A209849 -id:A209849 - cp's OEIS frontend

A079641 Matrix product of Stirling2-triangle A008277(n,k) and unsigned Stirling1-triangle |A008275(n,k)|.

1, 2, 1, 6, 6, 1, 26, 36, 12, 1, 150, 250, 120, 20, 1, 1082, 2040, 1230, 300, 30, 1, 9366, 19334, 13650, 4270, 630, 42, 1, 94586, 209580, 166376, 62160, 11900, 1176, 56, 1, 1091670, 2562354, 2229444, 952728, 220500, 28476, 2016, 72, 1, 14174522

Offset: 1

Vladeta Jovovic, Jan 30 2003

Triangle T(n,k), 1<=k<=n, 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, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 22 2011
Subtriangle of triangle in A129062. - Philippe Deléham, Feb 17 2013
Also the Bell transform of A000629. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 26 2016

			Triangle begins:
  1;
  2,1;
  6,6,1;
  26,36,12,1;
  150,250,120,20,1;
  1082,2040,1230,300,30,1;
  ...
Triangle (0,2,1,4,2,6,3,8,4,...) DELTA (1,0,1,0,1,0,1,0,1,...) 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. - _Philippe Deléham_, Dec 22 2011

Nick Early, Canonical Bases for Permutohedral Plates, arXiv:1712.08520 [math.CO], 2017.
Nick Early, Honeycomb tessellations and canonical bases for permutohedral blades, arXiv:1810.03246 [math.CO], 2018.
D. E. Knuth, Convolution polynomials, arXiv:math/9207221 [math.CA], 1992; The Mathematica J., 2 (1992), 67-78.

Cf. A000670 (row sums), A000629 (first column), A195204, A195205. A209849, A129062

# The function BellMatrix is defined in A264428.
# Adds (1, 0, 0, 0, ..) as column 0.
BellMatrix(n -> add((-1)^(n-k)*2^k*k!*combinat:-stirling2(n, k), k=0..n), 9); # Peter Luschny, Jan 26 2016

rows = 10;
t = Table[Sum[(-1)^(n-k)*2^k*k!*StirlingS2[n, k], {k,0,n}], {n, 0, rows}];
T[n_, k_] := BellY[n, k, t];
Table[T[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018 *)

T(n, k) = Sum_{i=k..n} A008277(n, i) * |A008275(i, k)|.
E.g.f.: (2-exp(x))^(-y). - Vladeta Jovovic, Nov 22 2003
From Peter Bala, Sep 12 2011: (Start)
The row generating polynomials R(n,x) begin R(1,x) = x, R(2,x) = 2*x + x^2, R(3,x) = 6*x + 6*x^2 + x^3 and satisfy the recurrence R(n+1,x) = x*(2*R(n,x+1) - R(n,x)). They form a sequence of binomial type polynomials. In particular, denoting R(n,x) by x^[n] to emphasize the analogies with the monomial polynomials x^n, we have the binomial expansion (x + y)^[n] = Sum_{k = 0..n} binomial(n,k)*x^[n-k]*y^[k].
There is a Dobinski-type formula: exp(-x)*Sum_{k >= 0} (-k)^[n] * x^k/k! = Bell(n,-x). The alternating n-th row entries (-1)^k * T(n,k) are the connection coefficients expressing the polynomial Bell(n,-x) as a linear combination of Bell(k,x), 1 <= k <= n. For example, the list of coefficients of R(4,x) is [26, 36, 12, 1] and we have Bell(4,-x) = -26*Bell(1,x) + 36*Bell(2,x) - 12*Bell(3,x) + Bell(4,x).
The row polynomials also satisfy an analog of the Bernoulli's summation formula for powers of integers, namely, Sum_{k = 1..n} k^[p] = 1/(p+1) * Sum_{k = 0..p} binomial(p+1,k) * B_k * n^[p+1-k], where B_k denotes the Bernoulli numbers. Compare with A195204 and A195205. (End)
Let D be the forward difference operator D(f(x)) = f(x+1) - f(x). Then the n-th row polynomial R(n,x) = 1/f(x) * (x*D)^n(f(x)) with f(x) = 2^x. Cf. A209849. Also cf. A008277, where the row polynomials are given by 1/f(x) * (x*d/dx)^n(f(x)), where now f(x) = exp(x). - Peter Bala, Mar 16 2012
Conjecture: o.g.f. as a continued fraction of Stieltjes type: 1/(1 - x*z/(1 - 2*z/(1 - (x + 1)*z/(1 - 4*z/(1 - (x + 2)*z/(1 - 6*z/(1 - (x + 3)*z/(1 - 8*z/(1 - ... ))))))))) =  1 + x*z + (2*x + x^2)*z^2 + (6*x + 6*x^2 + x^3)*z^3 + .... - Peter Bala, Dec 12 2024

