OFFSET
0,9
COMMENTS
The fractions A053382(n,m)/A053383(n,m) give the triangle of the coefficients of the Bernoulli polynomials:
1;
-1/2, 1;
1/6, -1, 1;
0, 1/2, -3/2, 1;
-1/30, 0, 1, -2, 1;
0, -1/6, 0, 5/3, -5/2, 1;
1/42, 0, -1/2, 0, 5/2, -3, 1;
The matrix inverse of this triangle defines coefficients of the umbral inverse Bernoulli polynomials B^{-1}(n,x) in row n:
1;
1/2, 1;
1/3, 1, 1;
1/4, 1, 3/2, 1;
1/5, 1, 2, 2, 1;
1/6, 1, 5/2, 10/3, 5/2, 1;
1/7, 1, 3, 5, 5, 3, 1;
1/8, 1, 7/2, 7, 35/4, 7, 7/2, 1;
1/9, 1, 4, 28/3, 14, 14, 28/3, 4, 1;
1/10, 1, 9/2, 12, 21, 126/5, 21, 12, 9/2, 1;
1/11, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1;
The current triangle T(n,m) is the numerator of the entry in row n and column m.
In the majority of cases, T(n,m) = A050169(n,m), but since we use the numerators of the reduced fractions, an integer factor may be missing in this equation.
Umbral composition (e.g., B(.,x)^k = B(k,x)) gives B^(-1)(n,B(.,x)) = x^n = B(n,B^(-1)(.,x)). - Tom Copeland, Aug 25 2015
FORMULA
"Palindromic:" T(n,m+1) = T(n,n-m). T(n,0)=1.
From Tom Copeland, Jun 18 2015: (Start)
The umbral inverse Bernoulli polynomials are Binv(n,x) = [(1+x)^(n+1)-x^(n+1)]/(n+1) with the e.g.f. e^(t*x) * (e^t-1)/t. (See A074909 for more details.) Therefore, T(n,k) is the numerator of the reduced fraction C(n+1,k)/(n+1) for 0 <= k < (n+1).
The reversed rows are presented as the diagonals of A258820.
T(n,k) = A258820(2n-k,n-k) = A003989(n+1,n+1-k) * n! / [ k! (n+1-k)! ], where A003989(j,k) = gcd(j,k). (End)
From Wolfdieter Lang, Aug 26 2015: (Start)
The following refers to the rational triangle TBinv with entries T(n,k)/A178340(n, m), n >= m >= 0.
The inverse of the Bernoulli triangle TB(n, m) with entries A196838(n,m)/A196839(n,m), n >= m >= 0, is the Sheffer triangle (z/(exp(z)-1),z). Therefore, the inverse triangle TBinv is the Sheffer triangle ((exp(z)-1)/z, z). This means that the e.g.f. of the sequence of column m of TBinv ((exp(x)-1)/x)*x^m/m! for m >= 0.
The e.g.f. of the row polynomials of TBinv, called Binv(n, x) = Sum_{m=0..n} TBinv(n,m)*x^m, is gBinv(z,x) = ((exp(z)-1)/z)*exp(x*z) (of the so-called Appell type).
The e.g.f. of the row sums is gBinv(x,1).
The e.g.f. of the alternating row sums is gBinv(x,-1) = (1 - exp(-x))/x.
The e.g.f. of the a-sequence of this Sheffer triangle is 1, and the e.g.f. of the z-sequence is (exp(x) - x -1)/((exp(x) -1)*x). This is the sequence 1/2, -1/12, 0, 1/120, 0, -1/252, 0, 1/240, 0, -1/132, .... For a- and z-sequences of Sheffer triangles and the corresponding recurrences see A006232.
The convolution property of the row polynomials Binv(n, x) is Binv(n, x+y) = Sum_{k=0..n} binomial(n, k)*Binv(n-k, x)*y^n (or with x and y exchanged).
The row polynomials satisfy (d/dx)Binv(n, x) = n*Binv(n-1, x), with Binv(0, x) = 1 (from Meixner's identity).
(End)
EXAMPLE
The triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13
0: 1
1: 1 1
2: 1 1 1
3: 1 1 3 1
4: 1 1 2 2 1
5: 1 1 5 10 5 1
6: 1 1 3 5 5 3 1
7: 1 1 7 7 35 7 7 1
8: 1 1 4 28 14 14 28 4 1
9: 1 1 9 12 21 126 21 12 9 1
10: 1 1 5 15 30 42 42 30 15 5 1
11: 1 1 11 55 165 66 77 66 165 55 11 1
12: 1 1 6 22 55 99 132 132 99 55 22 6 1
13: 1 1 13 26 143 143 429 1716 429 143 143 26 13 1
... reformatted. - Wolfdieter Lang, Aug 25 2015
MAPLE
nm := 15 : eM := Matrix(nm, nm) :
for n from 0 to nm-1 do for m from 0 to n do eM[n+1, m+1] :=coeff(bernoulli(n, x), x, m) ; end do: for m from n+1 to nm-1 do eM[n+1, m+1] := 0 ; end do: end do:
eM := LinearAlgebra[MatrixInverse](eM) :
for n from 1 to nm do for m from 1 to n do printf("%a, ", numer(eM[n, m])) ; end do: end do: # R. J. Mathar, Dec 21 2010
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 // Numerator (* Jean-François Alcover, Aug 09 2012 *)
PROG
(PARI) tabl(nn) = {for (n=0, nn, for (k=0, n, print1(numerator(binomial(n+1, k)/(n+1)), ", "); ); print(); ); } \\ after Tom Copeland comment; Michel Marcus, Jul 25 2015
CROSSREFS
KEYWORD
AUTHOR
Paul Curtz, May 24 2010
EXTENSIONS
Redefined based on reduced fractions by R. J. Mathar, Dec 21 2010
The term umbral was added by Tom Copeland, Aug 25 2015
STATUS
approved