A227830 -id:A227830 - cp's OEIS frontend

A120082 Numerators of expansion for Debye function for n=1: D(1,x).

1, -1, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -691, 0, 1, 0, -3617, 0, 43867, 0, -174611, 0, 77683, 0, -236364091, 0, 657931, 0, -3392780147, 0, 1723168255201, 0, -7709321041217, 0, 151628697551, 0, -26315271553053477373, 0, 154210205991661, 0, -261082718496449122051

Offset: 0

Wolfdieter Lang, Jul 20 2006

Denominators are found under A120083.
D(1,x) = (1/x)*integral_{t=0..x} t/(exp(t)-1) dt (note the factor x on the r.h.s. of  the Abramowitz-Stegun link). This is the e.g.f. for {Bernoulli(n)/(n+1)}A027641(n)/A227540(n).%20Thanks%20to%20_Peter%20Luschny">{n>=0}. See A027641(n)/A227540(n). Thanks to _Peter Luschny for asking me to revisit this sequence. - Wolfdieter Lang, Jul 15 2013
Also numerators of coefficients in expansion of x/(exp(x)-1). See A227830 for denominators. - N. J. A. Sloane, Aug 01 2013

			Rationals r(n): [1, -1/4, 1/36, 0, -1/3600, 0, 1/211680, 0, -1/10886400, ...].

M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 23.

Peter Luschny, Table of n, a(n) for n = 0..500
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 998, equ. 27.1.1 for n=1, with a factor x extracted.
Wolfdieter Lang, Rationals r(n).

Cf. A060054, A060055, A120083, A182918, A227540, A227830, A358625.

[Numerator(Bernoulli(n)/Factorial(n+1)): n in [0..50]]; // G. C. Greubel, May 01 2023

A120082 := proc(n) local b; if n = 0 then b := 1 ; elif n = 1 then b := -1/4 ; elif type(n, 'odd') then b := 0; else b := bernoulli(n)/(n+1)! ; fi; numer(b) ; end: # R. J. Mathar, Sep 03 2009
gf := (1 - x/4 + sum((bernoulli(2*k)/((2*k+1)*(2*k)!))*x^(2*k), k=0..infinity)):
a := proc(n) local ser; if n = 0 then return 1 fi; ser := series(gf, x, n+2):
numer(coeff(ser, x, n)) end: seq(a(n), n = 0..40); # Peter Luschny, Dec 02 2022

Table[Numerator[BernoulliB[n]/((n+1)!)], {n,0,50}] (* G. C. Greubel, May 01 2023 *)

def A120082(n): return numerator(bernoulli(n)/factorial(n+1))
[A120082(n) for n in range(51)] # G. C. Greubel, May 01 2023

a(n) = numerator(r(n)), with r(n) = [x^n] (1 - x/4 + Sum_{k>=0} (B(2*k)/((2*k+1)*(2*k)!))*x^(2*k)), |x| < 2*Pi. B(2*k) = A000367(k)/A002445(k) (Bernoulli numbers).
a(n) = numerator(B(n)/(n+1)!), n >= 0. See the above comment on the e.g.f. D(1,x). - Wolfdieter Lang, Jul 15 2013
Apart from the sign of a(1) this sequence differs from A358625 for the first time at n = 68. - Peter Luschny, Dec 02 2022

Edited after Andrey Zabolotskiy noticed an inconsistency by Peter Luschny, Dec 02 2022

A060054 Numerators of numbers appearing in the Euler-Maclaurin summation formula.

Original entry on oeis.org

-1, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -691, 0, 1, 0, -3617, 0, 43867, 0, -174611, 0, 77683, 0, -236364091, 0, 657931, 0, -3392780147, 0, 1723168255201, 0, -7709321041217, 0, 151628697551, 0, -26315271553053477373

Offset: 1

Wolfdieter Lang, Feb 16 2001

a(n+1) = numerator(-Zeta(-n)), n>=1, with Riemann's zeta function. a(1)=-1=-numerator(-Zeta(-0)). For denominators see A075180.
Comment from N. J. A. Sloane, Oct 15 2008: (Start)
It appears that essentially the same sequence of rational numbers arises when we expand 1/(exp(1/x)-1) for large x. Here is the result of applying Bruno Salvy's gdev Maple program (answering a question raised by Roger L. Bagula):
gdev(1/(exp(1/x)-1), x=infinity, 20);
x - 1/2 + (1/12)/x - (1/720)/x^3 + (1/30240)/x^5 - (1/1209600)/x^7 + (1/47900160)/x^9 - (691/1307674368000)/x^11 + (1/74724249600)/x^13 - (3617/10670622842880000)/x^15 + (43867/5109094217170944000)/x^17 - (174611/802857662698291200000)/x^19 + ... (End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 16 (3.6.28), p. 806 (23.1.30), p. 886 (25.4.7).

