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%I #30 Dec 12 2023 18:49:31
%S 3,1,8,7,2,1,15,7,13,3,1,1,24,23,22,21,4,3,2,1,35,17,11,8,31,5,2,1,1,
%T 1,48,47,46,45,44,43,6,5,4,3,2,1,63,31,61,15,59,29,57,7,3,5,1,3,1,1,
%U 80,79,26,77,76,25,74,73,8,7,2,5,4,1,2,1
%N Irregular triangle read by rows: numerators of the coefficients of polynomials J(2n-1,z) = Sum_(k=1,2, .. n) ((n+1)^2 - k + (n+1-k)*z^n)*z^(k-1)/k.
%C Irregular triangle of fractions:
%C 3, 1,
%C 8, 7/2, 2, 1/2,
%C 15, 7, 13/3, 3, 1, 1/3,
%C 24, 23/2, 22/3, 21/4, 4, 3/2, 2/3, 1/4,
%C 35, 17, 11, 8, 31/5, 5, 2, 1, 1/2, 1/5,
%C 48, 47/2, 46/3, 45/4, 44/5, 43/6, 6, 5/2, 4/3, 3/4, 2/5, 1/6.
%C etc.
%C First column: A005563; T(n, 1) = A005563(n).
%C Main diagonal: T(n, n) - n = n^2+1 = A002522(n).
%C The first upper diagonal is T(n, n+1) = n.
%C Consider TT(n, k) = k*T(n, k) for k = 1 to n:
%C 3,
%C 8, 7,
%C 15, 14, 13,
%C 24, 23, 22, 21,
%C etc.
%C Row sums: 3, 8+7, ... , are the positive terms of A059270; that is A059270(n).
%H Jean-François Alcover, <a href="/A270861/a270861.pdf">Roots of J(2n-1,z) lie close to two concentric circles (example)</a>
%e Irregular triangle:
%e 3, 1,
%e 8, 7, 2, 1,
%e 15, 7, 13, 3, 1, 1,
%e 24, 23, 22, 21, 4, 3, 2, 1,
%e 35, 17, 11, 8, 31, 5, 2, 1, 1, 1
%e 48, 47, 46, 45, 44, 43, 6, 5, 4, 3, 2, 1
%e etc.
%e Second half part by row: A112543.
%t row[n_] := CoefficientList[Sum[(((n + 1)^2 - k + (n + 1 - k)*z^n))*z^(k - 1)/k, {k, n}], z]; Table[row[n] // Numerator, {n, 1, 9}] // Flatten (* _Jean-François Alcover_, Apr 07 2016 *)
%Y Cf. A002260, A002378, A002522, A005563, A059270, A112543, A122197, A004736.
%K nonn,tabf,frac
%O 1,1
%A _Paul Curtz_, Mar 24 2016
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