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 1-4 of 4 results.

A307819 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 + 2*k*x + k*(k+4)*x^2).

Original entry on oeis.org

1, 1, 0, 1, -1, 0, 1, -2, -1, 0, 1, -3, 0, 5, 0, 1, -4, 3, 16, -5, 0, 1, -5, 8, 27, -56, -11, 0, 1, -6, 15, 32, -189, 48, 41, 0, 1, -7, 24, 25, -416, 567, 384, -29, 0, 1, -8, 35, 0, -725, 2176, 189, -1920, -125, 0, 1, -9, 48, -49, -1080, 5625, -4864, -11259, 3168, 365, 0
Offset: 0

Views

Author

Seiichi Manyama, May 05 2019

Keywords

Examples

			Square array begins:
   1,   1,     1,      1,      1,      1,      1, ...
   0,  -1,    -2,     -3,     -4,     -5,     -6, ...
   0,  -1,     0,      3,      8,     15,     24, ...
   0,   5,    16,     27,     32,     25,      0, ...
   0, -11,    48,    567,   2176,   5625,  11664, ...
   0,  41,   384,    189,  -4864, -24375, -74304, ...
   0, -29, -1920, -11259, -23552,   9375, 228096, ...
		

Crossrefs

Columns k=0..3 give A000007, (-1)^n * A098331, A116093, (-1)^n * A098340.
Main diagonal gives A307911.

Programs

  • Mathematica
    A[n_, k_] := (-k)^n*Hypergeometric2F1[(1-n)/2, -n/2, 1, -4/k]; A[0, ] = 1; A[, 0] = 0; Table[A[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, May 07 2019 *)

Formula

A(n,k) is the coefficient of x^n in the expansion of (1 - k*x - k*x^2)^n.
A(n,k) = Sum_{j=0..floor(n/2)} (-k)^(n-j) * binomial(n,j) * binomial(n-j,j) = Sum_{j=0..floor(n/2)} (-k)^(n-j) * binomial(n,2*j) * binomial(2*j,j).
n * A(n,k) = -k * (2*n-1) * A(n-1,k) - k * (k+4) * (n-1) * A(n-2,k).

A307855 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2*x + (1-4*k)*x^2).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 7, 1, 1, 1, 7, 13, 19, 1, 1, 1, 9, 19, 49, 51, 1, 1, 1, 11, 25, 91, 161, 141, 1, 1, 1, 13, 31, 145, 331, 581, 393, 1, 1, 1, 15, 37, 211, 561, 1441, 2045, 1107, 1, 1, 1, 17, 43, 289, 851, 2841, 5797, 7393, 3139, 1
Offset: 0

Views

Author

Seiichi Manyama, May 01 2019

Keywords

Examples

			Square array begins:
   1,   1,    1,    1,     1,     1,     1, ...
   1,   1,    1,    1,     1,     1,     1, ...
   1,   3,    5,    7,     9,    11,    13, ...
   1,   7,   13,   19,    25,    31,    37, ...
   1,  19,   49,   91,   145,   211,   289, ...
   1,  51,  161,  331,   561,   851,  1201, ...
   1, 141,  581, 1441,  2841,  4901,  7741, ...
   1, 393, 2045, 5797, 12489, 22961, 38053, ...
		

Crossrefs

Columns k=0..6 give A000012, A002426, A084601, A084603, A084605, A098264, A098265.
Main diagonal gives A187018.

Programs

  • Mathematica
    T[n_, k_] := Sum[If[k == j == 0, 1, k^j] * Binomial[n, j] * Binomial[n-j, j], {j, 0, Floor[n/2]}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 13 2021 *)

Formula

A(n,k) is the coefficient of x^n in the expansion of (1 + x + k*x^2)^n.
A(n,k) = Sum_{j=0..floor(n/2)} k^j * binomial(n,j) * binomial(n-j,j) = Sum_{j=0..floor(n/2)} k^j * binomial(n,2*j) * binomial(2*j,j).
D-finite with recurrence: n * A(n,k) = (2*n-1) * A(n-1,k) - (1-4*k) * (n-1) * A(n-2,k).

A307862 Coefficient of x^n in (1 + x - n*x^2)^n.