A102573 Triangle of coefficients of polynomials in Sum_{k=0..n} binomial(n,k) * k^r.

Original entry on oeis.org

1, 1, 3, 1, 5, -2, 1, 10, 15, -10, 1, 14, 31, -46, 16, 1, 21, 105, 35, -210, 112, 1, 27, 183, 97, -832, 860, -272, 1, 36, 378, 1008, -1575, -2436, 5292, -2448, 1, 44, 586, 2144, -3719, -10876, 31036, -26896, 7936, 1, 55, 990, 6270, 3465, -51513, 27720, 135300, -208560

Offset: 2

Eric W. Weisstein, Jan 15 2005

For a table of coefficients of these polynomials without factors removed see A209849. - Peter Bala, Mar 16 2012

			Triangle begins:
  1;
  1, 3;
  1, 5, -2;
  1, 10, 15, -10;
  1, 14, 31, -46, 16;
  ...
E.g. Sum_{k=0..n} binomial(n,k) * k^4 = 2^(n-4) * n * (n+1) * (n^2 + 5*n - 2).

E. Kilic, Y. T. Ulutas and N. Omur, Formulas for weighted binomial sums using the powers of terms of binary recurrences, Miskolc Mathematical Notes, Vol. 13 (2012), No. 1, pp. 53-65. - From N. J. A. Sloane, Dec 16 2012

Eric Weisstein's World of Mathematics, Binomial Sums

Cf. A209849.

A195204 Triangle of coefficients of a sequence of binomial type polynomials.

Original entry on oeis.org

2, 2, 4, 6, 12, 8, 26, 60, 48, 16, 150, 380, 360, 160, 32, 1082, 2940, 3120, 1680, 480, 64, 9366, 26908, 31080, 19040, 6720, 1344, 128, 94586, 284508, 351344, 236880, 96320, 24192, 3584, 256

Offset: 1

Peter Bala, Sep 13 2011

Define a polynomial sequence P_n(x) by means of the recursion
P_(n+1)(x) = x*(P_n(x)+ P_n(x+1)), with P_0(x) = 1.
The first few polynomials are
P_1(x) = 2*x, P_2(x) = 2*x*(2*x + 1),
P_3(x) = 2*x*(4*x^2 + 6*x + 3), P_4(x) = 2*x*(8*x^3+24*x^2+30*x+13).
The present table shows the coefficients of these polynomials (excluding P_0(x)) in ascending powers of x. The P_n(x) are a polynomial sequence of binomial type. In particular, if we denote P_n(x) by x^[n] then we have the analog of the binomial expansion
(x+y)^[n] = Sum_{k = 0..n} binomial(n,k)*x^[n-k]*y^[k].
There are further analogies between the x^[n] and the monomials x^n.
1) Dobinski-type formula
exp(-x)*Sum_{k >= 0} (-k)^[n]*x^k/k! = (-1)^n*Bell(n,2*x),
where the Bell (or exponential) polynomials are defined as
Bell(n,x) := Sum_{k = 1..n} Stirling2(n,k)*x^k.
Equivalently, the connection constants associated with the polynomial sequences {x^[n]} and {x^n} are (up to signs) the same as the connection constants associated with the polynomial sequences {Bell(n,2*x)} and {Bell(n,x)}. For example, the list of coefficients of x^[4] is [26,60,48,16] and a calculation gives
Bell(4,2*x) = -26*Bell(1,x) + 60*Bell(2,x) - 48*Bell(3,x) + 16*Bell(4,x).
2) Analog of Bernoulli's summation formula
Bernoulli's formula for the sum of the p-th powers of the first n positive integers is
Sum_{k = 1..n} k^p = (1/(p+1))*Sum_{k = 0..p} (-1)^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.
This generalizes to
2*Sum_{k = 1..n} k^[p] = 1/(p+1)*Sum_{k = 0..p} (-1)^k * binomial(p+1,k)*B_k*n^[p+1-k].
The polynomials P_n(x) belong to a family of polynomial sequences P_n(x,t) of binomial type, dependent on a parameter t, and defined recursively by P_(n+1)(x,t)= x*(P_n(x,t)+ t*P_n(x+1,t)), with P_0(x,t) = 1. When t = 0 we have P_n(x,0) = x^n, the monomial polynomials. The present table is the case t = 1. The case t = -2 is (up to signs) A079641. See also A195205 (case t = 2).
Triangle T(n,k) (1 <= k <= n), read by rows, given by (0, 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, ...) DELTA (2, 0, 2, 0, 2, 0, 2, 0, 2, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 22 2011
T(n,k) is the number of binary relations R on [n] with index = 1 containing exactly k strongly connected components (SCC's) and satisfying the condition that if (x,y) is in R then x and y are in the same SCC. - Geoffrey Critzer, Jan 17 2024

