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.

A129062 T(n, k) = [x^k] Sum_{k=0..n} Stirling2(n, k)*RisingFactorial(x, k), triangle read by rows, for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 6, 6, 1, 0, 26, 36, 12, 1, 0, 150, 250, 120, 20, 1, 0, 1082, 2040, 1230, 300, 30, 1, 0, 9366, 19334, 13650, 4270, 630, 42, 1, 0, 94586, 209580, 166376, 62160, 11900, 1176, 56, 1, 0, 1091670, 2562354, 2229444, 952728, 220500, 28476, 2016, 72, 1
Offset: 0

Views

Author

Wolfdieter Lang, May 04 2007

Keywords

Comments

Matrix product of Stirling2 with unsigned Stirling1 triangle.
For the subtriangle without column no. m=0 and row no. n=0 see A079641.
The reversed matrix product |S1|. S2 is given in A111596.
As a product of lower triangular Jabotinsky matrices this is a lower triangular Jabotinsky matrix. See the D. E. Knuth references given in A039692 for Jabotinsky type matrices.
E.g.f. for row polynomials P(n,x):=sum(a(n,m)*x^m,m=0..n) is 1/(2-exp(z))^x. See the e.g.f. for the columns given below.
A048993*A132393 as infinite lower triangular matrices. - Philippe Deléham, Nov 01 2009
Triangle T(n,k), read by rows, given by (0,2,1,4,2,6,3,8,4,10,5,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 19 2011.
Also the Bell transform of A000629. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle begins:
  1;
  0,    1;
  0,    2,    1;
  0,    6,    6,    1;
  0,   26,   36,   12,   1;
  0,  150,  250,  120,  20,  1;
  0, 1082, 2040, 1230, 300, 30,  1;
		

Crossrefs

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> polylog(-n,1/2), 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    rows = 9;
    t = Table[PolyLog[-n, 1/2], {n, 0, rows}]; T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 0, rows}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)
    p[n_] := Sum[StirlingS2[n, k] Pochhammer[x, k], {k, 0, n}];
    Table[CoefficientList[FunctionExpand[p[n]], x], {n, 0, 9}] // Flatten (* Peter Luschny, Jun 27 2019 *)
  • Sage
    def a_row(n):
        s = sum(stirling_number2(n,k)*rising_factorial(x,k) for k in (0..n))
        return expand(s).list()
    [a_row(n) for n in (0..9)] # Peter Luschny, Jun 28 2019

Formula

a(n,m) = Sum_{k=m..n} S2(n,k) * |S1(k,m)|, n>=0; S2=A048993, S1=A048994.
E.g.f. of column k (with leading zeros): (f(x)^k)/k! with f(x):= -log(1-(exp(x)-1)) = -log(2-exp(x)).
Sum_{0<=k<=n} T(n,k)*x^k = A153881(n+1), A000007(n), A000670(n), A005649(n) for x = -1,0,1,2 respectively. - Philippe Deléham, Nov 19 2011

Extensions

New name by Peter Luschny, Jun 27 2019

A209849 Triangle read by rows: coefficients of polynomials in Sum_{k = 0..t} k^n * binomial(t,k).

Original entry on oeis.org

1, 1, 1, 0, 3, 1, -2, 3, 6, 1, 0, -10, 15, 10, 1, 16, -30, -15, 45, 15, 1, 0, 112, -210, 35, 105, 21, 1, -272, 588, 28, -735, 280, 210, 28, 1, 0, -2448, 5292, -2436, -1575, 1008, 378, 36, 1, 7936, -18960, 4140, 20160, -14595, -1575, 2730, 630, 45, 1
Offset: 1

Views

Author

Peter Bala, Mar 15 2012

Keywords

Comments

Repeatedly applying the operator x*d/dx to (1 + x)^t (t a nonnegative integer) and evaluating at x = 1 yields Sum_{k = 0..t} k^n*binomial(t,k) = R(n,t)*2^(t-n), where R(n,t) is a polynomial in t for n = 1,2,.... The polynomial sequence {R(n,t)}_{n>=0} is of binomial type. The first few values are given in the example section below.
This triangle lists the coefficients of these polynomials in ascending powers of t (omitting R(0,t) = 1). A closely related triangle is A102573, which lists the coefficients of the polynomials R(n,t) after factors of t and t*(1 + t) have been removed.
This is the case m = 2 of a family of binomial type polynomials satisfying the recurrence R(n+1,t) = t*(m*(R(n,t) - R(n,t-1)) + R(n,t-1)) with R(0,t) = 1. Case m = 0 gives the falling factorials (A008275); Case m = -1 gives a signed version of A079641.