Original entry on oeis.org

1, 1, -3, -17, 49, 651, -1259, -38023, 26433, 2969299, 2225101, -289389891, -692529551, 33718183045, 143578976997, -4559187616649, -29119975483135, 699788001188403, 6188699469443869, -119828491083854707, -1404529670244379599, 22563726025297759345, 341997845736800473397
Offset: 0

Views

Author

Seiichi Manyama, May 02 2019

Keywords

Comments

Also coefficient of x^n in the expansion of 1/sqrt(1 - 2*x + (1+4*n)*x^2).

Crossrefs

Main diagonal of A307860.

Programs

  • Maple
    A307862:= n -> simplify(hypergeom([-n/2, (1-n)/2], [1], -4*n));
    seq(A307862(n), n = 0..30); # G. C. Greubel, May 31 2020
  • Mathematica
    a[n_]:= SeriesCoefficient[(1 +x -n*x^2)^n, {x,0,n}]; Table[a[n], {n,0,30}] (* G. C. Greubel, May 31 2020 *)
  • PARI
    {a(n) = polcoef((1+x-n*x^2)^n, n)}
    
  • PARI
    {a(n) = sum(k=0, n\2, (-n)^k*binomial(n, k)*binomial(n-k, k))}
    
  • PARI
    {a(n) = sum(k=0, n\2, (-n)^k*binomial(n, 2*k)*binomial(2*k, k))}
    
  • Sage
    [ hypergeometric([-n/2, (1-n)/2], [1], -4*n).simplify_hypergeometric() for n in (0..30)] # G. C. Greubel, May 31 2020

Formula

a(n) = Sum_{k=0..floor(n/2)} (-n)^k * binomial(n,k) * binomial(n-k,k).
a(n) = Sum_{k=0..floor(n/2)} (-n)^k * binomial(n,2*k) * binomial(2*k,k).
a(n) = n! * [x^n] exp(x) * BesselI(0,2*sqrt(-n)*x). - Ilya Gutkovskiy, May 31 2020
a(n) = Hypergeometric2F1(-n/2, (1-n)/2; 1; -4*n). - G. C. Greubel, May 31 2020

A308035 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 2/(1 - x + sqrt(1 - 2*x + (1+4*k)*x^2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, -1, -2, 1, 1, 1, -2, -5, -3, 1, 1, 1, -3, -8, -3, 1, 1, 1, 1, -4, -11, 1, 21, 11, 1, 1, 1, -5, -14, 9, 61, 51, 15, 1, 1, 1, -6, -17, 21, 121, 91, -41, -13, 1, 1, 1, -7, -20, 37, 201, 101, -377, -391, -77, 1
Offset: 0

Views

Author

Seiichi Manyama, May 10 2019

Keywords

Examples

			Square array begins:
   1,  1,   1,    1,     1,     1,     1, ...
   1,  1,   1,    1,     1,     1,     1, ...
   1,  0,  -1,   -2,    -3,    -4,    -5, ...
   1, -2,  -5,   -8,   -11,   -14,   -17, ...
   1, -3,  -3,    1,     9,    21,    37, ...
   1,  1,  21,   61,   121,   201,   301, ...
   1, 11,  51,   91,   101,    51,   -89, ...
   1, 15, -41, -377, -1203, -2729, -5165, ...
		

Crossrefs

Columns k=2..3 give (-1)^n * A091593, A308036.
Main diagonal gives A307947.

Programs

  • Mathematica
    T[n_, k_] := Sum[If[k == j == 0, 1, (-k)^j] * Binomial[n, 2*j] * CatalanNumber[j], {j, 0, Floor[n/2]}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 12 2021 *)

Formula

A(n,k) is the coefficient of x^n in the expansion of 1/(n+1) * (1 + x - k*x^2)^(n+1).
A(n,k) = Sum_{j=0..floor(n/2)} (-k)^j * binomial(n,j) * binomial(n-j,j)/(j+1) = Sum_{j=0..floor(n/2)} (-k)^j * binomial(n,2*j) * A000108(j).
(n+2) * A(n,k) = (2*n+1) * A(n-1,k) - (1+4*k) * (n-1) * A(n-2,k).
Showing 1-4 of 4 results.