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.

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

Offset: 1

Author

Paul Curtz, Mar 24 2016

Keywords

• nonn
• tabf
• frac

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).

Examples

			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.
		

Links

• Jean-François Alcover, Roots of J(2n-1,z) lie close to two concentric circles (example)

Crossrefs

Cf. A002260, A002378, A002522, A005563, A059270, A112543, A122197, A004736.

Programs

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 *)