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....2
..2|....2......4
..3|....6.....12......8
..4|...26.....60.....48.....16
..5|..150....380....360....160.....32
..6|.1082...2940...3120...1680....480.....64
..7|.9366..26908..31080..19040...6720...1344....128
...
Relation with rising factorials for row 4:
x^[4] = 16*x^4+48*x^3+60*x^2+26*x = 2^4*x*(x+1)*(x+2)*(x+3)-6*2^3*x*(x+1)*(x+2)+7*2^2*x*(x+1)-2*x, where [1,7,6,1] is the fourth row of the triangle of Stirling numbers of the second kind A008277.
Generalized Dobinski formula for row 4:
exp(-x)*Sum_{k >= 1} (-k)^[4]*x^k/k! = exp(-x)*Sum_{k >= 1} (16*k^4-48*k^3+60*k^2-26*k)*x^k/k! = 16*x^4+48*x^3+28*x^2+2*x = Bell(4,2*x).
Example of generalized Bernoulli summation formula:
2*(1^[2]+2^[2]+...+n^[2]) = 1/3*(B_0*n^[3]-3*B_1*n^[2]+3*B_2*n^[1]) =
n*(n+1)*(4*n+5)/3, where B_0 = 1, B_1 = -1/2, B_2 = 1/6 are Bernoulli numbers.
From _Philippe Deléham_, Dec 22 2011: (Start)
Triangle (0, 1, 2, 2, 4, 3, 6, ...) DELTA (2, 0, 2, 0, 2, ...) begins:
  1;
  0,    2;
  0,    2,     4;
  0,    6,    12,     8;
  0,   26,    60,    48,    16;
  0,  150,   380,   360,   160,   32;
  0, 1082,  2940,  3120,  1680,  480,   64;
  0, 9366, 26908, 31080, 19040, 6720, 1344, 128;
  ... (End)

Cf. A000629 (row sums), A000670 (one half row sums), A014307 (row polys. at x = 1/2), A079641, A195205, A209849.

# The function BellMatrix is defined in A264428.
# Adds (1,0,0,0, ..) as column 0.
BellMatrix(n -> (-1)^(n+1)*polylog(-n, 2), 10); # Peter Luschny, Jan 29 2016

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[(-1)^(#+1) PolyLog[-#, 2]&, rows];
Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 24 2018, after Peter Luschny *)