Examples

			Repeatedly applying the operator x*d/dx to (1 + x)^n and evaluating the result at x = 1 yields
Sum_{k = 0..n} k   * binomial(n,k) =  n                * 2^(n-1).
Sum_{k = 0..n} k^2 * binomial(n,k) = (n +   n^2)       * 2^(n-2).
Sum_{k = 0..n} k^3 * binomial(n,k) = (    3*n^2 + n^3) * 2^(n-3).
Triangle begins:
  n\k|    1     2     3     4     5     6     7     8
  = = = = = = = = = = = = = = = = = = = = = = = = = =
  1  |    1
  2  |    1     1
  3  |    0     3     1
  4  |   -2     3     6     1
  5  |    0   -10    15    10     1
  6  |   16   -30   -15    45    15     1
  7  |    0   112  -210    35   105    21     1
  8  | -272   588    28  -735   280   210    28     1
  ...
		

Crossrefs

Columns k=1..3 give A155585(n-1), A383165, A383166.

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
    g := n -> 2^n*euler(n,1): BellMatrix(g, 9); # Peter Luschny, Jan 21 2016
  • Mathematica
    BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
    rows = 12;
    M = BellMatrix[2^# EulerE[#, 1]&, rows];
    Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 24 2018, after Peter Luschny *)
  • PARI
    T(n, k) = sum(j=k, n, 2^(n-j)*stirling(n, j, 2)*stirling(j, k, 1)); \\ Seiichi Manyama, Apr 16 2025
  • Sage
    # uses[bell_matrix from A264428]
    g = lambda n: sum((-2)^(n-k)*factorial(k)*stirling_number2(n,k) for k in (0..n))
    bell_matrix(g, 9) # Peter Luschny, Jan 21 2016
    
  • Sage
    def a_row(n):
        s = sum(2^(n-k)*stirling_number2(n, k)*falling_factorial(x, k) for k in (0..n))
        return expand(s).list()[1:]
    for n in (1..10): print(a_row(n)) # Seiichi Manyama, Apr 16 2025
    

Formula

T(n,k) = Sum_{j = 0..n} (-1)^(n+k) * (-2)^(n-j) * Stirling2(n,j) * |Stirling1(j,k)|. [corrected by Seiichi Manyama, Apr 16 2025]
E.g.f.: F(x,t) := (1/2 + 1/2*exp(2*x))^t = (1 + tanh(-x))^(-t) = 1 + t*x + (t+t^2)*x^2/2! + (3*t^2+t^3)*x^3/3! + ... satisfies the delay differential equation d/dx(F(x,t)) = 2*F(x,t) - F(x,t-1).
Recurrence for row polynomials R(n,t): R(n+1,t) = t*(2*R(n,t) - R(n,t-1)) with R(0,t) = 1.
Let D be the backward difference operator D(f(x)) = f(x) - f(x-1). Then (x*D)^n(2^x) = 2^(x-n)*R(n,x). Cf. A079641.
Discrete Dobinski-type relation: R(n,x) = 1/2^x*Sum_{k = 0..inf} (2*k)^n*x*(x - 1)*...*(x - k + 1)/k!, valid for x = 0,1,2,.... and n >= 1.
Other Dobinski-type relations: exp(-x)*Sum_{k = 0..inf} R(n,k)*x^k/k! = n-th row polynomial of A075497.
exp(-x)*Sum_{k = 0..inf} R(n,k+1)*x^k/k! = n-th row polynomial of A154602.
i^(-n)*exp(i*x)*Sum_{k = 0..inf} R(n,-k)*(-i*x)^k/k! = n-th row polynomial of A059419 where i = sqrt(-1).
Writing x^[n] in place of R(n,x) we have the analog of the Bernoulli summation formula for powers of integers: Sum_{k = 1..n-1} k^[p] = 1/(p + 1)*Sum_{k = 0..p} 2^k*binomial(p+1,k)*B_k*n^[p+1-k], where B_k = [1,-1/2,1/6,0,-1/30,...] is the sequence of Bernoulli numbers.
n-th row sum R(n,1) equals 2^(n-1). Alternating row sums R(n,-1) starting [-1,0,2,0,-16,0,272,...] are signed tangent numbers - see A009006 and A155585.
R(n+1,2) = 2^n + 4^n = A063376(n).
Triangle as a product of lower triangular arrays equals A075497*A008275.
The triangle of connection constants between the polynomials (x + 1)^[n] and x^[n] appears to be A119468 = (P^2 + 1)/2, where P denotes Pascal's triangle.
Also the Bell transform of the sequence 2^n*E(n,1), E(n,x) the Euler polynomials (A155585). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 21 2016
From Peter Bala, Jun 26 2016: (Start)
With row and column numbering starting at 0:
E.g.f. is exp(x)/cosh(x)*((1 + exp(2*x))/2)^t = 1 + (1 + t)*x + (3*t + t^2)*x^2/2! + (-2 + 3*t + 6*t^2 + t^3)*x^3/3! + ....
Exponential Riordan array [d/dx(f(x)), f(x)] belonging to the Derivative subgroup of the Riordan group, where f(x) = log((1 + exp(2*x))/2) and df/dx = exp(x)/cosh(x) is the e.g.f. for A155585. (End)
T(n,k) = [x^k] Sum_{k=0..n} 2^(n-k) * Stirling2(n,k) * FallingFactorial(x,k). - Seiichi Manyama, Apr 16 2025
E.g.f. of column k (with leading zeros): f(x)^k / k! with f(x) = log(1 + (exp(2*x) - 1)/2). - Seiichi Manyama, Apr 18 2025

