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Triangle read by rows, denominators of Jakob Bernoulli's "Sums of Powers" triangle.
5

%I #22 Feb 04 2023 14:15:23

%S 1,2,2,3,2,6,4,2,4,5,2,3,-30,6,2,12,-12,7,2,2,-6,42,8,2,12,-24,12,9,2,

%T 3,-15,9,-30,10,2,4,-10,2,-20,11,2,6,-1,1,-2,66,12,2,12,-8,6,-8,12,13,

%U 2,1,-6,7,-10,3,-2730,14,2,12,-60,28,-20,12,-420,15,2,6,-30,18,-10,6,-90,6,16,2,4,-24,12,-16,12,-24,4

%N Triangle read by rows, denominators of Jakob Bernoulli's "Sums of Powers" triangle.

%C Let the triangle = T. Row sums = 1. Row sums of n-th binomial transform of T = powers of (n-1). Then multiply the results by the partial sum operator, (1; 1,1; 1,1,1; ...) to obtain Bernoulli's "Sums of Powers".

%C Inserting zeros to account for (n+1) terms per row, right border = Bernoulli numbers: (A106458): (1, 1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, ...).

%D Jakob Bernoulli, "Ars conjectandi", posthumously published in 1713, in which Bernoulli gives the table "Summae Potestatum (Sums of Powers) [cf. Young, p. 86].

%D Robert M. Young, "Excursions in Calculus", MAA, 1992.

%H Seiichi Manyama, <a href="/A159688/b159688.txt">Rows n = 0..200, flattened</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Faulhaber%27s_formula">Faulhaber's formula</a>.

%e Let row 0 = 1; followed by the corrected table, giving denominators:

%e 1;

%e 2, 2;

%e 3, 2, 6;

%e 4, 2, 4;

%e 5, 2, 3, -30;

%e 6, 2, 12, -12;

%e 7, 2, 2, -6, 42;

%e 8, 2, 12, -24, 12;

%e 9, 2, 3, -15, 9, -30;

%e 10, 2, 4, -10, 2, -20;

%e 11, 2, 6, -1, 1, -2, 66;

%e ...

%e The complete triangle with row 0 = 1, along with numerators:

%e 1;

%e 1/2, 1/2;

%e 1/3, 1/2, 1/6;

%e 1/4, 1/2, 1/4;

%e 1/5, 1/2, 1/3, -1/30;

%e 1/6, 1/2, 5/12, -1/12;

%e 1/7, 1/2, 1/2, -1/6, 1/42;

%e 1/8, 1/2, 7/12, -7/14, 1/12;

%e 1/9, 1/2, 2/3, -7/15, 1/2, -3/20;

%e 1/10, 1/2, 3/4, -7/10, 1/2, -3/20;

%e 1/11, 1/2, 5/6, -1/1, 1/1, -1/2, 5/66;

%e ...

%t f[n_, x_] := f[n, x] = ((x+1)^(n+1) - 1)/(n+1) - Sum[Binomial[n+1, k]*f[k, x], {k, 0, n-1}]/(n+1); f[0, x_] := x; row[n_] := CoefficientList[f[n, x], x] // Reverse // (Sign[#]*Denominator[#])& // DeleteCases[#,0]&; Table[row[n], {n, 0, 15}] // Flatten (* _Jean-François Alcover_, Dec 29 2012 *)

%Y Cf. A106458.

%K tabf,sign

%O 0,2

%A _Gary W. Adamson_, Apr 19 2009

%E Extended to 15 rows by _Jean-François Alcover_, Dec 29 2012