Vincenzo Librandi, Table of n, a(n) for n = 1..600
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 16 (3.6.28), p. 806 (23.1.30), p. 886 (25.4.7).
Zhanna Kuznetsova and Francesco Toppan, Classification of minimal Z_2 X Z_2-graded Lie (super)algebras and some applications, arXiv:2103.04385 [math-ph], 2021.

Denominators of nonzero numbers give A060055.
Cf. A001067 (numerator of B(2*k)/(2*k)).
Cf. A075180.
Cf. also A120082/A227830.
Cf. A242179, A106831, A000142.

a060054 n = a060054_list !! n
a060054_list = -1 : map (numerator . sum) (tail $ zipWith (zipWith (%))
   (zipWith (map . (*)) a000142_list a242179_tabf) a106831_tabf)
-- Reinhard Zumkeller, Jul 04 2014

a[m_] := Sum[(-2)^(-k-1) k! StirlingS2[m,k],{k,0,m}]/(2^(m+1)-1); Table[Numerator[a[i]], {i,0,30}] (* Peter Luschny, Apr 29 2009 *)

a(n):=num((-1)^n*sum(binomial(n+k-1,n-1)*sum((j!*(-1)^(j)*binomial(k,j)*stirling1(n+j,j))/(n+j)!,j,1,k),k,1,n)); /* Vladimir Kruchinin, Feb 03 2013 */

a(n) = numerator(b(n)) with b(1) = -1/2; b(2*k+1) = 0, k >= 1; b(2*k) = B(2*k)/(2*k)! (B(2*n) = B_{2n} Bernoulli numbers: numerators A000367, denominators A002445)

A321331 Triangle read by rows: T(n, k) = (k+1)*S2(n+1, k+1), for n >= k >= 0, and S2 = A048993 (Stirling2).

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 14, 18, 4, 1, 30, 75, 40, 5, 1, 62, 270, 260, 75, 6, 1, 126, 903, 1400, 700, 126, 7, 1, 254, 2898, 6804, 5250, 1596, 196, 8, 1, 510, 9075, 31080, 34755, 15876, 3234, 288, 9, 1, 1022, 27990, 136420, 212625, 136962, 41160, 6000, 405, 10, 1, 2046, 85503, 583000, 1233650, 1076922, 447909, 95040, 10395, 550, 11

Offset: 0

Wolfdieter Lang, Dec 03 2018

This lower triangular matrix T is the inverse of the triangular matrix with elements Narumi[-1](n,m)/(m+1) = S1(n+1, m+1)/(n+1), with the Narumi triangle for parameter a = -1, and S1 = A048994 (Stirling1), i.e., Sum_{k=m..n} T(n, k) * S1(k+1, m+1)/(k+1) = delta_{n,m} (Kronecker symbol).
This triangle arises from the inverse of the rational Sheffer matrix Narumi[-1] = (log(1+x)/x, log(1+x) (such special Sheffer matrices (g(x), x*g(x)) define elements of the Narumi subgroup). The inverse matrix is (Narumi[-1])^(-1) = ((exp(x) - 1)/x, exp(x) - 1).
In order to have an integer matrix one takes T(n, k) := (n+1)*(Narumi[-1])^(-1)(n, k) = (k+1)*S2(n+1, k+1). The connection to S2 = A048993 results from the general relation between each Narumi-type matrix N = (g(x), x*g(x)) and its associated Sheffer matrix J = (1, x*g(x)) (this is of the Jabotinsky-type), i.e., N(n, m) = (m+1)*J(n+1, m+1)/(n+1), or with the row polynomials Npol(n, x) = (1/(n+1))*(d/dx)Jpol(n+1, x).
The signed triangle (-1)^(n-k)*A028421(n, k) (with upper diagonals filled with zeros) gives the integer matrix Nscaled with elements (n+1)*Narumi[-1](n,k). This inverse of Nscaled has the rational elements (Narumi[-1])^(-1)(k, m)/(m+1) = (1/(k+1))*S2(k+1, m+1).
The a- and z- sequence for the Sheffer matrix (Narumi[-1])^(-1) (see A006232 for a link on these sequences) have e.g.f.s Ea(x) = x/log(1 + x) and Ez(x) = 1/log(1 + x) - 1/x, hence a(n) = A006232(n)/A006233(n) and z(n) = A006232(n+1)/A075178(n), for n >= 0. This leads to the recurrence for T(n, k) given in the formula section.
The Boas-Buck-type column recurrence (see the link, also for references) uses the sequence with o.g.f. GBB(y) = exp(y)/(exp(y) - 1) - 1/y, with BB(n) = (-1)^(n+1)*A060054(n+1 ) / A227830(n+1), for n >= 1. For the recurrence see the formula section.
The Meixner-type identity (see the Meixner link) for the row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, derived from the one for the Narumi[-1]^(-1) row polynomials is Sum_{k=1..n} (-1)^{k+1}*(1/k)*(d/dx)^k R(n, x)/(n+1) = R(n-1, x), for n >= 1, and R(0, x) = 1. Here d/dx is a differentiation operator.
The Roman-type recurrence for the row polynomials (see the reference, Corollary 3.7.2. p. 50) becomes, with the z-sequence from above: R(n, x) = ((n+1)/n)*{(x + 1/2)*1 + (x - z(1))*d/dx - Sum_{k=2..n-1} (1/k!)*z(k)*(d/dx)^k}*R(n-1, x), for n >= 1, and R(0, x) = 1.
The triangle is the exponential Riordan square (cf. A321620) of exp(x)-1 with an additional main diagonal of zeros. - Peter Luschny, Jan 03 2019

