1)*D[1/(1+x^2), {x, n}]]], x, k]; Flatten[Table[a[n, k], {n, 0, 10}, {k, 0, n}]]

T(n,k) = if((n+k)%2, 0, (-1)^((n+k)/2)*n!*binomial(n+1, k)) \\ Andrew Howroyd, Aug 08 2024

T(n, k) = (-1)^((n+k)/2)*n!*binomial(n+1, k) for n + k even;
T(n, k) = 0 for n + k odd.
E.g.f.: A(x,t) = Sum_{n>=0} Sum_{k=0..n} T(n,k)*x^n*t^k/n! = 1/(1 + 2*x*t + x^2*(1+t^2)). - Fabian Pereyra, Aug 08 2024
From Fabian Pereyra, Sep 11 2024: (Start)
T(n,k) = -n*(n-1)*T(n-2,k) - 2*n*T(n-1,k-1) - n*(n-1)*T(n-2,k-2), with T(0,0) = 1, T(n,k) = 0 if k<0 or k>n.
Let p(n,x) the n-th polynomial in x defined by: p(n,x) = Sum_{k=0..n} T(n,k)*x^k.
Then, the p(n,x) satisfy:
p(n,x) = -2*n*x*p(n-1,x) - n*(n-1)*(1+x^2)*p(n-2,x).
p'(n,x) = -n*(n+1)*p(n-1,x).
(1+x^2)*p''(n,x) - 2*n*x*p'(n,x) + n*(n+1)*p(n,x) = 0.
Integral_{x=-inf..inf} p(n,x)*p(m,x)*(1/(1+x^2))^(max(n,m)+1) dx = n!*(n+1)!*pi* delta(n,m), where delta(n,m) is the Kronecker delta. (End)
Sum_{k=0..n} abs(T(n,k)) = A000165(n). - Alois P. Heinz, Sep 18 2024

Edited by Dean Hickerson, Nov 28 2002

A076257 Coefficients of the polynomials in the numerator of 1/(1+x^2) and its successive derivatives, starting with the coefficient of the highest power of x.

Original entry on oeis.org

1, -2, 0, 6, 0, -2, -24, 0, 24, 0, 120, 0, -240, 0, 24, -720, 0, 2400, 0, -720, 0, 5040, 0, -25200, 0, 15120, 0, -720, -40320, 0, 282240, 0, -282240, 0, 40320, 0, 362880, 0, -3386880, 0, 5080320, 0, -1451520, 0, 40320, -3628800, 0, 43545600, 0, -91445760, 0, 43545600, 0

Offset: 0

Mohammad K. Azarian, Nov 05 2002

Denominator of n-th derivative is (1+x^2)^(n+1), whose coefficients are the binomial coefficients, A007318.

			The coefficients of the numerators starting with the coefficient of the highest power of x are 1; -2,0; 6,0,-2; -24,0,24,0; ...

Cf. A076256, A076741, A076743.

a[n_, k_] := Coefficient[Expand[Together[(1+x^2)^(n+1)*D[1/(1+x^2), {x, n}]]], x, k]; Flatten[Table[a[n, k], {n, 0, 10}, {k, n, 0, -1}]]

For 0<=k<=n, let a(n, k) be the coefficient of x^k in the numerator of the n-th derivative of 1/(1+x^2). If n+k is even, a(n, k) = (-1)^((n+k)/2)*n!*binomial(n+1, k); if n+k is odd, a(n, k)=0.

Edited by Dean Hickerson, Nov 28 2002

A076741 Nonzero coefficients of the polynomials in the numerator of 1/(1+x^2) and its successive derivatives, starting with the constant term.

Original entry on oeis.org

1, -2, -2, 6, 24, -24, 24, -240, 120, -720, 2400, -720, -720, 15120, -25200, 5040, 40320, -282240, 282240, -40320, 40320, -1451520, 5080320, -3386880, 362880, -3628800, 43545600, -91445760, 43545600, -3628800, -3628800, 199584000, -1197504000

Offset: 0

Mohammad K. Azarian, Nov 11 2002

Denominator of n-th derivative is (1+x^2)^(n+1), whose coefficients are the binomial coefficients, A007318.

			The nonzero coefficients of the numerators starting with the constant term are: 1; -2; -2,6; 24,-24; ...

Roland Zumkeller, Formal global optimization with Taylor models, IJCAR (Ulrich Furbach and Natara jan Shankar, eds.), Lecture Notes in Computer Science, vol. 4130, Springer, 2006, pp. 408-422.

Roland Zumkeller, Formal global optimization with Taylor models, Preprint, 2006.
Roland Zumkeller, Formal global optimization with Taylor models, Thesis 2006.

Cf. A076256, A076257, A076743.

a[n_, k_] := Coefficient[Expand[Together[(1+x^2)^(n+1)*D[1/(1+x^2), {x, n}]]], x, k]; Select[Flatten[Table[a[n, k], {n, 0, 10}, {k, 0, n}]], #!=0&]

For 0<=k<=n, let a(n, k) be the coefficient of x^k in the numerator of the n-th derivative of 1/(1+x^2). If n+k is even, a(n, k) = (-1)^((n+k)/2)*n!*binomial(n+1, k); if n+k is odd, a(n, k)=0.

Edited by Dean Hickerson, Nov 28 2002

A131980 A coefficient tree from the list partition transform relating A000129, A000142, A000165, A110327, and A110330.

