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

0,2

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.

Links

Table of n, a(n) for n=0..78.

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

Example

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

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 *)

Crossrefs

Cf. A178252.

Sequence in context: A335124 A367579 A262891 * A173261 A084296 A209235

Adjacent sequences: A178337 A178338 A178339 * A178341 A178342 A178343

Keyword

nonn,easy,frac,tabl

Author

Paul Curtz, May 25 2010

Status

approved