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

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

Author

Peter Bala, Mar 15 2012

Keywords

• sign
• easy
• tabl

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
  ...
		

Links

• Seiichi Manyama, Rows n = 1..140, flattened

Crossrefs

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

Cf. A008275, A009006 (alt. row sums), A059419, A063376, A075497, A079641, A102573, A119468, A129062, A154602, A176668, A195204, A383140.

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