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 1-3 of 3 results.

A168518 Expansion of g.f. (1/2)*( a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi(x, -n-1, 1) + c*2^(n+1)*(1-x)^(n+1)*LerchPhi(x, -n, 1/2) ), where a = -4, b = 2, and c = 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 12, 1, 1, 51, 51, 1, 1, 170, 514, 170, 1, 1, 521, 3646, 3646, 521, 1, 1, 1552, 22247, 49472, 22247, 1552, 1, 1, 4591, 125565, 534995, 534995, 125565, 4591, 1, 1, 13590, 677776, 5058698, 9506078, 5058698, 677776, 13590, 1, 1, 40341, 3560448, 43870968, 140136690, 140136690, 43870968, 3560448, 40341, 1
Offset: 0

Views

Author

Roger L. Bagula, Nov 28 2009

Keywords

Examples

			Triangle begins as:
  1;
  1,     1;
  1,    12,       1;
  1,    51,      51,        1;
  1,   170,     514,      170,         1;
  1,   521,    3646,     3646,       521,         1;
  1,  1552,   22247,    49472,     22247,      1552,        1;
  1,  4591,  125565,   534995,    534995,    125565,     4591,       1;
  1, 13590,  677776,  5058698,   9506078,   5058698,   677776,   13590,     1;
  1, 40341, 3560448, 43870968, 140136690, 140136690, 43870968, 3560448, 40341, 1;

Links

Crossrefs

Cf. A142460, A155491, A155495, A157273, A166343.
Cf. A168517, A168549, A168551, A168552.

Programs

  • Mathematica
    p[x_, n_, a_, b_, c_]= (1/2)*(a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi[x,-n-1,1] + c*2^(n+1)*(1-x)^(n+1)*LerchPhi[x,-n,1/2]);
Table[CoefficientList[p[x,n,-4,2,2], x], {n,0,10}]//Flatten (* modified by G. C. Greubel, Mar 31 2022 *)
  • Sage
    def A168518(n,k,a,b,c): return (1/2)*( a*binomial(n,k) + sum( (-1)^(k-j)*(b*binomial(n+2, k-j)*(j+1)^(n+1) + 2*c*binomial(n+1,k-j)*(2*j+1)^n) for j in (0..k)) )
flatten([[A168518(n,k,-4,2,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 31 2022

Formula

G.f.: (1/2)*( a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi(x, -n-1, 1) + c*2^(n+1)*(1 - x)^(n+1)*LerchPhi(x, -n, 1/2) ), where a = -4, b = 2, and c = 2.
From G. C. Greubel, Mar 31 2022: (Start)
T(n, k) = (1/2)*( a*binomial(n,k) + sum( (-1)^(k-j)*(b*binomial(n+2, k-j)*(j+1)^(n+1) + 2*c*binomial(n+1,k-j)*(2*j+1)^n) for j in (0..k)) ), with a = -4, b = 2, and c = 2.
T(n, n-k) = T(n, k). (End)

Extensions

Edited by G. C. Greubel, Mar 31 2022

A155467 Triangle T(n, k) = Eulerian(n+1, k)*Binomial(n+1, k)/(k+1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 22, 22, 1, 1, 65, 220, 65, 1, 1, 171, 1510, 1510, 171, 1, 1, 420, 8337, 21140, 8337, 420, 1, 1, 988, 40068, 218666, 218666, 40068, 988, 1, 1, 2259, 175296, 1852914, 3935988, 1852914, 175296, 2259, 1, 1, 5065, 717600, 13655760, 55034868, 55034868, 13655760, 717600, 5065, 1
Offset: 0

Views

Author

Roger L. Bagula, Jan 22 2009

Keywords

Comments

The sequence substitutes Eulerian numbers for the binomial in a triangle of Narayana numbers A001263,

Examples

			Triangle begins as:
  1;
  1,    1;
  1,    6,      1;
  1,   22,     22,        1;
  1,   65,    220,       65,        1;
  1,  171,   1510,     1510,      171,        1;
  1,  420,   8337,    21140,     8337,      420,        1;
  1,  988,  40068,   218666,   218666,    40068,      988,      1;
  1, 2259, 175296,  1852914,  3935988,  1852914,   175296,   2259,    1;
  1, 5065, 717600, 13655760, 55034868, 55034868, 13655760, 717600, 5065, 1;

Links

Crossrefs

Cf. A001263 (m=0), this sequence (m=1), A155491 (m=3), A155493 (m=4).
Cf. A001263, A008292, A099765 (row sums).