Original entry on oeis.org

1, 2, 6, 2, 24, 24, 120, 240, 24, 720, 2400, 720, 5040, 25200, 15120, 720, 40320, 282240, 282240, 40320, 362880, 3386880, 5080320, 1451520, 40320, 3628800, 43545600, 91445760, 43545600, 3628800, 39916800, 598752000, 1676505600, 1197504000, 199584000, 3628800

Offset: 0

Tom Copeland, Oct 30 2007, Nov 29 2007, Nov 30 2007

Construct the infinite array of polynomials
a(0,t) = 1
a(1,t) = 2
a(2,t) = 6 + 2 t
a(3,t) = 24 + 24 t
a(4,t) = 120 + 240 t + 24 t^2
a(5,t) = 720 + 2400 t + 720 t^2
a(6,t) = 5040 + 25200 t + 15120 t^2 + 720 t^3
This array is the reciprocal array of the following array b(n,t) under the list partition transform and its associated operations described in A133314.
b(0,t) = 1, b(1,t) = -2, b(2,t) = -2*(t-1), b(n,t) = 0 for n>2.
Then A000165(n) = a(n,1).
Lower triangular matrix A110327 = binomial(n,k)*a(n-k,2).
n! * A000129(n+1) = a(n,2) = A110327(n,0).
A110330 = matrix inverse of binomial(n,k)*a(n-k,2) = binomial(n,k)*b(n-k,2).
A000142(n+1) = a(n,0).
From Peter Bala, Sep 09 2013: (Start)
Let {P(n,x)}n>=0 be a polynomial sequence. Koutras has defined generalized Eulerian numbers associated with the sequence P(n,x) as the coefficients A(n,k) in the expansion of P(n,x) in a series of factorials of degree n, namely P(n,x) = Sum_{k=0..n} A(n,k)* binomial(x+n-k,n). The choice P(n,x) = x^n produces the classical Eulerian numbers of A008292. Let now P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial polynomial. Then the present table is the generalized Eulerian numbers associated with the polynomial sequence P(n,2*x). See A228955 for the generalized Eulerian numbers associated with the polynomial sequence P(n,2*x + 1). (End)

			Triangle begins as:
        1;
        2;
        6,        2;
       24,       24;
      120,      240,       24;
      720,     2400,      720;
     5040,    25200,    15120,      720;
    40320,   282240,   282240,    40320;
   362880,  3386880,  5080320,  1451520,   40320;
  3628800, 43545600, 91445760, 43545600, 3628800;

M. V. Koutras, Eulerian numbers associated with sequences of polynomials, The Fibonacci Quarterly, 32 (1994), 44-57.

Cf. A228955.

Flat(List([0..10], n-> List([0..Int(n/2)], k-> Factorial(n)*Binomial(n+1, 2*k+1) ))); # G. C. Greubel, Dec 30 2019

[Factorial(n)*Binomial(n+1, 2*k+1): k in [0..Floor(n/2)], n in [0..10]]; // G. C. Greubel, Dec 30 2019

for n from 0 to 10 do
seq( n!*binomial(n+1,2*k+1), k = 0..floor(n/2) )
end do; # Peter Bala, Sep 09 2013

Table[n!*Binomial[n+1, 2*k+1], {n,0,10}, {k,0,Floor[n/2]}]//Flatten (* G. C. Greubel, Dec 30 2019 *)

T(n,k) = n!*binomial(n+1, 2*k+1);
for(n=0,10, for(k=0, n\2, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 30 2019

[[factorial(n)*binomial(n+1, 2*k+1) for k in (0..floor(n/2))] for n in (0..10)] # G. C. Greubel, Dec 30 2019

E.g.f. for the polynomials b(.,t), introduced above, is 1 - 2x - (t-1) * x^2; therefore e.g.f. for the polynomials a(.,t), which are the row polynomials of this array, is 1 / ( 1 - 2x - (t-1) * x^2 ) = (t-1) / ( t - ( 1 + x*(t-1) )^2 ).
Also, a(n,t) = (1 - t*u^2)^(n+1) (D_u)^n [ 1 / (1 - t*u^2) ] with eval. at u = 1/t. Compare A076743.
a(n,t) = n!*Sum_{k>=0} binomial(n+1,2k+1) * t^k = n!*Sum_{k>=0} A034867(n,k) * t^k.
Additional relations are given by formulas in A133314.
From Peter Bala, Sep 09 2013: (Start)
Recurrence equation: T(n+1,k) = (n+2 +2*k)T(n,k) + (n +2 -2*k)T(n,k-1).
Let P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial.
T(n,k) = Sum_{j=0..k+1} (-1)^(k+1-j)*binomial(n+1,k+1-j)*P(n,2*j) for n >= 1.
The row polynomial a(n,t) satisfies t*a(n,t)/(1 - t)^(n+1) = Sum_{j>=1} P(n,2*j)*t^j. For example, for n = 3 we have t*(24 + 24*t)/(1 - t)^4 = 2*3*4*t + (4*5*6)*t^2 + (6*7*8)*t^3 + ..., while for n = 4 we have t*(120 + 240*t + 24*t^2)/(1 - t)^5 = (2*3*4*5)*t + (4*5*6*7)*t^2 + (6*7*8*9)*t^3 + .... (End)

Removed erroneous and duplicate statements. - Tom Copeland, Dec 03 2013

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A076256 Coefficients of the polynomials in the numerator of 1/(1+x^2) and its successive derivatives, starting with the constant term.

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A076257 Coefficients of the polynomials in the numerator of 1/(1+x^2) and its successive derivatives, starting with the coefficient of the highest power of x.

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A076741 Nonzero coefficients of the polynomials in the numerator of 1/(1+x^2) and its successive derivatives, starting with the constant term.

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A131980 A coefficient tree from the list partition transform relating A000129, A000142, A000165, A110327, and A110330.

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