E.g.f.: F(x,z) := (exp(z)/(2-exp(z)))^x = Sum_{n>=0} P_n(x)*z^n/n!
= 1 + 2*x*z + (2*x+4*x^2)*z^2/2! + (6*x+12*x^2+8*x^3)*z^3/3! + ....
The generating function F(x,z) satisfies the partial differential equation d/dz(F(x,z)) = x*F(x,z) + x*F(x+1,z) and hence the row polynomials P_n(x) satisfy the recurrence relation
P_(n+1)(x)= x*(P_n(x) + P_n(x+1)), with P_0(x) = 1.
In what follows we change notation and write x^[n] for P_n(x).
Relation with the factorial polynomials:
For n >= 1,
x^[n] = Sum_{k = 1..n} (-1)^(n-k)*Stirling2(n,k)*2^k*x^(k),
and its inverse formula
2^n*x^(n) = Sum_{k = 1..n} |Stirling1(n,k)|*x^[k],
where x^(n) denotes the rising factorial x*(x+1)*...*(x+n-1).
Relation with the Bell polynomials:
The alternating n-th row entries (-1)^(n+k)*T(n,k) are the connection coefficients expressing the polynomial Bell(n,2*x) as a linear combination of Bell(k,x), 1 <= k <= n.
The delta operator:
The sequence of row polynomials is of binomial type. If D denotes the derivative operator d/dx then the delta operator D* for this sequence of binomial type polynomials is given by
D* = D/2 - log(cosh(D/2)) = log(2*exp(D)/(exp(D)+1))
= (D/2) - (D/2)^2/2! + 2*(D/2)^4/4! - 16*(D/2)^6/6! + 272*(D/2)^8/8! - ...,
where [1,2,16,272,...] is the sequence of tangent numbers A000182.
D* is the lowering operator for the row polynomials
(D*)x^[n] = n*x^[n-1].
Associated Bernoulli polynomials:
Generalized Bernoulli polynomial GB(n,x) associated with the polynomials x^[n] may be defined by
GB(n,x) := ((D*)/(exp(D)-1))x^[n].
They satisfy the difference equation
GB(n,x+1) - GB(n,x) = n*x^[n-1]
and have the expansion
GB(n,x) = -(1/2)*n*x^[n-1] + (1/2)*Sum_{k = 0..n} binomial(n,k) * B_k * x^[n-k], where B_k denotes the ordinary Bernoulli numbers.
The first few polynomials are
GB(0,x) = 1/2, GB(1,x) = x-3/4, GB(2,x) = 2*x^2-2*x+1/12,
GB(3,x) = 4*x^3-3*x^2-x, GB(4,x) = 8*x^4-4*x^2-4*x-1/60.
It can be shown that
1/(n+1)*(d/dx)(GB(n+1,x)) = Sum_{i = 0..n} 1/(i+1) * Sum_{k = 0..i} (-1)^k *binomial(i,k)*(x+k)^[n].
This generalizes a well-known formula for Bernoulli polynomials.
Relations with other sequences:
Row sums: A000629(n) = 2*A000670(n). Column 1: 2*A000670(n-1). Row polynomials evaluated at x = 1/2: {P_n(1/2)}n>=0 = [1,1,2,7,35,226,...] = A014307.
T(n,k) = A184962(n,k)*2^k. - Philippe Deléham, Feb 17 2013
Also the Bell transform of A076726. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 29 2016
Conjecture: o.g.f. as a continued fraction of Stieltjes type: 1/(1 - 2*x*z/(1 - z/(1 - 2*(x + 1)*z/(1 - 2*z/(1 - 2*(x + 2)*z/(1 - 3*z/(1 - 2*(x + 3)*z/(1 - 4*z/(1 - ... ))))))))). - Peter Bala, Dec 12 2024

a(1) added by Philippe Deléham, Dec 22 2011

A383140 Triangle read by rows: the coefficients of polynomials (1/3^(m-n)) * Sum_{k=0..m} k^n * 2^(m-k) * binomial(m,k) in the variable m.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 2, 6, 1, 0, -6, 20, 12, 1, 0, -30, 10, 80, 20, 1, 0, 42, -320, 270, 220, 30, 1, 0, 882, -1386, -770, 1470, 490, 42, 1, 0, 954, 7308, -15064, 2800, 5180, 952, 56, 1, 0, -39870, 101826, -39340, -61992, 29820, 14364, 1680, 72, 1, 0, -203958, -40680, 841770, -666820, -86940, 139440, 34020, 2760, 90, 1

Offset: 0

Seiichi Manyama, Apr 17 2025

			f_n(m) = (1/3^(m-n)) * Sum_{k=0..m} k^n * 2^(m-k) * binomial(m,k).
f_0(m) = 1.
f_1(m) =    m.
f_2(m) =  2*m +   m^2.
f_3(m) =  2*m + 6*m^2 + m^3.
Triangle begins:
  1;
  0,   1;
  0,   2,    1;
  0,   2,    6,   1;
  0,  -6,   20,  12,   1;
  0, -30,   10,  80,  20,  1;
  0,  42, -320, 270, 220, 30, 1;
  ...

