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.

A168523 Triangle of coefficients of g.f. a*(1+x)^n + b*(1-x)^(n+2)*polylog(-n-1, x)/x + 2^n*c*(1-x)^(n+1)*LerchPhi(x, -n, 1/2), with a = -1, b = 1, c = 1.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 31, 31, 1, 1, 98, 290, 98, 1, 1, 289, 1974, 1974, 289, 1, 1, 836, 11719, 25944, 11719, 836, 1, 1, 2419, 64929, 275307, 275307, 64929, 2419, 1, 1, 7046, 346192, 2573466, 4831134, 2573466, 346192, 7046, 1, 1, 20677, 1804144, 22163080, 70723522, 70723522, 22163080, 1804144, 20677, 1
Offset: 0

Views

Author

Roger L. Bagula, Nov 28 2009

Keywords

Examples

			Triangle begins as:
  1;
  1,     1;
  1,     8,       1;
  1,    31,      31,        1;
  1,    98,     290,       98,        1;
  1,   289,    1974,     1974,      289,        1;
  1,   836,   11719,    25944,    11719,      836,        1;
  1,  2419,   64929,   275307,   275307,    64929,     2419,       1;
  1,  7046,  346192,  2573466,  4831134,  2573466,   346192,    7046,     1;
  1, 20677, 1804144, 22163080, 70723522, 70723522, 22163080, 1804144, 20677, 1;
		

Crossrefs

Programs

  • Mathematica
    T[n_, a_, b_, c_]:= CoefficientList[Series[a*(1+x)^n + b*(1-x)^(n+2)* PolyLog[-n-1, x]/x + 2^n*c*(1-x)^(n+1)*LerchPhi[x, -n, 1/2], {x,0,30}], x];
    Table[T[n,-1,1,1], {n, 0, 12}]//Flatten (* modified by G. C. Greubel, Mar 19 2022 *)
  • Sage
    m=12
    def LerchPhi(x,s,a): return sum( x^j/(j+a)^s for j in (0..3*m) )
    def p(n,x,a,b,c): return a*(1+x)^n + b*(1-x)^(n+2)*polylog(-n-1, x)/x + 2^n*c*(1-x)^(n+1)*LerchPhi(x, -n, 1/2)
    def T(n,k,a,b,c): return ( p(n,x,a,b,c) ).series(x, n+1).list()[k]
    flatten([[T(n,k,-1,1,1) for k in (0..n)] for n in (0..m)]) # G. C. Greubel, Mar 19 2022

Formula

From G. C. Greubel, Mar 19 2022: (Start)
G.f.: a*(1+x)^n + b*(1-x)^(n+2)*polylog(-n-1, x)/x + 2^n*c*(1-x)^(n+1)*LerchPhi(x, -n, 1/2), with a = -1, b = 1, c = 1.
T(n, n-k) = T(n, k). (End)

Extensions

Edited by G. C. Greubel, Mar 19 2022

A168525 Triangle of coefficients of g.f. a*(1+x)^n + b*(1-x)^(n+2)*polylog(-n-1, x)/x + 2^n*c*(1-x)^(n+1)*LerchPhi(x, -n, 1/2), with a = 65/2, b = -162/2, c = 135/2.

Original entry on oeis.org

19, 19, 19, 19, 146, 19, 19, 759, 759, 19, 19, 3154, 10374, 3154, 19, 19, 11543, 89398, 89398, 11543, 19, 19, 39210, 615669, 1394444, 615669, 39210, 19, 19, 127303, 3747297, 16267301, 16267301, 3747297, 127303, 19, 19, 401858, 21201472, 160611806, 302914330, 160611806, 21201472, 401858, 19
Offset: 0

Views

Author

Roger L. Bagula, Nov 28 2009

Keywords

Examples

			Triangle begins as:
  19;
  19,     19;
  19,    146,       19;
  19,    759,      759,        19;
  19,   3154,    10374,      3154,        19;
  19,  11543,    89398,     89398,     11543,        19;
  19,  39210,   615669,   1394444,    615669,     39210,       19;
  19, 127303,  3747297,  16267301,  16267301,   3747297,   127303,     19;
  19, 401858, 21201472, 160611806, 302914330, 160611806, 21201472, 401858, 19;
		

