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 1-5 of 5 results.

A104509 Matrix inverse of triangle A104505, which is the right-hand side of triangle A084610 of coefficients in (1 + x - x^2)^n.

Original entry on oeis.org

1, 1, -1, 3, -2, 1, 4, -6, 3, -1, 7, -12, 10, -4, 1, 11, -25, 25, -15, 5, -1, 18, -48, 60, -44, 21, -6, 1, 29, -91, 133, -119, 70, -28, 7, -1, 47, -168, 284, -296, 210, -104, 36, -8, 1, 76, -306, 585, -699, 576, -342, 147, -45, 9, -1, 123, -550, 1175, -1580, 1485, -1022, 525, -200, 55, -10, 1, 199, -979, 2310, -3454, 3641
Offset: 0

Views

Author

Paul D. Hanna, Mar 11 2005

Keywords

Comments

Riordan array ( (1 + x^2)/(1 - x - x^2), -x/(1 - x - x^2) ) belonging to the hitting time subgroup of the Riordan group (see Peart and Woan). - Peter Bala, Jun 29 2015
The sums of absolute values along steep diagonals in this triangle are: 1, 1, 3, 4 + |-1|, 7 + |-2|, 11 + |-6|, 18 + |-12| + 1, ... and these are the tribonacci numbers A000213 that begin with 1, 1, 1, 3. To see this, replace the y in the g.f. A(x,y) = (1 + x^2)/(1-x-x^2 + x*y) with y=-x^2, multiply by x, and add 1, to obtain the g.f. (1 - x^2)/(1-x-x^2-x^3) for A000213. - Noah Carey and Greg Dresden, Nov 02 2021

Examples

			Rows begin:
   1;
   1,   -1;
   3,   -2,   1;
   4,   -6,   3,   -1;
   7,  -12,  10,   -4,   1;
  11,  -25,  25,  -15,   5,   -1;
  18,  -48,  60,  -44,  21,   -6,   1;
  29,  -91, 133, -119,  70,  -28,   7,  -1;
  47, -168, 284, -296, 210, -104,  36,  -8, 1;
  76, -306, 585, -699, 576, -342, 147, -45, 9, -1; ...

Links

Crossrefs

Leftmost column is A000204 (Lucas numbers). Other columns include: A045925, A067988. Row sums are: {1,0,2,0,2,0,2,...}. Absolute row sums form: A099425. Antidiagonal sums are: {1,1,2,2,2,2,2,...}. Absolute antidiagonal sums are: A084214.
Cf. A104505, A084610, A000213.

Programs

  • Maple
    S:= series((1 + x^2)/(1-x-x^2 + x*y),x, 20):
for n from 0 to 19 do R[n]:= coeff(S,x,n) od:
seq(seq(coeff(R[n],y,j),j=0..n), n=0..19); # Robert Israel, Jun 30 2015
  • Mathematica
    nmax = 11;
T[n_, k_] := Coefficient[(1 + x - x^2)^n, x, n + k];
M = Table[T[n, k], {n, 0, nmax}, {k, 0, nmax}] // Inverse;
Table[M[[n+1, k+1]], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 27 2019 *)
  • PARI
    { T(n,k) = my(X=x+x*O(x^n),Y=y+y*O(y^k)); polcoeff(polcoeff((1 + X^2)/(1-X-X^2 + X*Y),n,x),k,y); }
  • PARI
    { tabl(nn) = my(m = matrix(nn, nn, n, k, n--; k--; if((nMichel Marcus, Jun 30 2015
  • PARI
    { A104509(n,k) = if(n==0, k==0, (-1)^k * sum(i=0, (n-k)\2, n/(n-i) * binomial(n-k-i,i) * binomial(n-i,k) )); } \\ Max Alekseyev, Oct 11 2021

Formula

For n>=1, a(n,k) = (-1)^k * Sum_{i=0..[(n-k)/2]} n/(n-i) * binomial(n-i,i) * binomial(n-2*i,k) = (-1)^k * Sum_{i=0..[(n-k)/2]} n/(n-i) * binomial(n-k-i,i) * binomial(n-i,k). - Max Alekseyev, Oct 11 2021
G.f.: A(x, y) = (1 + x^2)/(1-x-x^2 + x*y).
G.f. for column k: g_k(x) = -(x^2+1)*x^k/(x^2+x-1)^(k+1). - Robert Israel, Jun 30 2015
G.f. for row n>=1 is the Lucas polynomial L_n(1-x). - Max Alekseyev, Oct 11 2021

A104507 Row sums of triangle A104505, which is equal to the right-hand side of the triangle A084610 of coefficients in (1+x-x^2)^n.

Original entry on oeis.org

1, 0, -2, -3, 2, 15, 19, -28, -134, -129, 353, 1254, 791, -4238, -11818, -3123, 49162, 110007, -17783, -554458, -996323, 690932, 6096792, 8624747, -12287153, -65419110, -69285296, 178655307, 684550946, 483569751, -2354830741, -6970706252, -2324044054, 29195280375, 68793790705
Offset: 0

Views

Author

Paul D. Hanna, Mar 11 2005

Keywords

Crossrefs

Cf. A104505, A1086223, A246437, A370616, A386548.