A383149 Triangle T(n,k), n >= 0, 0 <= k <= n, read by rows, where T(n,k) = (-1)^k * [m^k] (1/2^(m-n)) * Sum_{k=0..m} k^n * (-1)^m * 3^(m-k) * binomial(m,k).

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 12, 9, 1, 0, 66, 75, 18, 1, 0, 480, 690, 255, 30, 1, 0, 4368, 7290, 3555, 645, 45, 1, 0, 47712, 88536, 52290, 12705, 1365, 63, 1, 0, 608016, 1223628, 831684, 249585, 36120, 2562, 84, 1, 0, 8855040, 19019664, 14405580, 5073012, 915705, 87696, 4410, 108, 1
Offset: 0

Views

Author

Seiichi Manyama, Apr 18 2025

Keywords

Examples

			f_0(m) = 1.
f_1(m) =      -m.
f_2(m) =    -3*m +     m^2.
f_3(m) =   -12*m +   9*m^2 -     m^3.
f_4(m) =   -66*m +  75*m^2 -  18*m^3 +    m^4.
f_5(m) =  -480*m + 690*m^2 - 255*m^3 + 30*m^4 - m^5.
Triangle begins:
  1;
  0,     1;
  0,     3,     1;
  0,    12,     9,     1;
  0,    66,    75,    18,     1;
  0,   480,   690,   255,    30,    1;
  0,  4368,  7290,  3555,   645,   45,  1;
  0, 47712, 88536, 52290, 12705, 1365, 63, 1;
  ...
		

Crossrefs

Columns k=0..3 give A000007, A123227(n-1), A383163, A383164.
Row sums give A122704.

Programs

  • PARI
    T(n, k) = sum(j=k, n, 2^(n-j)*stirling(n, j, 2)*abs(stirling(j, k, 1)));
    
  • Sage
    def a_row(n):
        s = sum(2^(n-k)*stirling_number2(n, k)*rising_factorial(x, k) for k in (0..n))
        return expand(s).list()
    for n in (0..9): print(a_row(n))

Formula

f_n(m) = (1/2^(m-n)) * Sum_{k=0..m} k^n * (-1)^m * 3^(m-k) * binomial(m,k).
T(n,k) = [m^k] f_n(-m).
T(n,k) = Sum_{j=k..n} 2^(n-j) * Stirling2(n,j) * |Stirling1(j,k)|.
T(n,k) = [x^k] Sum_{k=0..n} 2^(n-k) * Stirling2(n,k) * RisingFactorial(x,k).
Sum_{k=0..n} (-1)^k * T(n,k) = f_m(1) = -2^(n-1) for n > 0.
E.g.f. of column k (with leading zeros): g(x)^k / k! with g(x) = -log(1 - (exp(2*x) - 1)/2).
Showing 1-3 of 3 results.