Eric Weisstein's World of Mathematics, Falling Factorial

Columns k=0..1 give A000007, A179929(n-1).
Row sums give A133494.
Alternating row sums give A212846.
Cf. A027471, A383136, A383137, A383138, A383139.
Cf. A129062, A209849.

T(n, k) = sum(j=k, n, 3^(n-j)*stirling(n, j, 2)*stirling(j, k, 1));

def a_row(n):
    s = sum(3^(n-k)*stirling_number2(n, k)*falling_factorial(x, k) for k in (0..n))
    return expand(s).list()
for n in (0..10): print(a_row(n))

T(n,k) = Sum_{j=k..n} 3^(n-j) * Stirling2(n,j) * Stirling1(j,k).
T(n,k) = [x^k] Sum_{k=0..n} 3^(n-k) * Stirling2(n,k) * FallingFactorial(x,k).
E.g.f. of column k (with leading zeros): g(x)^k / k! with g(x) = log(1 + (exp(3*x) - 1)/3).

A176668 Triangle T(n,k) read by rows: coefficient [x^k] of the polynomial sum_{k=0..infinity} (2k+1)^nbinomial(x,k) / 2^x.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 8, 21, 10, 1, 1, 5, 45, 55, 15, 1, 1, 7, 30, 185, 120, 21, 1, 1, 70, -77, 245, 595, 231, 28, 1, 1, 72, 490, -756, 1435, 1596, 406, 36, 1, 1, -1311, 3762, -546, -2625, 6111, 3738, 666, 45, 1, 1, -1309, -11325, 35130, -20895, -1743, 20685

Offset: 0

Roger L. Bagula, Apr 23 2010

Row sums are A007051(n).

Exponential Riordan array [exp(x), log((exp(2x)+1)/2)]=[exp(x),x+log(cosh(x))]. - Paul Barry Jan 10 2011

			1;
1, 1;
1, 3, 1;
1, 6, 6, 1;
1, 8, 21, 10, 1;
1, 5, 45, 55, 15, 1;
1, 7, 30, 185, 120, 21, 1;
1, 70, -77, 245, 595, 231, 28, 1;
1, 72, 490, -756, 1435, 1596, 406, 36, 1;
1, -1311, 3762, -546, -2625, 6111, 3738, 666, 45, 1;
1, -1309, -11325, 35130, -20895, -1743, 20685, 7890, 1035, 55, 1;
Production matrix begins
1, 1,
0, 2, 1,
0, -1, 3, 1,
0, 1, -3, 4, 1,
0, -1, 4, -6, 5, 1,
0, 1, -5, 10, -10, 6, 1,
0, -1, 6, -15, 20, -15, 7, 1,
0, 1, -7, 21, -35, 35, -21, 8, 1,
0, -1, 8, -28, 56, -70, 56, -28, 9, 1
- _Paul Barry_ Jan 10 2011

Cf. A007051, A009393.

A176668 := proc(n,k) sum( (2*l+1)^n*binomial(x,l),l=0..infinity) ; simplify(%/2^x) ; coeftayl(%,x=0,k) ; end proc: # R. J. Mathar, Jan 15 2011

p[x_, n_] = Sum[(2*k + 1)^n*Binomial[x, k], {k, 0, Infinity}]/2^x ;
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];
Flatten[%]

From Peter Bala, Mar 16 2012. (Start)
The row polynomials of this triangle may be obtained by applying the operator x*d/dx repeatedly to x*(1+x^2)^n = sum {k = 0..n} binomial(n,k)*x^(2*k+1) and evaluating the result at x = 1. The first few results are:
sum {k = 0..n} (2*k+1)*binomial(n,k) = (n+1)*2^n
sum {k = 0..n} (2*k+1)^2*binomial(n,k) = (n^2+3*n^+1)*2^n
sum {k = 0..n} (2*k+1)^3*binomial(n,k) = (n^3+6*n^2+6*n+1)*2^n.
Compare with A209849. (End)

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

Seiichi Manyama, Apr 18 2025

			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;
  ...

