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%I #14 Apr 11 2013 17:16:05
%S 1,16,8,118,91,27,560,496,280,64,2003,1878,1366,637,125,5888,5672,
%T 4672,2944,1216,216,14988,14645,12917,9542,5446,2071,343,34176,33664,
%U 30920,25088,17088,9088,3256,512,71445,70716,66620,57359,43535
%N Rectangular array: (row n) = b**c, where b(h) = h^3, c(h) = (n-1+h)^3, n>=1, h>=1, and ** = convolution.
%C Principal diagonal: A213559
%C Antidiagonal sums: A213560
%C Row 1, (1,8,27,...)**(1,8,27,...): A145216
%C For a guide to related arrays, see A213500.
%H Clark Kimberling, <a href="/A213558/b213558.txt">Antidiagonals n = 1..60, flattened</a>
%H Henri Muehle, <a href="http://arxiv.org/abs/1301.1654">Proper Mergings of Stars and Chains are Counted by Sums of Antidiagonals in Certain Convolution Arrays -- The Details</a>, arXiv preprint arXiv:1301.1654, 2013.
%F T(n,k) = 8*T(n,k-1) - 28*T(n,k-2) + 56*T(n,k-3) - 70*T(n,k-4) + 56*T(n,k-5) - 28*T(n,k-6) + 8*T(n,k-7) - T(n,k-8).
%F G.f. for row n: f(x)/g(x), where f(x) = n^3 + ((n + 1)^3)*x + (-8*n^3 + 6*n^2 + 12*n + 8)*x^2 + (8*n^3 - 18*n^2 + 18)*x^3 - ((n - 2)^3)*x^4 - ((n + 1)^3)*x^5 and g(x) = (1 - x)^8.
%e Northwest corner (the array is read by falling antidiagonals):
%e 1.....16.....118....560.....2003
%e 8.....91.....496....1878....5672
%e 27....280....1366...4672....12917
%e 64....637....2944...9542....25088
%e 125...1216...5446...17088...43535
%t b[n_] := n^3; c[n_] := n^3
%t t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
%t TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
%t Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
%t r[n_] := Table[t[n, k], {k, 1, 60}] (* A213558 *)
%t d = Table[t[n, n], {n, 1, 40}] (* A213559 *)
%t s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
%t s1 = Table[s[n], {n, 1, 50}] (* A213560 *)
%Y Cf. A213500.
%K nonn,tabl,easy
%O 1,2
%A _Clark Kimberling_, Jun 17 2012
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