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

%I A196837 #56 Oct 24 2024 14:42:04
%S A196837 1,2,-3,3,-12,11,4,-30,70,-50,5,-60,255,-450,274,6,-105,700,-2205,
%T A196837 3248,-1764,7,-168,1610,-7840,20307,-26264,13068,8,-252,3276,-22680,
%U A196837 89796,-201852,236248,-109584,9,-360,6090,-56700,316365,-1077300,2171040,-2345400,1026576,10,-495,10560,-127050,946638,-4510275,13667720,-25228500,25507152,-10628640
%N A196837 Coefficient table of numerator polynomials of o.g.f.s for partial sums of powers of positive integers.
%C A196837 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).
%C A196837 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
%C A196837   exp(x)*(exp(n*x)-1)/(exp(x)-1) = Sum_{j=1..n} exp(j*x)
%C A196837   (it is trivial that the sum is the e.g.f.).
%C A196837   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.
%C A196837 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).
%C A196837 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).
%C A196837 From _Wolfdieter Lang_, Oct 12 2011: (Start)
%C A196837 With the formulas given below one finds for n >= 1, k >= 0, Sum_{j=1..n} j^k =
%C A196837   Sum_{m=0..min(k,n-1)} ((n-m)*S1(n+1, n-m+1)*S2(k+n-m, n)),
%C A196837   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.
%C A196837 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.
%C A196837   See also the given link below, eqs. (11) and (12). (End)
%H A196837 José L. Cereceda, <a href="https://arxiv.org/abs/2301.02141">A refinement of Lang's formula for the sum of powers of integers</a>, arXiv:2301.02141 [math.NT], 2023.
%H A196837 Wolfdieter Lang, <a href="/A196837/a196837.pdf">Proofs and first 15 row polynomials</a>, 2011.
%F A196837 a(n,m) = [x^m] P(n,x), m=0..n-1, with the row polynomials defined by
%F A196837   (Sum_{j=1..n} 1/(1-j*x))*Product_{j=1..n} (1-j*x) (see the comment given above).
%F A196837 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.
%F A196837 From _Wolfdieter Lang_, Oct 12 2011: (Start)
%F A196837 The row polynomial P(n,x) is therefore
%F A196837 Sum_{j=1..n} (Product_{k=1..n omitting k=j} (1-k*x)), n >= 1. This leads to:
%F A196837 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.
%F A196837 (End)
%F A196837 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
%e A196837 n\m  0    1    2     3     4      5...
%e A196837 1    1
%e A196837 2    2   -3
%e A196837 3    3  -12   11
%e A196837 4    4  -30   70   -50
%e A196837 5    5  -60  255  -450   274
%e A196837 6    6 -105  700 -2205  3248  -1764
%e A196837 ...
%e A196837 n=4 (A001551=2*A196836): the row polynomial factorizes into 2*(2-5*x)*(1-5*x+5*x^2).
%e A196837 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
%e A196837   Sum_{j=1..5} 1/(1-j*x), and this is
%e A196837   (5 - 60*x + 255*x^2 - 450*x^3 + 274*x^4)/Product_{j=1..5} (1-j*x).
%e A196837 n=6 (A001553): the row polynomial factorizes into
%e A196837      (2 - 7*x)*(3 - 42*x + 203*x^2 - 392*x^3 + 252*x^4).
%e A196837 Sums of powers of the first n positive integers in terms of S2:
%e A196837 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).
%e A196837 From _Wolfdieter Lang_, Oct 12 2011: (Start)
%e A196837 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.
%e A196837 a(3,2) = +(sigma_2(2,3) + sigma_2(1,3) + sigma_2(1,2)) =
%e A196837   2*3 + 1*3 + 1*2 = 11 = +1*sigma_2(1,2,3) = +1*|S1(4,4-2)|.
%e A196837 S1,S2 formula for sums of powers with n=4, k=3:
%e A196837 A001551(3) = Sum_{j=1..n} j^3 = 1*4*350 - 3*10*65 + 2*35*10 - 1*50*1 = 100. (End)
%t A196837 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_ *)
%o A196837 (Python)
%o A196837 from itertools import count, islice
%o A196837 from sympy.functions.combinatorial.numbers import stirling
%o A196837 def A196837_T(n,m): return (n-m)*stirling(n+1,n+1-m,kind=1,signed=True)
%o A196837 def A196837_gen(): # generator of terms
%o A196837     return (A196837_T(n,m) for n in count(1) for m in range(n))
%o A196837 A196837_list = list(islice(A196837_gen(),40)) # _Chai Wah Wu_, Oct 24 2024
%Y A196837 Cf. A103438, A093556/A093557 (for sums of powers).
%K A196837 sign,easy,tabl
%O A196837 1,2
%A A196837 _Wolfdieter Lang_, Oct 10 2011