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.

A110165 Riordan array (1/sqrt(1-6x+5x^2),(1-3x-sqrt(1-6x+5x^2))/(2x)).

Original entry on oeis.org

1, 3, 1, 11, 6, 1, 45, 30, 9, 1, 195, 144, 58, 12, 1, 873, 685, 330, 95, 15, 1, 3989, 3258, 1770, 630, 141, 18, 1, 18483, 15533, 9198, 3801, 1071, 196, 21, 1, 86515, 74280, 46928, 21672, 7210, 1680, 260, 24, 1, 408105, 356283, 236736, 119154, 44982, 12510, 2484, 333, 27, 1
Offset: 0

Views

Author

Paul Barry, Jul 14 2005

Keywords

Comments

Columns include A026375, A026376 and A026377. Inverse is A110168. Rows sums are A110166. Diagonal sums are A110167.
From Peter Bala, Jan 09 2022: (Start)
This Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = (1 - 3*x - sqrt(1 - 6*x + 5*x^2))/(2*x) and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = 1 + 3*x + x^2. In general the (n,k)-th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)

Examples

			Rows begin
    1;
    3,   1;
   11,   6,   1;
   45,  30,   9,   1;
  195, 144,  58,  12,   1;
  873, 685, 330,  95,  15,   1;
Production array begins:
  3, 1;
  2, 3, 1;
  0, 1, 3, 1;
  0, 0, 1, 3, 1;
  0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 1, 3, 1;
  0, 0, 0, 0, 0, 1, 3, 1;
  ... - _Philippe Deléham_, Feb 08 2014

Links

Programs

  • Maple
    seq(seq( coeff((x^2 + 3*x + 1)^n, x, n-k), k = 0..n ), n = 0..10); # Peter Bala, Jan 09 2022
  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
RiordanArray[1/Sqrt[1-6#+5#^2]&, (1-3#-Sqrt[1-6#+5#^2])/(2#)&, 10] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

Formula

Number triangle T(n, k) = Sum_{j = 0..n} C(n, j)C(2j, j+k).
T(n,0) = 3*T(n-1,0) + 2*T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k > 0, T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 24 2014

A387237 Expansion of 1/((1-x) * (1-5*x))^(5/2).

Original entry on oeis.org

1, 15, 145, 1155, 8260, 55188, 351960, 2170080, 13042095, 76827465, 445335891, 2547479025, 14412134100, 80773641900, 449065521300, 2479190589180, 13603361708775, 74238475926825, 403197150223175, 2180369322394725, 11744998515662720, 63044308615576200, 337323759106291100
Offset: 0

Views

Author

Seiichi Manyama, Aug 23 2025

Keywords

Links

Crossrefs

Cf. A026375, A385563, A387238.
Cf. A026377, A374508.

Programs

  • Magma
    R := PowerSeriesRing(Rationals(), 34); f := 1/((1-x) * (1-5*x))^(5/2); coeffs := [ Coefficient(f, n) : n in [0..33] ]; coeffs; // Vincenzo Librandi, Aug 24 2025
  • Mathematica
    CoefficientList[Series[1/((1-x)*(1-5*x))^(5/2),{x,0,33}],x] (* Vincenzo Librandi, Aug 24 2025 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/((1-x)*(1-5*x))^(5/2))

Formula

n*a(n) = (6*n+9)*a(n-1) - 5*(n+3)*a(n-2) for n > 1.
a(n) = (-1)^n * Sum_{k=0..n} 5^k * binomial(-5/2,k) * binomial(-5/2,n-k).
a(n) = Sum_{k=0..n} (-4)^k * binomial(-5/2,k) * binomial(n+4,n-k).
a(n) = Sum_{k=0..n} 4^k * 5^(n-k) * binomial(-5/2,k) * binomial(n+4,n-k).
a(n) = (binomial(n+4,2)/6) * Sum_{k=0..floor(n/2)} 3^(n-2*k) * binomial(n+2,n-2*k) * binomial(2*k+2,k) = (binomial(n+4,2)/6) * A026377(n+2).
a(n) = (-1)^n * Sum_{k=0..n} 6^k * (5/6)^(n-k) * binomial(-5/2,k) * binomial(k,n-k).

