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.

A283439 Hankel transform of A033434.

Original entry on oeis.org

1, -3, -9, -6, 10, 25, 15, -21, -49, -28, 36, 81, 45, -55, -121, -66, 78, 169, 91, -105, -225, -120, 136, 289, 153, -171, -361, -190, 210, 441, 231, -253, -529, -276, 300, 625, 325, -351, -729, -378, 406, 841, 435, -465, -961, -496, 528, 1089, 561, -595, -1225
Offset: 0

Views

Author

Paul Barry, Mar 07 2017

Keywords

Comments

a(n) modulo 2 is A131719(n+2).

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - 5*x + 2*x^3 - 3*x^4 + 2*x^5 - x^6)/((1 + x)*(1 - x + x^2)^3) ,{x, 0, 35}], x] (* Indranil Ghosh, Mar 08 2017 *)
  • PARI
    print(Vec((1 - 5*x + 2*x^3 - 3*x^4 + 2*x^5 - x^6)/((1 + x)*(1 - x + x^2)^3) + O(x^36))); \\ Indranil Ghosh, Mar 08 2017

Formula

G.f.: (1 - 5*x + 2*x^3 - 3*x^4 + 2*x^5 - x^6)/((1 + x)*(1 - x + x^2)^3).
a(3*k) = (-1)^k*(k + 1)*(2*k + 1).
a(3*k + 1) = -(-1)^k*(k + 1)*(2*k + 3).
a(3*k + 2) = -(-1)^k*(k + 3)^2.

Extensions

More terms from Indranil Ghosh, Mar 08 2017

A059346 Difference array of Catalan numbers A000108 read by antidiagonals.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 1, 2, 3, 5, 3, 4, 6, 9, 14, 6, 9, 13, 19, 28, 42, 15, 21, 30, 43, 62, 90, 132, 36, 51, 72, 102, 145, 207, 297, 429, 91, 127, 178, 250, 352, 497, 704, 1001, 1430, 232, 323, 450, 628, 878, 1230, 1727, 2431, 3432, 4862, 603, 835, 1158, 1608, 2236, 3114
Offset: 0

Views

Author

N. J. A. Sloane, Jan 27 2001

Keywords

Examples

			Array starts:
      1       1       2       5      14      42     132     429
      0       1       3       9      28      90     297    1001
      1       2       6      19      62     207     704    2431
      1       4      13      43     145     497    1727    6071
      3       9      30     102     352    1230    4344   15483
      6      21      72     250     878    3114   11139   40143
     15      51     178     628    2236    8025   29004  105477
     36     127     450    1608    5789   20979   76473  280221
     91     323    1158    4181   15190   55494  203748  751422
    232     835    3023   11009   40304  148254  547674 2031054
    603    2188    7986   29295  107950  399420 1483380 5527750
Triangle starts:
  1;
  0,  1;
  1,  1,  2;
  1,  2,  3,  5;
  3,  4,  6,  9, 14;
		

Crossrefs

Top row is A000108, leading diagonals give A005043, A001006, A005554.
Row sums are A106640.

Programs

  • Maple
    T := (n,k) -> (-1)^(n-k)*binomial(2*k,k)*hypergeom([k-n,k+1/2], [k+2], 4)/(k+1): seq(seq(simplify(T(n,k)), k=0..n), n=0..10);
    # Peter Luschny, Aug 16 2012, updated May 25 2021
  • Mathematica
    max = 11; t = Table[ Differences[ Table[ CatalanNumber[k], {k, 0, max}], n], {n, 0, max}]; Flatten[ Table[t[[n-k+1, k]], {n, 1, max}, {k, 1, n}]] (* Jean-François Alcover, Nov 15 2011 *)
  • Sage
    def T(n, k) :
        if k > n : return 0
        if n == k : return binomial(2*n, n)/(n+1)
        return T(n-1, k) - T(n, k+1)
    A059346 = lambda n,k: (-1)^(n-k)*T(n, k)
    for n in (0..5): [A059346(n,k) for k in (0..n)] # Peter Luschny, Aug 16 2012

Formula

T(n, k) = (-1)^(n-k)*binomial(2*k,k)/(k+1)*hypergeometric([k-n, k+1/2],[k+2], 4). - Peter Luschny, Aug 16 2012

