login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
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. 7
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 (list; table; graph; refs; listen; history; text; internal format)
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.

LINKS

Table of n, a(n) for n=0..77.

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 } = [x/(e^x+1)] [e^((n+1)x) -1 ]/x] = [ [2x/(e^x+1)] e^((n+1)x) -  [2x/(e^x+1)] /2x = (1/2) sum(k >=0, [Eul(k+1,n+1) - Eul(k+1,0)]/(k+1) * 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)

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

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.

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

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy .

Last modified June 22 23:15 EDT 2017. Contains 288633 sequences.