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Triangle read by rows, a generalization of the Bernoulli numbers, the numerators for n>=0 and 0<=k<=n.
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%I #20 Jul 19 2018 08:23:29

%S 1,0,1,0,-1,1,0,1,-4,0,0,-1,47,-10,-1,0,1,-221,205,-209,0,0,-1,953,

%T -5495,10789,-427,1,0,1,-3953,123445,-8646163,177093,-22807,0,0,-1,

%U 16097,-2534735,22337747,-356249173,3440131,-46212,-1

%N Triangle read by rows, a generalization of the Bernoulli numbers, the numerators for n>=0 and 0<=k<=n.

%C The diagonal B(n, n) gives the Bernoulli numbers B_n = B_n(1). The formula is due to L. Kronecker and the generalization to Fukuhara, Kawazumi and Kuno.

%H S. Fukuhara, N. Kawazumi and Y. Kuno, <a href="https://arxiv.org/abs/1505.04840">Generalized Kronecker formula for Bernoulli numbers and self-intersections of curves on a surface</a>, arXiv:1505.04840 [math.NT], 2015.

%H L. Kronecker, <a href="http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002158752">Über die Bernoullischen Zahlen</a>, J. Reine Angew. Math. 94 (1883), 268-269.

%F B(n, k) = Sum_{j=0..k}(((-1)^(j-n)/(j+1))*binomial(k+1, j+1)*Sum_{i=0..j}(i^n*(j-i+1)^(k-n))) if n >= 1 and B(0, 0) = 1.

%F B_n = B(n, n) = Sum_{j=0..n}((-1)^(n-j)/(j+1))*binomial(n+1,j+1)*(Sum_{i=0..j}i^n).

%F T(n, k) = numerator(B(n, k)).

%e The triangle T(n, k) begins:

%e [0], 1

%e [1], 0, 1

%e [2], 0, -1, 1

%e [3], 0, 1, -4, 0

%e [4], 0, -1, 47, -10, -1

%e [5], 0, 1, -221, 205, -209, 0

%e [6], 0, -1, 953, -5495, 10789, -427, 1

%e [7], 0, 1, -3953, 123445, -8646163, 177093, -22807, 0

%e [8], 0, -1, 16097, -2534735, 22337747, -356249173, 3440131, -46212, -1

%e The rational triangle B(n, k) begins:

%e [0], 1

%e [1], 0, 1/2

%e [2], 0, -1/2, 1/6

%e [3], 0, 1/2, -4/3, 0

%e [4], 0, -1/2, 47/12, -10/3, -1/30

%e [5], 0, 1/2, -221/24, 205/9, -209/20, 0

%e [6], 0, -1/2, 953/48, -5495/54, 10789/80, -427/10, 1/42

%e [7], 0, 1/2, -3953/96, 123445/324, -8646163/8640, 177093/200, -22807/105, 0

%p B := (n, k) -> `if`(n = 0, 1, add(((-1)^(j-n)/(j+1))*binomial(k+1, j+1)*add(i^n*(j-i+1)^(k-n), i=0..j), j=0..k)):

%p for n from 0 to 8 do seq(numer(B(n,k)), k=0..n) od;

%t B[0, 0] = 1; B[n_, k_] := Sum[(-1)^(j-n)/(j+1)*Binomial[k+1, j+1]* Sum[i^n*(j-i+1)^(k-n) , {i, 0, j}] , {j, 0, k}];

%t Table[B[n, k] // Numerator, {n, 0, 8}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 19 2018, from Maple *)

%Y Cf. A292901 (denominators), B(n, n) = A164555(n)/A027642(n), A215083.

%K sign,tabl,frac

%O 0,9

%A _Peter Luschny_, Oct 01 2017