			The triangle T(n, k) begins:
  n\k  0    1     2      3       4       5      6     7     8   9 10 ...
  ----------------------------------------------------------------------
  0:   1
  1:   1    2
  2:   1    6     3
  3:   1   14    18      4
  4:   1   30    75     40       5
  5:   1   62   270    260      75       6
  6:   1  126   903   1400     700     126      7
  7:   1  254  2898   6804    5250    1596    196     8
  8:   1  510  9075  31080   34755   15876   3234   288     9
  9:   1 1022 27990 136420  212625  136962  41160  6000   405  10
  10:  1 2046 85503 583000 1233650 1076922 447909 95040 10395 550 11
  ...
Recurrence (from Stirling2): T(4, 2) = 3*(T(3, 2) + T(3, 1)/2) = 3*(18 + 14/2) = 75.
Recurrence (from a- and z-sequence): T(4, 0) = 5*((1/2)*T(3, 0) - (1/12)*T(3, 1) + (1/12)*T(3, 2) - (19/120)*T(3, 3)) = 5*(1/2 - 14/12 + 18/12 - 4*19/120) = 1; T(4,2) = (5/2)*(1*1*T(3, 1) + 2*(1/2)*T(3, 2) + 3*(-1/6)* T(3, 3)) = (5/2)*(14 + 18 - 2) = 75.
Recurrence for column k=2 (Boas-Buck-type): T(4, 2) = (5!*3/2)*((1/3!)*(1/12)*T(2, 2) + (1/4!)*(1/2)*T(3, 2)) = (5!*3/2)*((1/72)*3 + (1/48)*18) = 75.
Meixner identity for the row polynomials, for n = 3: {d/dx  - (1/2)*(d/dx)^2 + (1/3)*(d/dx)^3)}*R(3, x)/4) = ((14 - 36/2 + 24/3) + (36 - 24/2)*x + 12*x^2)/4 = (1 + 6*x + 3*x^2) = R(2, x).
Roman type recurrence for row polynomials: R(n, 3) = (3/2)*{(x + 1/12)*(1 + 6*x + 3*x^2) + (x - (-1/2))*(6 + 6*x) - (1/2!)*(1/12)*6} = 1 + 14*x + 18*x^2 + 4*x^3.

Steven Roman, The umbral calculus, Academic Press, 1984.

Wolfdieter Lang, On the Generating Functions of the Boas-Buck Sequences for the Inverse of Riordan and Sheffer Matrices
J. Meixner, Orthogonale Polynomsysteme mit einer besonderen Gestalt der erzeugenden Funktion, J. Lond. Math. Soc. 9 (1934), 6-13.

Cf. A006232, A006233, A060054, A028421, A048993, A048994, A075178, A227830.