Eric Weisstein's World of Mathematics, Rising Factorial

Columns k=0..3 give A000007, A123227(n-1), A383163, A383164.
Row sums give A122704.
Cf. A001787, A178987, A383150, A383151, A383152, A383155.
Cf. A129062, A209849, A383140.

T(n, k) = sum(j=k, n, 2^(n-j)*stirling(n, j, 2)*abs(stirling(j, k, 1)));

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))

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).

A383166 Expansion of e.g.f. log(1 + (exp(2*x) - 1)/2)^3 / 6.

Original entry on oeis.org

0, 0, 0, 1, 6, 15, -15, -210, 28, 5292, 4140, -208560, -369864, 11847264, 33630688, -917280000, -3642944640, 92903375616, 479824306944, -11926470604800, -76477342307840, 1892813347934208, 14591875555074048, -363945109924577280, -3293838565260693504, 83374884181664563200

Offset: 0

Seiichi Manyama, Apr 18 2025

sign

Column k=3 of A209849.

Cf. A383164.

a(n) = sum(k=3, n, 2^(n-k)*stirling(n, k, 2)*stirling(k, 3, 1));

a(n) = Sum{k=3..n} 2^(n-k) * Stirling2(n,k) * Stirling1(k,3).

A349706 Square array T(n,k) = Sum_{j=0..k} binomial(k,j) * j^n for n and k >= 0, read by ascending antidiagonals.

Original entry on oeis.org

1, 0, 2, 0, 1, 4, 0, 1, 4, 8, 0, 1, 6, 12, 16, 0, 1, 10, 24, 32, 32, 0, 1, 18, 54, 80, 80, 64, 0, 1, 34, 132, 224, 240, 192, 128, 0, 1, 66, 342, 680, 800, 672, 448, 256, 0, 1, 130, 924, 2192, 2880, 2592, 1792, 1024, 512, 0, 1, 258, 2574, 7400, 11000, 10752, 7840, 4608, 2304, 1024

Offset: 0

Michel Marcus, Nov 26 2021

			Square array begins:
  1 2  4   8   16    32
  0 1  4  12   32    80
  0 1  6  24   80   240
  0 1 10  54  224   800
  0 1 18 132  680  2880
  0 1 34 342 2192 11000

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Renate Golombek, Aufgabe 1088, El. Math., 49 (1994), 126-127.
Simsek Yilmaz, New families of special numbers for computing negative order Euler numbers and related numbers and polynomials, Applicable Analysis and Discrete Mathematics 2018 Volume 12, Issue 1, Pages: 1-35. See B(n,k).

Rows n=0..7 give A000079, A001787, A001788, A058645, A058649, A059338, A056468, A084641.
Main diagonal gives A072034.
Cf. A209849.

T[n_, k_] := Sum[Binomial[k, j] * If[j == n == 0, 1, j^n], {j, 0, k}]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 26 2021 *)

T(n,k) = sum(j=0, k, binomial(k,j)*j^n);

cp's OEIS Frontend

A079641 Matrix product of Stirling2-triangle A008277(n,k) and unsigned Stirling1-triangle |A008275(n,k)|.

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A102573 Triangle of coefficients of polynomials in Sum_{k=0..n} binomial(n,k) * k^r.

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A195204 Triangle of coefficients of a sequence of binomial type polynomials.

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A383140 Triangle read by rows: the coefficients of polynomials (1/3^(m-n)) * Sum_{k=0..m} k^n * 2^(m-k) * binomial(m,k) in the variable m.

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A176668 Triangle T(n,k) read by rows: coefficient [x^k] of the polynomial sum_{k=0..infinity} (2*k+1)^n*binomial(x,k) / 2^x.

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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).

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A383166 Expansion of e.g.f. log(1 + (exp(2*x) - 1)/2)^3 / 6.

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A349706 Square array T(n,k) = Sum_{j=0..k} binomial(k,j) * j^n for n and k >= 0, read by ascending antidiagonals.

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A176668 Triangle T(n,k) read by rows: coefficient [x^k] of the polynomial sum_{k=0..infinity} (2k+1)^nbinomial(x,k) / 2^x.