Crossrefs

Programs

  • Mathematica
    T[n_, a_, b_, c_]:= CoefficientList[Series[a*(1+x)^n + b*(1-x)^(n+2)* PolyLog[-n-1, x]/x + 2^n*c*(1-x)^(n+1)*LerchPhi[x, -n, 1/2], {x,0,30}], x];
    Table[T[n, 65/2, -162/2, 135/2], {n,0,12}]//Flatten (* modified by G. C. Greubel, Mar 19 2022 *)
  • Sage
    m=12
    def LerchPhi(x,s,a): return sum( x^j/(j+a)^s for j in (0..3*m) )
    def p(n,x,a,b,c): return a*(1+x)^n + b*(1-x)^(n+2)*polylog(-n-1, x)/x + 2^n*c*(1-x)^(n+1)*LerchPhi(x, -n, 1/2)
    def T(n,k,a,b,c): return ( p(n,x,a,b,c) ).series(x, n+1).list()[k]
    flatten([[T(n,k,65/2, -162/2, 135/2) for k in (0..n)] for n in (0..m)]) # G. C. Greubel, Mar 19 2022

Formula

From G. C. Greubel, Mar 19 2022: (Start)
G.f.: a*(1+x)^n + b*(1-x)^(n+2)*polylog(-n-1, x)/x + 2^n*c*(1-x)^(n+1)*LerchPhi(x, -n, 1/2), with a = 65/2, b = -162/2, c = 135/2.
T(n, n-k) = T(n, k). (End)

Extensions

Edited by G. C. Greubel, Mar 19 2022

A141697 T(n,k) = (q*Sum_{j=0..k+1} (-1)^j*binomial(n+1, j)*(k+1-j)^n - p*binomial(n-1, k))/2 where p=12 and q=14.

Original entry on oeis.org

1, 1, 1, 1, 16, 1, 1, 59, 59, 1, 1, 158, 426, 158, 1, 1, 369, 2054, 2054, 369, 1, 1, 804, 8247, 16792, 8247, 804, 1, 1, 1687, 29925, 109123, 109123, 29925, 1687, 1, 1, 3466, 102088, 617302, 1092910, 617302, 102088, 3466, 1, 1, 7037, 334664, 3185840, 9171722, 9171722, 3185840, 334664, 7037, 1
Offset: 1

Views

Author

Roger L. Bagula, Sep 11 2008

Keywords

Comments

Row n is made of coefficients from 7*(1 - x)^(n+1) * polylog(-n,x)/x - 6*(1 + x)^(n-1). - Thomas Baruchel, Jun 03 2018

Examples

			Triangle begins:
  1;
  1,    1;
  1,   16,      1;
  1,   59,     59,       1;
  1,  158,    426,     158,       1;
  1,  369,   2054,    2054,     369,       1;
  1,  804,   8247,   16792,    8247,     804,       1;
  1, 1687,  29925,  109123,  109123,   29925,    1687,      1;
		

Crossrefs

Cf. Eulerian numbers (A008292) and Pascal's triangle (A007318).
Cf. A141696.

Programs

  • Magma
    [ 7*(&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) - 6*Binomial(n-1,k): k in [0..n-1], n in [1..10]]; // G. C. Greubel, Nov 13 2019
    
  • Maple
    T:= proc(n, k): 7*add((-1)^j*binomial(n+1, j)*(k-j+1)^n, j = 0..k+1) - 6*binomial(n-1, k); end proc; seq(seq(T(n,k), k=0..n-1), n=1..10); # G. C. Greubel, Nov 13 2019
  • Mathematica
    i=12; l=14; Table[Table[(l*Sum[(-1)^j*Binomial[n+1, j](k+1-j)^n, {j, 0, k+1}] - i*Binomial[n-1, k])/2, {k,0,n-1}], {n,10}]//Flatten
  • PARI
    T(n,k) = 7*sum(j=0, k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n) - 6* binomial(n-1,k);
    for(n=1,10, for(k=0,n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Jun 03 2018
    
  • PARI
    row(n) = Vec(7*(1 - x)^(n+1)*polylog(-n,x)/x - 6*(1 + x)^(n-1)); \\ Michel Marcus, Jun 08 2018
    
  • Sage
    [[ 7*sum( (-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1)) - 6*binomial(n-1,k) for k in (0..n-1)] for n in (1..10)] # G. C. Greubel, Nov 13 2019

Formula

p=12; q=14; T(n,k) = (q*Sum_{j=0..k+1} (-1)^j*binomial(n+1, j)*(k+1-j)^n - p*binomial(n-1, k))/2.
a(n) = 3*A168524(n) - 2*A154337(n). - Thomas Baruchel, Jun 08 2018

Extensions

Edited by G. C. Greubel, Nov 13 2019
Showing 1-3 of 3 results.