0,2

We make a few remarks about the general array M(a) := (I - a*P)^-1,

where a <> 1, and its connection with weighted sums of powers of

positive integers. The present case corresponds to -M(2).

The array M(a) begins

/

| 1/(1-a)

| a/(1-a)^2............... 1/(1-a)

| (a+a^2)/(1-a)^3......... 2*a/(1-a)^2........ 1/(1-a)

| (a+4*a^2+a^3)/(1-a)^4... 3*(a+a^2)/(1-a)^3.. 3*a/(1-a)^2... 1/(1-a)

| ...

\

In the first column the numerator polynomials are the Eulerian

polynomials A_n(a). See A008292.

The e.g.f. for this array is

(1)... exp(x*t)/(1-a*exp(t)) = 1/(1-a) + [a/(1-a)^2 + x/(1-a)]*t

+ [(a+a^2)/(1-a)^3 + 2*a*x/(1-a)^2 + x^2/(1-a)]*t^2/2! + ....

The row generating polynomials P_m(x) of the array M(a), which,

of course, depend on a, have properties similar to those of the

Bernoulli polynomials. They form an Appell sequence and may be

expressed in terms of the Eulerian polynomials as

(2)... P_m(x) = sum {k=0..m} binomial(m,k)*A_k(a)/(1-a)^(k+1)*x^(m-k).

As a Newton series we have

(3)... P_m(x) = 1/(1-a)*sum {j = 0..m} sum {k = j..m}(a/(1-a))^j * k!

*Stirling2(m,k)*binomial(x,k-j).

The proof of this result in the particular case a = -1 given in

[Roman, p.100] can be easily generalised to a proof of (3).

A result equivalent to (3) is

(4)... P_m(x) = 1/(1-a)*sum {j = 0..m} sum {k = 0..j}(a/(1-a))^j

*(-1)^(j-k)*comb(j,k)*(x + k)^m,

which in turn leads to the infinite series expansion

(5)... P_m(x) = sum {k = 0..inf} a^k*(x + k)^m,

provided |a| < 1. See [Nelsen].

The polynomials P_m(x) satisfy the difference equation

(6)... P_m(x) - a*P_m(x + 1) = x^m (recall a <> 1),

which leads easily to the evaluation of the weighted sums of powers

of integers

(7)... sum {k = 0..n-1} a^k*k^m = P_m(0) - a^n*P_m(n).

for m = 0,1,2,... and a <> 1.

More generally we have

(8)... sum {k = 0..n-1} a^k*(x + k)^m = P_m(x) - a^n*P_m(x + n).

for m = 0,1,2,... and a <> 1.

In the remaining case a = 1 the sums are evaluated in terms of the

Bernoulli polynomials.

The most well-studied case is when a = -1. The row polynomials of the

array M(-1) are then one half of the Euler polynomials E_m(x), which

may be used to evaluate the alternating sums of powers of integers

(9)... 2*sum {k = 1..n-1} (-1)^k*k^m = E_m(0) - (-1)^n*E_m(n).

G. F. C. De Bruyn, Formulas for a + a^2*2^p + a^3*3^p + ... + a^n*n^p. Fibonacci Quart. 33 (1995), no. 2, 98-103.

R. B. Nelsen, Problem E3062: Amer. Math. Monthly, Vol. 94, No. 4 (Apr., 1987), 376-377.

S. Roman, The Umbral Calculus, Dover Publications.

TABLE ENTRIES

(1)... T(n,k) = (-1)^(n+k)*binomial(n,k)*A000629(n-k).

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

*Stirling2(n-k+1,j+1).

GENERATING FUNCTION

(3)... exp(x*t)/(2*exp(t)-1) = 1 + (-2+x)*t + (6-4*x+x^2)*t^2/2!

+ ....

PROPERTIES OF ROW POLYNOMIALS

The row generating polynomials R_n(x) form an Appell sequence. The

first few values are R_0(x) = 1, R_1(x) = x-2, R_3(x) = x^2-4*x+6

and R_4(x) = x^3-6*x^2+18*x-26.

They may be recursively computed by means of

(4)... R_n(x) = x^n - 2*sum {k = 0..n-1} binomial(n,k)*R_k(x).

Explicit formulas are

(5)... R_n(x) = sum {j = 0..n} sum {k = j..n} (-2)^j*k!*Stirling2(n,k)

* binomial(x,k-j),

(6)... R_n(x) = (-1)^n*sum {j = 0..n} sum {k = j..n} k!*Stirling2(n,k)

*binomial(-x+1,k-j),

as well as

(7)... R_n(x) = sum {j = 0..n} sum {k = 0..j} 2^j*(-1)^k*comb(j,k)

*(x + k)^n.

Other expansions include

(8)... R_n(x) = sum {k = 0..n} binomial(n,k)*(-1)^k*A000670(k)

*(x-1)^(n-k),

(9)... R_n(x) = sum {k = 0..n} binomial(n,k)*(-1/2)^k*A080253(k)

*(x-1/2)^(n-k)

and

(10)... R_n(x) = sum {k = 0..n} binomial(n,k)*(-1)^k*A007047(k)

*(x+1)^(n-k).

SUMS OF POWERS OF INTEGERS

The row polynomials satisfy the difference equation

(11)... 2*R_n(x+1) - R_n(x) = x^n,

and so may be used to evaluate the weighted sum of powers of integers

(12)... sum {k = 0..n-1} 2^k*k^m = 2^n*R_m(n) - R_m(0).

For example, m = 3 gives

(13)... sum {k = 0..n-1} 2^k*k^3 = 2^n*(n^3-6*n^2+18*n-26) + 26.

More generally we have

(14)... sum {k = 0..n-1} 2^k*(x + k)^m = 2^n*R_m(x + n) - R_m(x).

RELATIONS WITH OTHER SEQUENCES

(15)... Row sums [1,-1,3,-13,75,...] = (-1)^n*A000670(n).

(16)... Alt. row sums [1,-3,11,-51,299,...] = (-1)^n*A007047(n).

(17)... Column 0: (-1)^n*A000629(n).

(18)... (-2)^n*R_n(1/2) = A080253(n).

(19)... R_n(1-x) = (-1)^n*P_n(x),

where P_n(x) are the row generating polynomials of A154921.

This provides the connection between (12) and the result

(20)... sum {k = 0..n-1} (1/2)^k*k^m = 2*P_m(0) - (1/2)^(n-1)*P_m(n).

Triangle begins

====================================================

n\k|.....0......1......2......3......4......5......6

====================================================

0..|.....1

1..|....-2......1

2..|.....6.....-4......1

3..|...-26.....18.....-6......1

4..|...150...-104.....36.....-8......1

5..|.-1082....750...-260.....60....-10......1

6..|..9366..-6492...2250...-520.....90....-12......1

...

#A162312

with(combinat):

T := (n, k) -> (-1)^(n+k)*binomial(n, k)

*add( j!*stirling2(n-k+1, j+1), j = 0..n):

for n from 0 to 9 do

seq(T(n, k), k = 0..n);

end do;

A000629, A000670, A007047, A008292, A080253, A154921.

Sequence in context: A132159 A112356 A135885 this_sequence A141715 A098697 A021466

Adjacent sequences: A162309 A162310 A162311 this_sequence A162313 A162314 A162315

easy,sign,tabl

Peter Bala (pbala(AT)talktalk.net), Jul 01 2009