A387239 a(n) = Sum_{k=0..n} binomial(n+3,k+3) * binomial(2*k+6,k+6).

Original entry on oeis.org

1, 12, 95, 630, 3801, 21672, 119154, 639180, 3369795, 17543196, 90476100, 463291920, 2359240975, 11961944400, 60440659640, 304543085040, 1531044995355, 7682898791700, 38494752520175, 192632866196694, 962948703201331, 4809438625979592, 24002988378037350, 119719958370912900
Offset: 0

Views

Author

Seiichi Manyama, Aug 23 2025

Keywords

Links

Crossrefs

Cf. A026376, A026377.
Cf. A126190.

Programs

  • Magma
    [&+[Binomial(n+3,k+3) * Binomial(2*k+6,k+6): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 24 2025
  • Mathematica
    Table[Sum[Binomial[n+3,k+3]* Binomial[2*k+6, k+6],{k,0,n}],{n,0,25}] (* Vincenzo Librandi, Aug 24 2025 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n+3, k+3)*binomial(2*k+6, k+6));

Formula

n*(n+6)*a(n) = (n+3) * (3*(2*n+5)*a(n-1) - 5*(n+2)*a(n-2)) for n > 1.
a(n) = Sum_{k=0..floor(n/2)} 3^(n-2*k) * binomial(n+3,n-2*k) * binomial(2*k+3,k).
a(n) = [x^n] (1+3*x+x^2)^(n+3).
E.g.f.: exp(3*x) * BesselI(3, 2*x), with offset 3.

A128727 Triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n having k DDU and LDU's.

Original entry on oeis.org

1, 1, 3, 9, 1, 27, 9, 81, 54, 2, 243, 270, 30, 729, 1215, 270, 5, 2187, 5103, 1890, 105, 6561, 20412, 11340, 1260, 14, 19683, 78732, 61236, 11340, 378, 59049, 295245, 306180, 85050, 5670, 42, 177147, 1082565, 1443420, 561330, 62370, 1386, 531441
Offset: 0

Views

Author

Emeric Deutsch, Mar 31 2007

Keywords

Comments

A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of steps in it.
Row n has ceiling(n/2) terms (n >= 1).
Row sums yield A002212.
Apparently a(n) = A126177(n-1). - Georg Fischer, Oct 28 2018

Examples

			T(5,2)=2 because we have UU(DDU)U(DDU)D and UUU(DDU)(DDU)D (the 2 subwords are shown between parentheses).
Triangle starts:
    1;
    1;
    3;
    9,    1;
   27,    9;
   81,   54,    2;
  243,  270,   30;
  729, 1215,  270,    5;

Links

  • E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203

Crossrefs

Cf. A000108, A002212, A003409, A026377, A126177.

Programs

  • Maple
    T:=(n,k)->3^(n-1-2*k)*binomial(n,k)*binomial(n-k,k+1)/n: 1; for n from 1 to 13 do seq(T(n,k),k=0..floor((n-1)/2)) od; # yields sequence in triangular form

Formula

T(n,0) = 3^(n-1).
T(2k+1,k) = binomial(2k,k)/(k+1) (the Catalan numbers, A000108).
T(2k,k-1) = 3binomial(2k-1,k) = A003409(k).
Sum_{k>=0} k*T(n,k) = A026377(n-1).
T(n,k) = (1/n)*3^(n-1-2k)*binomial(n,k)*binomial(n-k,k+1).
G.f.: G = G(t,z) satisfies tzG^2 - (1 - 3z + 2tz)G + 1 - 2z + tz = 0.
Showing 1-4 of 4 results.

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