A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1

Offset

0,8

Comments

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:

(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;

(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);

(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).

More generally, for polynomial sequences,

(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,

where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).

Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.

Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.

Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).

From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.

Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

Links

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

J. Adams, On the groups J(x)-II, Topology, Vol. 3, p. 137-171, Pergamon Press, (1965).

MathOverflow, Cyclotomic Polynomials in Combinatorics

Mathoverflow, Inequality for Laguerre polynomials

Formula

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).

E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).

O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).

Associated operator identities:

With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),

A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!

= Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)

B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)

= Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).

Associated binomial identities:

A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)

= Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)

B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)

= Sum_{k=0..n} (-1)^k bin(-s-1+k,k)

= Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).

In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.

From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.

With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):

Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.

From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.

U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014

From Tom Copeland, Dec 26 2015: (Start)

With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) - (x / (e^x-1)) ] / x = Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.

With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials.

(End)

As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016

From Tom Copeland, Sep 05 2016:

Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.

The quantum integers [n+1]_q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n) (1 - q^(2(n+1)) / (1 - q^2) = q^(-n)*Sum_{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)

T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019

a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Example

   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1

Maple

A239473 := proc(n, k)
    add(binomial(j, k)*(-1)^(j+k), j=k..n) ;
end proc; # R. J. Mathar, Jul 21 2016

Mathematica

Table[Sum[(-1)^(j+k)*Binomial[j, k], {j, 0, n}], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)

Prog

(PARI) for(n=0, 10, for(k=0, n, print1(sum(j=0, n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018
(Magma) [[(&+[(-1)^(j+k)*Binomial(j, k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
(Sage)
Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Crossrefs

For column 2: A001057, A004526, A008619, A140106.

Column 3: A002620, A087811.

Column 4: A002623, A173196.

Column 5: A001752.

Column 6: A001753.

Cf. Bottomley's cross-references in A059260.

Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.

Cf. A049310, A112468, A341091.

Sequence in context: A238453 A066287 A059260 * A135229 A257543 A081372

Adjacent sequences: A239470 A239471 A239472 * A239474 A239475 A239476

Keyword

sign,tabl,easy

Author

Tom Copeland, Mar 19 2014

Extensions

Inverse array added by Tom Copeland, Mar 26 2014

Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

Status

approved