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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
 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, 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, 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 (list; table; graph; refs; listen; history; text; internal format)
 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 LINKS 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 Cf. A178340 (denominators). Cf. A074909, A258820, A003989, A053382, A053383. Sequence in context: A065372 A015137 A136148 * A325498 A224850 A269976 Adjacent sequences:  A178249 A178250 A178251 * A178253 A178254 A178255 KEYWORD nonn,tabl,frac 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

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Last modified September 21 19:59 EDT 2019. Contains 327282 sequences. (Running on oeis4.)