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 A178252 Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n. 4

%I

%S 1,1,1,1,1,1,1,1,3,1,1,1,2,2,1,1,1,5,10,5,1,1,1,3,5,5,3,1,1,1,7,7,35,

%T 7,7,1,1,1,4,28,14,14,28,4,1,1,1,9,12,21,126,21,12,9,1,1,1,5,15,30,42,

%U 42,30,15,5,1,1,1,11,55,165,66,77,66,165,55,11,1,1,1,6,22,55,99,132,132,99,55,22,6,1

%N Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n.

%C The fractions A053382(n,m)/A053383(n,m) give the triangle of the coefficients of the Bernoulli polynomials:

%C 1;

%C -1/2, 1;

%C 1/6, -1, 1;

%C 0, 1/2, -3/2, 1;

%C -1/30, 0, 1, -2, 1;

%C 0, -1/6, 0, 5/3, -5/2, 1;

%C 1/42, 0, -1/2, 0, 5/2, -3, 1;

%C The matrix inverse of this triangle defines coefficients of the umbral inverse Bernoulli polynomials B^{-1}(n,x) in row n:

%C 1;

%C 1/2, 1;

%C 1/3, 1, 1;

%C 1/4, 1, 3/2, 1;

%C 1/5, 1, 2, 2, 1;

%C 1/6, 1, 5/2, 10/3, 5/2, 1;

%C 1/7, 1, 3, 5, 5, 3, 1;

%C 1/8, 1, 7/2, 7, 35/4, 7, 7/2, 1;

%C 1/9, 1, 4, 28/3, 14, 14, 28/3, 4, 1;

%C 1/10, 1, 9/2, 12, 21, 126/5, 21, 12, 9/2, 1;

%C 1/11, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1;

%C The current triangle T(n,m) is the numerator of the entry in row n and column m.

%C 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.

%C 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

%F "Palindromic:" T(n,m+1) = T(n,n-m). T(n,0)=1.

%F From _Tom Copeland_, Jun 18 2015: (Start)

%F 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).

%F The reversed rows are presented as the diagonals of A258820.

%F 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)

%F From _Wolfdieter Lang_, Aug 26 2015: (Start)

%F The following refers to the rational triangle TBinv with entries T(n,k)/A178340(n, m), n >= m >= 0.

%F 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.

%F 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).

%F The e.g.f. of the row sums is gBinv(x,1).

%F The e.g.f. of the alternating row sums is gBinv(x,-1) = (1 - exp(-x))/x.

%F 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.

%F 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).

%F The row polynomials satisfy (d/dx)Binv(n, x) = n*Binv(n-1, x), with Binv(0, x) = 1 (from Meixner's identity).

%F (End)

%e The triangle T(n,k) begins:

%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13

%e 0: 1

%e 1: 1 1

%e 2: 1 1 1

%e 3: 1 1 3 1

%e 4: 1 1 2 2 1

%e 5: 1 1 5 10 5 1

%e 6: 1 1 3 5 5 3 1

%e 7: 1 1 7 7 35 7 7 1

%e 8: 1 1 4 28 14 14 28 4 1

%e 9: 1 1 9 12 21 126 21 12 9 1

%e 10: 1 1 5 15 30 42 42 30 15 5 1

%e 11: 1 1 11 55 165 66 77 66 165 55 11 1

%e 12: 1 1 6 22 55 99 132 132 99 55 22 6 1

%e 13: 1 1 13 26 143 143 429 1716 429 143 143 26 13 1

%e ... reformatted. - _Wolfdieter Lang_, Aug 25 2015

%p nm := 15 : eM := Matrix(nm,nm) :

%p 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:

%p eM := LinearAlgebra[MatrixInverse](eM) :

%p 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

%t 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 *)

%o (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

%Y Cf. A178340 (denominators).

%Y Cf. A074909, A258820, A003989, A053382, A053383.

%K nonn,tabl,frac

%O 0,9

%A _Paul Curtz_, May 24 2010

%E Redefined based on reduced fractions by _R. J. Mathar_, Dec 21 2010

%E The term umbral was added by _Tom Copeland_, Aug 25 2015

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Last modified January 20 08:07 EST 2020. Contains 331081 sequences. (Running on oeis4.)