Programs

  • Mathematica
    (* First program *)
Needs["Combinatorica`"]
T[n_, k_]:= Eulerian[n+1, k]*Binomial[n+1, k]/(k+1);
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* Roger L. Bagula, Apr 14 2010 *)
(* Second program *)
t[n_, k_, m_]:= t[n,k,m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1,k-1,m] + (m*k -(m -1))*t[n-1,k,m]];
T[n_, k_, m_]:= Binomial[n+1,k]*t[n+1,k+1,m]/(k+1);
Table[T[n,k,1], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 01 2022 *)
  • Sage
    @CachedFunction
def t(n,k,m):
    if (k==1 or k==n): return 1
    else: return (m*(n-k)+1)*t(n-1,k-1,m) + (m*k-m+1)*t(n-1,k,m)
def T(n,k,m): return binomial(n+1,k)*t(n+1,k+1,m)/(k+1)
flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 01 2022

Formula

T(n, k) = binomial(n+1, k)*t(n, k, m)/(k+1), where t(n,k,m) = (m*(n-k)+1)*t(n-1,k-1,m) + (m*k-m+1)*t(n-1,k,m), t(n,1,m) = t(n,n,m) = 1, and m = 1.
Sum_{k=0..n} T(n, k) = A099765(n+2).
T(n, k) = Eulerian(n+1, k)*Binomial(n+1, k)/(k+1). - Roger L. Bagula, Apr 14 2010
From G. C. Greubel, Apr 01 2022: (Start)
T(n, k) = binomial(n+1, k)*A008292(n+1, k+1)/(k+1).
T(n, n-k) = T(n, k).
T(n, 1) = A003469(n). (End)

Extensions

Edited by G. C. Greubel, Apr 01 2022

A155493 Triangle T(n, k) = binomial(n+1, k)*A142459(n+1, k+1)/(k+1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 15, 1, 1, 118, 118, 1, 1, 770, 3540, 770, 1, 1, 4671, 67810, 67810, 4671, 1, 1, 27321, 1039689, 3085355, 1039689, 27321, 1, 1, 156220, 14006244, 99524810, 99524810, 14006244, 156220, 1, 1, 878868, 173788752, 2602528824, 6090918372, 2602528824, 173788752, 878868, 1
Offset: 0

Views

Author

Roger L. Bagula, Jan 23 2009

Keywords

Examples

			Triangle begins as:
  1;
  1,      1;
  1,     15,         1;
  1,    118,       118,          1;
  1,    770,      3540,        770,          1;
  1,   4671,     67810,      67810,       4671,          1;
  1,  27321,   1039689,    3085355,    1039689,      27321,         1;
  1, 156220,  14006244,   99524810,   99524810,   14006244,    156220,      1;
  1, 878868, 173788752, 2602528824, 6090918372, 2602528824, 173788752, 878868, 1;

Links

Crossrefs

Cf. A001263 (m=0), A155467 (m=1), A155491 (m=3), this sequence (m=4).
Cf. A142459.

Programs

  • Mathematica
    t[n_, k_, m_]:= t[n,k,m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1,k-1,m] + (m*k -(m -1))*t[n-1,k,m]];
T[n_, k_, m_]:= Binomial[n+1,k]*t[n+1,k+1,m]/(k+1);
Table[T[n,k,4], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Apr 01 2022 *)
  • Sage
    @CachedFunction
def t(n,k,m):
    if (k==1 or k==n): return 1
    else: return (m*(n-k)+1)*t(n-1,k-1,m) + (m*k-m+1)*t(n-1,k,m)
def T(n,k,m): return binomial(n+1,k)*t(n+1,k+1,m)/(k+1)
flatten([[T(n,k,4) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 01 2022

Formula

T(n, k) = binomial(n+1, k)*t(n, k, m)/(k+1), where t(n,k,m) = (m*(n-k)+1)*t(n-1,k-1,m) + (m*k-m+1)*t(n-1,k,m), t(n,1,m) = t(n,n,m) = 1, and m = 4.
From G. C. Greubel, Apr 01 2022: (Start)
T(n, k) = binomial(n+1, k)*A142459(n+1, k+1)/(k+1).
T(n, n-k) = T(n, k). (End)

Extensions

Edited by G. C. Greubel, Apr 01 2022
Showing 1-3 of 3 results.

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