Programs

  • Mathematica
    CoefficientList[Series[(x/((1 - x)) + 1/((-Sqrt[5 x^2 - 2 x + 1] + x + 1)) x (1 - (5 x - 1)/(Sqrt[5 x^2 - 2 x + 1]))), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 05 2015 *)
  • Maxima
    a(n):=sum((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j),j,0,n/2); /* Vladimir Kruchinin, Oct 04 2015 */
  • PARI
    a(n)=sum(k=0,n,polcoeff((1+x-x^2)^n,n+k))
  • PARI
    a(n) = sum(k=0, n/2, (-1)^k*binomial(n,k)*binomial(n-k-1,n-2*k));
vector(40, n, a(n-1)) \\ Altug Alkan, Oct 04 2015

Formula

G.f.: (x/((1-x))+1/((-sqrt(5*x^2-2*x+1)+x+1))*x*(1-(5*x-1)/(sqrt(5*x^2-2*x+1)))). - Vladimir Kruchinin, Oct 04 2015
a(n) = Sum_{j=0..n/2}((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j)). - Vladimir Kruchinin, Oct 04 2015
From Peter Bala, Jul 24 2025: (Start)
a(n) = [x^n] (1 - x^2/(1 - x))^n. Cf. A246437.
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all primes p and all positive integers n and k.
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 - x^2 - x^3 + x^4 + 4*x^5 + ... is the g.f. of A108623.(End)

A104508 Absolute row sums of triangle A104505, which is equal to the right-hand side of the triangle A084610 of coefficients in (1+x-x^2)^n.

Original entry on oeis.org

1, 2, 4, 9, 20, 47, 103, 198, 512, 1161, 1985, 5590, 13005, 21684, 59294, 142273, 270858, 634187, 1526399, 3256372, 6779657, 16020364, 37758202, 79702477, 164647757, 424655100, 940639154, 1709638551, 4675803604, 10774789429
Offset: 0

Views

Author

Paul D. Hanna, Mar 11 2005

Keywords

Crossrefs

Cf. A104505.

Programs

  • PARI
    a(n)=sum(k=0,n,abs(polcoeff((1+x-x^2)^n,n+k)))

A104506 Column 1 of triangle A104505, which is equal to the right-hand side of the triangle A084610 of coefficients in (1 + x - x^2)^n.

Original entry on oeis.org

0, -1, -2, 0, 8, 15, -6, -77, -120, 117, 770, 946, -1728, -7735, -6930, 22800, 76960, 42245, -282150, -751640, -125800, 3341205, 7145710, -2002725, -38228232, -65418925, 55550014, 424605078, 566938400, -936604097, -4587287310
Offset: 0

Views

Author

Paul D. Hanna, Mar 11 2005

Keywords

Links

Crossrefs

Cf. A104505, A007440, A104507, A104508, A104509.

Programs

  • PARI
    {a(n)=polcoeff(((1-x)/sqrt(1-2*x+5*x^2+x^2*O(x^n))-1)/(2*x),n)}

Formula

G.f.: ((1-x)/sqrt(1-2*x+5*x^2) - 1)/(2*x).
a(n) = (-1)^n*n*A007440(n) (reversion of g.f. for Fibonacci numbers).
a(n) = -Sum_{k=0..floor(n/2)} C(n, k)*C(n-k, k+1)*(-1)^k. - Paul Barry, May 02 2005
E.g.f.: -exp(x)Bessel_I(1,2*i*x)/i, i=sqrt(-1). - Paul Barry, Feb 10 2006
-(n-1)*(n+1)*a(n) + n*(2*n-1)*a(n-1) - 5*n*(n-1)*a(n-2) = 0. - R. J. Mathar, Aug 17 2017

A120616 Generalized Riordan array (1/sqrt(1+4x^2),(1-sqrt(1+4x^2))/(2x)).

Original entry on oeis.org

1, 0, -1, -2, 0, 1, 0, 3, 0, -1, 6, 0, -4, 0, 1, 0, -10, 0, 5, 0, -1, -20, 0, 15, 0, -6, 0, 1, 0, 35, 0, -21, 0, 7, 0, -1, 70, 0, -56, 0, 28, 0, -8, 0, 1, 0, -126, 0, 84, 0, -36, 0, 9, 0, -1, -252, 0, 210, 0, -120, 0, 45, 0, -10, 0, 1
Offset: 0

Views

Author

Paul Barry, Jun 17 2006

Keywords

Comments

Product by A007318 is A104505.

Examples

			Triangle begins
     1;
     0,   -1;
    -2,    0,   1;
     0,    3,   0,  -1;
     6,    0,  -4,   0,    1;
     0,  -10,   0,   5,    0,  -1;
   -20,    0,  15,   0,   -6,   0,  1;
     0,   35,   0, -21,    0,   7,  0, -1;
    70,    0, -56,   0,   28,   0, -8,  0,   1;
     0, -126,   0,  84,    0, -36,  0,  9,   0, -1;
  -252,    0, 210,   0, -120,   0, 45,  0, -10,  0, 1;

Links

Programs

  • Mathematica
    T[n_, k_] := Binomial[n, (n + k)/2]*(-1)^((n + k)/2) (1 + (-1)^(n + k))/2; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 22 2025 *)

Formula

Number triangle T(n,k)=C(n,(n+k)/2)(-1)^((n+k)/2)(1+(-1)^(n+k))/2.
abs(T(n,k)) = A108044(n,k).
Showing 1-5 of 5 results.

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