login
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
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).
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 *)
KEYWORD
nonn,tabf,frac
AUTHOR
Paul Curtz, Mar 24 2016
STATUS
approved