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.

A062139 Coefficient triangle of generalized Laguerre polynomials n!*L(n,2,x) (rising powers of x).

Original entry on oeis.org

1, 3, -1, 12, -8, 1, 60, -60, 15, -1, 360, -480, 180, -24, 1, 2520, -4200, 2100, -420, 35, -1, 20160, -40320, 25200, -6720, 840, -48, 1, 181440, -423360, 317520, -105840, 17640, -1512, 63, -1, 1814400, -4838400, 4233600
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The row polynomials s(n,x) := n!*L(n,2,x) = Sum_{m=0..n} a(n,m)*x^m have e.g.f. exp(-z*x/(1-z))/(1-z)^3. They are Sheffer polynomials satisfying the binomial convolution identity s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,x)*p(n-k,y), with polynomials p(n,x) = Sum_{m=1..n} |A008297(n,m)|*(-x)^m, n >= 1 and p(0,x)=1 (for Sheffer polynomials see A048854 for S. Roman reference).
This unsigned matrix is embedded in the matrix for n!*L(n,-2,-x). Introduce 0,0 to each unsigned row and then add 1,-1,1 to the array as the first two rows to generate n!*L(n,-2,-x). - Tom Copeland, Apr 20 2014
The unsigned n-th row reverse polynomial equals the numerator polynomial of the finite continued fraction 1 - x/(1 + (n+1)*x/(1 + n*x/(1 + n*x/(1 + ... + 2*x/(1 + 2*x/(1 + x/(1 + x/(1)))))))). Cf. A089231. The denominator polynomial of the continued fraction is the (n+1)-th row polynomial of A144084. An example is given below. - Peter Bala, Oct 06 2019

Examples

			Triangle begins:
     1;
     3,    -1;
    12,    -8,    1;
    60,   -60,   15,   -1;
   360,  -480,  180,  -24,  1;
  2520, -4200, 2100, -420, 35, -1;
  ...
2!*L(2,2,x) = 12 - 8*x + x^2.
Unsigned row 3 polynomial in reverse form as the numerator of a continued fraction: 1 - x/(1 + 4*x/(1 + 3*x/(1 + 3*x/(1 + 2*x/(1 + 2*x/(1 + x/(1 + x))))))) = (60*x^3 + 60*x^2 + 15*x + 1)/(24*x^4 + 96*x^3 + 72*x^2 + 16*x + 1). - _Peter Bala_, Oct 06 2019

Links

Crossrefs

For m=0..5 the (unsigned) columns give A001710, A005990, A005461, A062193-A062195. The row sums (signed) give A062197, the row sums (unsigned) give A052852.
Cf. A021009, A062137-A062140, A066667, A089231, A144084.

Programs

  • Maple
    with(PolynomialTools):
p := n -> (n+2)!*hypergeom([-n],[3],x)/2:
seq(CoefficientList(simplify(p(n)), x), n=0..9); # Peter Luschny, Apr 08 2015
  • Mathematica
    Flatten[Table[((-1)^m)*n!*Binomial[n+2,n-m]/m!,{n,0,8},{m,0,n}]] (* Indranil Ghosh, Feb 24 2017 *)
  • PARI
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(((-1)^k)*n!*binomial(n+2, n-k)/k!, ", ");); print(););} \\ Michel Marcus, May 06 2014
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n, 2)); \\ Michel Marcus, Feb 06 2021
  • Python
    import math
f=math.factorial
def C(n,r):return f(n)//f(r)//f(n-r)
i=0
for n in range(16):
    for m in range(n+1):
        i += 1
        print(i,((-1)**m)*f(n)*C(n+2,n-m)//f(m)) # Indranil Ghosh, Feb 24 2017
  • Python
    from functools import cache
@cache
def T(n, k):
    if k < 0 or k > n: return 0
    if k == n: return (-1)**n
    return (n + k + 2) * T(n-1, k) - T(n-1, k-1)
for n in range(7): print([T(n,k) for k in range(n + 1)])
# Peter Luschny, Mar 25 2024

Formula

T(n, m) = ((-1)^m)*n!*binomial(n+2, n-m)/m!.
E.g.f. for m-th column sequence: ((-x/(1-x))^m)/(m!*(1-x)^3), m >= 0.
n!*L(n,2,x) = (n+2)!*hypergeom([-n],[3],x)/2. - Peter Luschny, Apr 08 2015
From Werner Schulte, Mar 24 2024: (Start)
T(n, k) = (n+k+2) * T(n-1, k) - T(n-1, k-1) with initial values T(0, 0) = 1 and T(i, j) = 0 if j < 0 or j > i.
T = T^(-1), i.e., T is matrix inverse of T. (End)

A062194 Fifth column sequence of triangle A062139 (generalized a=2 Laguerre).