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Feb 16 2001

A237124 Triangle of numbers related to Catalan numbers (A000108).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 4, 3, 1, 9, 11, 8, 4, 1, 28, 33, 24, 13, 5, 1, 90, 104, 76, 43, 19, 6, 1, 297, 339, 249, 145, 69, 26, 7, 1, 1001, 1133, 836, 497, 248, 103, 34, 8, 1, 3432, 3861, 2860, 1727, 891, 394, 146, 43, 9, 1, 11934, 13364, 9932, 6071, 3211, 1484, 593, 199, 53, 10, 1
Offset: 0

Views

Author

Philippe Deléham, Feb 03 2014

Keywords

Comments

Riordan array (1 +x +x^2*C(x)^3, x*C(x)) where C(x) is the g.f. of A000108.
Diagonal sums are A000108(n).
Row sums are T(n+1,1).
T(n,0) = A071724(n-1).
T(n,1) = A220902(n), n>=2.
T(n,2) = A228404(n-2), n>=4.
T(n+3,3) = A033434(n).
T(n,n) = 1.
T(n+1,n) = n+1.
T(n+2,n) = A034856(n+1).

Examples

			Triangle begins:
      1;
      1,     1;
      1,     2,    1;
      3,     4,    3,    1;
      9,    11,    8,    4,    1;
     28,    33,   24,   13,    5,    1;
     90,   104,   76,   43,   19,    6,   1;
    297,   339,  249,  145,   69,   26,   7,   1;
   1001,  1133,  836,  497,  248,  103,  34,   8,  1;
   3432,  3861, 2860, 1727,  891,  394, 146,  43,  9,  1;
  11934, 13364, 9932, 6071, 3211, 1484, 593, 199, 53, 10, 1;
  ...
		

Crossrefs

Cf. A000108.

Programs

  • Mathematica
    b[n_, k_]:= Binomial[2*n-k+1, n-k];
    T[n_, k_]:= If[n<3, Binomial[n, k], b[n, k] -2*b[n, k+1] -b[n, k+2] +3*b[n, k+3] - 2*b[n, k+4]];
    Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 08 2021 *)
  • Sage
    def b(n,k): return binomial(2*n-k+1, n-k)
    def T(n,k): return binomial(n,k) if (n<3) else b(n,k) -2*b(n, k+1) -b(n, k+2) +3*b(n, k+3) -2*b(n, k+4)
    flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 08 2021

Formula

From Peter Bala, Feb 18 2018: (Start)
T(n,k) = C(2*n+1-k, n-k) - 2*C(2*n-k, n-k-1) - C(2*n-1-k, n-k-2) + 3*C(2*n-2-k, n-k-3) - 2*C(2*n-3-k, n-k-4), for n > 2, otherwise C(n, k).
The n-th row polynomial of the row reverse triangle equals the n-th degree Taylor polynomial of the function (1 - x^2 + x^3)*(1 - 2*x)/(1 - x)^2 * 1/(1 - x)^n about 0. For example, for n = 4, (1 - x^2 + x^3)*(1 - 2*x)/(1 - x)^2 * 1/(1 - x)^4 = 1 + 4*x + 8*x^2 + 11*x^3 + 9*x^4 + O(x^5), giving (9, 11, 8, 4, 1) as row 4. (End)

A228339 Fourth differences of Catalan numbers (A000108).

Original entry on oeis.org

3, 9, 30, 102, 352, 1230, 4344, 15483, 55626, 201246, 732564, 2681223, 9861342, 36428320, 135100620, 502841295, 1877678370, 7032454470, 26410804020, 99437742720, 375260126904, 1419223506516, 5378236459328, 20419260060462, 77659985060772, 295844249258796, 1128738495397128, 4312685074680465, 16500218817839274, 63209983347693924
Offset: 0

Views

Author

N. J. A. Sloane, Aug 29 2013

Keywords

Crossrefs

Programs

  • Mathematica
    Differences[Table[CatalanNumber[n], {n, 0, 30}], 4] (* Amiram Eldar, Jul 10 2023 *)

Formula

From Amiram Eldar, Jul 10 2023: (Start)
a(n) = 9*(9*n^4 + 54*n^3 + 135*n^2 + 122*n + 40) * n! * binomial(2*n, n)/(n+5)!.
Sum_{n>=0} a(n)/4^n = 38. (End)
Showing 1-4 of 4 results.