A178340 Triangle T(n,m) read by rows: denominator of the coefficient [x^m] of the umbral inverse Bernoulli polynomial B^{-1}(n,x).

1, 2, 1, 3, 1, 1, 4, 1, 2, 1, 5, 1, 1, 1, 1, 6, 1, 2, 3, 2, 1, 7, 1, 1, 1, 1, 1, 1, 8, 1, 2, 1, 4, 1, 2, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 10, 1, 2, 1, 1, 5, 1, 1, 2, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 1, 2, 3, 4, 1, 1, 1, 4, 3, 2, 1, 13

Offset: 0

Author

Paul Curtz, May 25 2010

Keywords

• nonn
• easy
• frac
• tabl

Comments

This is the triangle of denominators associated with the numerators of A178252.

(Unlike the coefficients of the Bernoulli Polynomials, the coefficients of the umbral inverse Bernoulli polynomials are all positive.)

Usually T(n,m) = A003989(n-m+1,m) for m>=1, but since we are tabulating denominators of reduced fractions here, this formula may be wrong by a cancelling integer factor.

Examples

			The triangle T(n,m) begins:
n\m  0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
0:   1
1:   2 1
2:   3 1 1
3:   4 1 2 1
4:   5 1 1 1 1
5:   6 1 2 3 2 1
6:   7 1 1 1 1 1 1
7:   8 1 2 1 4 1 2 1
8:   9 1 1 3 1 1 3 1 1
9:  10 1 2 1 1 5 1 1 2 1
10: 11 1 1 1 1 1 1 1 1 1  1
11: 12 1 2 3 4 1 1 1 4 3  2  1
12: 13 1 1 1 1 1 1 1 1 1  1  1  1
13: 14 1 2 1 2 1 2 7 2 1  2  1  2  1
14: 15 1 1 3 1 5 3 1 1 3  5  1  3  1  1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015
-------------------------------------------------
The rational triangle TinvB(n,m):= A178252(n,m) / T(n,m) begins:
n\m    0 1   2    3    4     5    6  7   8  9 10
0:     1
1:   1/2 1
2:   1/3 1   1
3    1/4 1 3/2    1
4:   1/5 1   2    2    1
5:   1/6 1 5/2 10/3  5/2     1
6:   1/7 1   3    5    5     3    1
7:   1/8 1 7/2    7 35/4     7  7/2  1
8:   1/9 1   4 28/3   14    14 28/3  4   1
9:  1/10 1 9/2   12   21 126/5   21 12 9/2  1
10: 1/11 1   5   15   30    42   42 30  15  5  1
... - _Wolfdieter Lang_, Aug 25 2015
Recurrence from the Sheffer a-sequence:
Tinv(3,2) = (3/2)*TinvB(2,1) = (3/2)*1 = 3/2.
From the z-sequence: Tinv(3,0) = 3*Sum_{j=0..2} z_j*TinvB(2,j) = 3*((1/2)*(1/3) -(1/12)*1 + 0*1) = 3*(1/6 - 1/12) = 1/4. - _Wolfdieter Lang_, Aug 25 2015
		

Crossrefs

Cf. A178252.

Programs

Mathematica

max = 13; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes]; Table[ Take[inv[[n]], n], {n, 1, max}] // Flatten // Denominator (* Jean-François Alcover_, Aug 09 2012 *)

Formula

T(n,0) = n+1.

Recurrence for the rational triangle

TinvB(n,m):= A178252(n,m) / T(n,m) from the Sheffer a-sequence, which is 1, (repeat 0), see the comment under A178252: TinvB(n,m) = (n/m)*TinvB(n-1,m-1), for n >= m >= 1. From the z-sequence: TinvB(n,0) = n*Sum_{j=0..n-1} z_j * TinvB(n-1,j), n >= 1, TinvB(0,0) = 1. - Wolfdieter Lang, Aug 25 2015