Original entry on oeis.org

1, 35, 840, 17640, 352800, 6985440, 139708800, 2854051200, 59935075200, 1298593296000, 29088489830400, 674324082432000, 16183777978368000, 402104637462528000, 10339833534750720000, 275039572024369152000
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Links

Crossrefs

Cf. A001710, A005461, A005990, A062193.
Cf. A001113, A001620, A091725, A099285.

Programs

  • GAP
    List([0..15],n->Factorial(n+4)*Binomial(n+6,6)/Factorial(4)); # Muniru A Asiru, Jul 01 2018
  • Magma
    [Factorial(n+4)*Binomial(n+6, 6)/Factorial(4): n in [0..20]]; // G. C. Greubel, May 12 2018
  • Mathematica
    Table[(n+4)!*Binomial[n+6,6]/4!, {n, 0, 20}] (* G. C. Greubel, May 12 2018 *)
  • PARI
    { f=6; for (n=0, 100, f*=n + 4; write("b062194.txt", n, " ", f*binomial(n + 6, 6)/24) ) } \\ Harry J. Smith, Aug 02 2009
  • Sage
    [binomial(n,6)*factorial (n-2)/factorial (4) for n in range(6, 22)] # Zerinvary Lajos, Jul 07 2009

Formula

E.g.f.: (1 + 24*x + 90*x^2 + 80*x^3 + 15*x^4)/(1-x)^11.
a(n) = A062139(n+4, 4).
a(n) = (n+4)!*binomial(n+6, 6)/4!.
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..n} binomial(k,j) * Stirling1(n,k) * Stirling2(j,i) * x^(k-j) then a(n-4) = (-1)^n*f(n,4,-7), (n >= 4). - Milan Janjic, Mar 01 2009
From Amiram Eldar, May 06 2022: (Start)
Sum_{n>=0} 1/a(n) = 336*(gamma - Ei(1)) - 96*e + 3524/5, where gamma = A001620, Ei(1) = A091725, and e = A001113.
Sum_{n>=0} (-1)^n/a(n) = 3264*(gamma - Ei(-1)) - 1920/e - 9464/5, where Ei(-1) = -A099285. (End)

A062195 Sixth (unsigned) column sequence of triangle A062139 (generalized a=2 Laguerre).

Original entry on oeis.org

1, 48, 1512, 40320, 997920, 23950080, 570810240, 13699445760, 333923990400, 8310997094400, 211930425907200, 5548723878297600, 149353151057510400, 4135933413900288000, 117874102296158208000
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Links

Crossrefs

Cf. A001710, A005990, A005461, A062139, A062193, A062194.
Cf. A001113, A001620, A091725, A099285.

Programs

  • Magma
    [Factorial(n+5)*Binomial(n+7, 7)/Factorial(5): n in [0..20]]; // G. C. Greubel, May 12 2018
  • Mathematica
    Table[(n+5)!*Binomial[n+7, 7]/5!, {n, 0, 20}] (* G. C. Greubel, May 12 2018 *)
  • PARI
    { f=24; for (n=0, 100, f*=n + 5; write("b062195.txt", n, " ", f*binomial(n + 7, 7)/120) ) } \\ Harry J. Smith, Aug 02 2009

Formula

E.g.f.: N(2;5, x)/(1-x)^13 with N(2;5, x) := Sum_{k=0..5} A062196(5, k)*x^k = 1+35*x+210*x^2+350*x^3+175*x^4+21*x^5.
a(n) = A062139(n+5, 5).
a(n) = (n+5)!*binomial(n+7, 7)/5!.
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j) * Stirling1(n,k) * Stirling2(j,i) * x^(k-j) then a(n-5) = (-1)^(n-1)*f(n,5,-8), (n>=5). - Milan Janjic, Mar 01 2009
From Amiram Eldar, May 06 2022: (Start)
Sum_{n>=0} 1/a(n) = 1295*(Ei(1) - gamma) + 2170*e - 22813/3, where Ei(1) = A091725, gamma = A001620, and e = A001113.
Sum_{n>=0} (-1)^n/a(n) = 36575*(gamma - Ei(-1)) - 21700/e - 63455/3, where Ei(-1) = -A099285. (End)
Showing 1-3 of 3 results.

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