A196837 Coefficient table of numerator polynomials of o.g.f.s for partial sums of powers of positive integers.

1, 2, -3, 3, -12, 11, 4, -30, 70, -50, 5, -60, 255, -450, 274, 6, -105, 700, -2205, 3248, -1764, 7, -168, 1610, -7840, 20307, -26264, 13068, 8, -252, 3276, -22680, 89796, -201852, 236248, -109584, 9, -360, 6090, -56700, 316365, -1077300, 2171040, -2345400, 1026576, 10, -495, 10560, -127050, 946638, -4510275, 13667720, -25228500, 25507152, -10628640

Offset: 1

Author

Wolfdieter Lang, Oct 10 2011

Keywords

• sign
• easy
• tabl

Comments

The k-th power of the positive integers has partial sums Sum_{j=1..n} j^k given as column number n >= 1, in the array A103438 (not in the triangle; see the example array given there; note that 0^0 has been set to 0 there).

The o.g.f. of column number n >= 1 of the array A103438 is obtained via Laplace transformation from the e.g.f. which is given there as

exp(x)*(exp(n*x)-1)/(exp(x)-1) = Sum_{j=1..n} exp(j*x)

(it is trivial that the sum is the e.g.f.).

The o.g.f. is, therefore, Sum_{j=1..n} 1/(1-j*x), which is rewritten as P(n,x)/Product_{j=1..n} (1-j*x). This defines the row polynomials P(n,x) of the present triangle. See the link for details.

This e.g.f. - o.g.f. connection proves some conjectures by Simon Plouffe. See the o.g.f. Maple programs under, e.g., A001551(n=4) and A001552 (n=5).

This triangle organizes the sum of powers of the first n positive integers in terms of the column no. n of the Stirling2 numbers A048993 (see the formula and example given below, as well as the link).

From Wolfdieter Lang, Oct 12 2011: (Start)

With the formulas given below one finds for n >= 1, k >= 0, Sum_{j=1..n} j^k =

Sum_{m=0..min(k,n-1)} ((n-m)*S1(n+1, n-m+1)*S2(k+n-m, n)),

with the Stirling numbers S1 from A048994 and S2 from A048993 (this formula I did not (yet) find in the literature). See the link for the proof.

For two other formulas expressing these sums of k-th powers of the first n positive integers in terms of the row no. k of Stirling2 numbers and binomials in n see the D. E. Knuth reference given under A093556, p. 285.

See also the given link below, eqs. (11) and (12). (End)

Examples

			n\m  0    1    2     3     4      5...
1    1
2    2   -3
3    3  -12   11
4    4  -30   70   -50
5    5  -60  255  -450   274
6    6 -105  700 -2205  3248  -1764
...
n=4 (A001551=2*A196836): the row polynomial factorizes into 2*(2-5*x)*(1-5*x+5*x^2).
n=5: 1^k + 2^k + 3^k + 4^k + 5^k, k>=0, (A001552) has as e.g.f. Sum_{j=1..5} exp(j*x). The o.g.f. is
  Sum_{j=1..5} 1/(1-j*x), and this is
  (5 - 60*x + 255*x^2 - 450*x^3 + 274*x^4)/Product_{j=1..5} (1-j*x).
n=6 (A001553): the row polynomial factorizes into
     (2 - 7*x)*(3 - 42*x + 203*x^2 - 392*x^3 + 252*x^4).
Sums of powers of the first n positive integers in terms of S2:
n=4: A001551(k) = 4*S2(k+4,4) - 30*S2(k+3,4) + 70*S2(k+2,4) - 50*S2(k+1,4), k >= 0. E.g., k=3: 4*350 - 30*65 + 70*10 - 50*1 = 100 = A001551(3).
From _Wolfdieter Lang_, Oct 12 2011: (Start)
Row polynomial for n=3: P(3,x) = (1-2*x)*(1-3*x) + (1-1*x)*(1-3*x) + (1-1*x)*(1-2*x) = 3 - 12*x + 11*x^2.
a(3,2) = +(sigma_2(2,3) + sigma_2(1,3) + sigma_2(1,2)) =
  2*3 + 1*3 + 1*2 = 11 = +1*sigma_2(1,2,3) = +1*|S1(4,4-2)|.
S1,S2 formula for sums of powers with n=4, k=3:
A001551(3) = Sum_{j=1..n} j^3 = 1*4*350 - 3*10*65 + 2*35*10 - 1*50*1 = 100. (End)
		

Links

• José L. Cereceda, A refinement of Lang's formula for the sum of powers of integers, arXiv:2301.02141 [math.NT], 2023.
• Wolfdieter Lang, Proofs and first 15 row polynomials, 2011.

Crossrefs

Cf. A103438, A093556/A093557 (for sums of powers).

Programs

Mathematica

a[n_, m_] := (n-m)*StirlingS1[n+1, n+1-m]; Flatten[ Table[ a[n, m], {n, 1, 10}, {m, 0, n-1}] ] (* Jean-François Alcover, Dec 02 2011, after Wolfdieter Lang *)

Python

from itertools import count, islice
from sympy.functions.combinatorial.numbers import stirling
def A196837_T(n,m): return (n-m)*stirling(n+1,n+1-m,kind=1,signed=True)
def A196837_gen(): # generator of terms
    return (A196837_T(n,m) for n in count(1) for m in range(n))
A196837_list = list(islice(A196837_gen(),40)) # Chai Wah Wu, Oct 24 2024

Formula

a(n,m) = [x^m] P(n,x), m=0..n-1, with the row polynomials defined by

(Sum_{j=1..n} 1/(1-j*x))*Product_{j=1..n} (1-j*x) (see the comment given above).

Sum_{j=1..n} j^k = Sum_{m=0..n-1} a(n,m)*S2(k+n-m,n), n >= 1, k >= 0, with the Stirling2 triangle A048993.

From Wolfdieter Lang, Oct 12 2011: (Start)

The row polynomial P(n,x) is therefore

Sum_{j=1..n} (Product_{k=1..n omitting k=j} (1-k*x)), n >= 1. This leads to:

a(n,m) = (n-m)*S1(n+1, n+1-m), n-1 >= m >= 0, with the (signed) Stirling1 numbers A048994. For the proof see the link.

(End)

A similar polynomial occurs in the expansion of 1/(n+x)^2 as a series with factorials in the denominator: 1/(n+x)^2 = -Sum_{k>=1} n!/(n+k+1)! * P(k,1/x) x^(k-1). - Matt Majic, Nov 01 2019