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 A270861 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. 1
 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, 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, 80, 79, 26, 77, 76, 25, 74, 73, 8, 7, 2, 5, 4, 1, 2, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Irregular triangle of fractions: 3,     1, 8,   7/2,    2,  1/2, 15,    7, 13/3,    3,    1,  1/3, 24, 23/2, 22/3, 21/4,    4,  3/2, 2/3, 1/4, 35,   17,   11,    8, 31/5,    5,   2,   1, 1/2, 1/5, 48, 47/2, 46/3, 45/4, 44/5, 43/6,   6, 5/2, 4/3, 3/4, 2/5, 1/6. etc. First column: A005563; T(n, 1) = A005563(n). Main diagonal: T(n, n) - n = n^2+1 = A002522(n). The first upper diagonal is T(n, n+1) = n. Consider TT(n, k) = k*T(n, k) for k = 1 to n: 3, 8, 7, 15, 14, 13, 24, 23, 22, 21, etc. Row sums: 3, 8+7, ... , are the positive terms of A059270; that is A059270(n). LINKS Jean-François Alcover, Roots of J(2n-1,z) lie close to two concentric circles (example) EXAMPLE Irregular triangle: 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, 1 48, 47, 46, 45, 44, 43, 6, 5, 4, 3, 2, 1 etc. Second half part by row: A112543. MATHEMATICA 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 *) CROSSREFS Cf. A002260, A002378, A002522, A005563, A059270, A112543, A122197, A004736. Sequence in context: A077897 A308737 A257142 * A208656 A242440 A245651 Adjacent sequences:  A270858 A270859 A270860 * A270862 A270863 A270864 KEYWORD nonn,tabf,frac AUTHOR Paul Curtz, Mar 24 2016 STATUS approved

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Last modified December 12 07:31 EST 2019. Contains 329948 sequences. (Running on oeis4.)