%I #130 Mar 28 2024 23:53:02
%S 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,
%T 4,-12,22,-24,16,-6,1,1,-4,16,-34,46,-40,22,-7,1,0,5,-20,50,-80,86,
%U -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
%N 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.
%C 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:
%C (A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
%C (B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
%C (C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
%C More generally, for polynomial sequences,
%C (D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
%C where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
%C 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.
%C Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
%C Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
%C 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.
%C Equals A112468 with the first column of ones removed. - _Georg Fischer_, Jul 26 2023
%H G. C. Greubel, <a href="/A239473/b239473.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H J. Adams, <a href="http://dx.doi.org/10.1016/0040-9383(65)90040-6">On the groups J(x)-II</a>, Topology, Vol. 3, p. 137-171, Pergamon Press, (1965).
%H MathOverflow, <a href="http://mathoverflow.net/questions/82560/cyclotomic-polynomials-in-combinatorics">Cyclotomic Polynomials in Combinatorics</a>
%H Mathoverflow, <a href="http://mathoverflow.net/questions/160686/inequality-for-laguerre-polynomials">Inequality for Laguerre polynomials</a>
%F T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
%F E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
%F O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
%F Associated operator identities:
%F 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),
%F A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
%F = Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
%F B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
%F = Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
%F Associated binomial identities:
%F A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
%F = Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
%F B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
%F = Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
%F = Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
%F 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.
%F 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.
%F With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
%F 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.
%F 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.
%F 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
%F From _Tom Copeland_, Dec 26 2015: (Start)
%F 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.
%F 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.
%F (End)
%F 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
%F From _Tom Copeland_, Sep 05 2016:
%F 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.
%F 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)
%F T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - _Peter Luschny_, Jul 09 2019
%F a(n) = -n + Sum_{k=0..n} A341091(k). - _Thomas Scheuerle_, Jun 17 2022
%e 1
%e 0 1
%e 1 -1 1
%e 0 2 -2 1
%e 1 -2 4 -3 1
%e 0 3 -6 7 -4 1
%e 1 -3 9 -13 11 -5 1
%e 0 4 -12 22 -24 16 -6 1
%e 1 -4 16 -34 46 -40 22 -7 1
%e 0 5 -20 50 -80 86 -62 29 -8 1
%e 1 -5 25 -70 130 -166 148 -91 37 -9 1
%p A239473 := proc(n,k)
%p add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
%p end proc; # _R. J. Mathar_, Jul 21 2016
%t Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* _G. C. Greubel_, Feb 06 2018 *)
%o (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
%o (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
%o (Sage)
%o Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
%o for n in (0..10): print(Trow(n)) # _Peter Luschny_, Jul 09 2019
%Y For column 2: A001057, A004526, A008619, A140106.
%Y Column 3: A002620, A087811.
%Y Column 4: A002623, A173196.
%Y Column 5: A001752.
%Y Column 6: A001753.
%Y Cf. Bottomley's cross-references in A059260.
%Y Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.
%Y Cf. A049310, A112468, A341091.
%K sign,tabl,easy
%O 0,8
%A _Tom Copeland_, Mar 19 2014
%E Inverse array added by _Tom Copeland_, Mar 26 2014
%E Formula re Euler polynomials corrected by _Tom Copeland_, Mar 08 2024