Flat(List([0..10],n->List([0..n],k->(k+1)*Stirling2(n+1,k+1)))); # Muniru A Asiru, Dec 03 2018

T:=(n,k)->(k+1)*Stirling2(n+1,k+1): seq(seq(T(n,k),k=0..n),n=0..10); # Muniru A Asiru, Dec 03 2018

T[n_, k_] := (k+1) * StirlingS2[n+1, k+1];  Table[T[n, k], {n,0,10}, {k, 0, n}] //Flatten (* Amiram Eldar, Dec 03 2018 *)

T(n, k) = (k+1)*stirling(n+1, k+1, 2) \\ Thomas Scheuerle, Nov 10 2023

# uses[riordan_square from A321620]
riordan_square(exp(x) - 1, 10, True) # Peter Luschny, Jan 03 2019

T(n, k) = (k+1)*A048993(n+1, k+1), with A048993 = Stirling2, for n >= k >= 0, and 0 otherwise.
T(n, k) = (n+1)*(Narumi[a=-1])^(-1)(n, k), with the Narumi[a=-1] matrix with entries (-1)^(n-k)*A028421(n, k)/(n+1).
E.g.f. for column k sequence: E(k, x) = (x*d/dx + 1)*EN(k, x), where EN(k, x) = (exp(x) - 1)^(k+1)/(x*k!) is the e.g.f. for the (Narumi[a=-1])^(-1) columns. Hence E(k, x) = exp(x)*(exp(x) - 1)*(k+1)/k!, for k >= 0.
E.g.f. for (ordinary) row polynomials R(n, x): Epol(z, x) = exp(z)*exp(x*(exp(z) - 1))*(1 + x*(exp(z) - 1)).
Recurrence (from Stirling2): T(n, k) = 0 for n < k; T(n, 0) = (k + 1)*T(n-1, k), for n <= 1, T(0, 0) = 1; T(n, k) = (k+1)*(T(n-1, k) + T(n-1, k-1)/k), for n >= 1, k >= 1.
Recurrence (from a- and z-sequence, see above): a = {1, 1/2, -1/6, 1/4, -19/30, 9/4, ...}, z = {1/2, -1/12, 1/12, -19/120, 9/20, -863/504, ...}.
T(n, k) = 0, for n < k; T(n, 0) = (n+1)*Sum_{j=0..n-1} z(j)*T(n-1, j), for n >= 1, with T(0, 0) = 1; T(n, k) = ((n+1)/k)*Sum_{j=0..n-m} binomial(k-1+j, j)*a(j)*T(n-1, k-1+j).
Recurrence for column k, from the Boas-Buck-type sequence BB(n) = (-1)^(n+1)*A060054(n+1)/A227830(n+1), for n >= 0; BB = {1/2, 1/12, 0, -1/720, 0, 1/30240, 0, -1/1209600, ...}: T(n, k) = 0, for  n < k; T(n, n) = n+1, for n >= 0; T(n, k) = ((n+1)!*(k+1)/(n-k))*Sum_{j=k..n-1} (1/(j+1)!)*BB(n-(j+1))*T(j, k), for n >= 0 and k = 0, 1, ..., n-1.
T(n, k) = Stirling2(n+2, k+1) - Stirling2(n+1, k). - Peter Luschny, May 26 2020

A358111 The multiplicative inverse of the coefficients of the factorially normalized Bernoulli polynomials (provided they do not vanish, otherwise by convention 0).

Original entry on oeis.org

1, -2, 1, 12, -2, 2, 0, 12, -4, 6, -720, 0, 24, -12, 24, 0, -720, 0, 72, -48, 120, 30240, 0, -1440, 0, 288, -240, 720, 0, 30240, 0, -4320, 0, 1440, -1440, 5040, -1209600, 0, 60480, 0, -17280, 0, 8640, -10080, 40320, 0, -1209600, 0, 181440, 0, -86400, 0, 60480, -80640, 362880

Offset: 0

Peter Luschny, Oct 30 2022

The factorially normalized Bernoulli polynomials are defined inductively by:

b(0, x) = 1, (d/dx) b(n, x) = b(n-1, x), and Integral_{x=0..1} b(n, x) = 0.

			0]        1;
1]       -2,        1;
2]       12,       -2,     2;
3]        0,       12,    -4,      6;
4]     -720,        0,    24,    -12,     24;
5]        0,     -720,     0,     72,    -48,   120;
6]    30240,        0, -1440,      0,    288,  -240,   720;
7]        0,    30240,     0,  -4320,      0,  1440, -1440,   5040;
8] -1209600,        0, 60480,      0, -17280,     0,  8640, -10080, 40320;
9]        0, -1209600,     0, 181440,      0, -86400,    0,  60480,-80640, 362880;

N. E. Nörlund, Vorlesungen über Differenzenrechnung, Springer, 1924. (page 31)

Yassine El Maazouz and Jim Pitman, The Bernoulli clock: probabilistic and combinatorial interpretations of the Bernoulli polynomials by circular convolution, arXiv:2210.02027 [math.PR], Oct. 2022.

Cf. A227830 (column 0), A196838/A196839 (Bernoulli polynomials), A000142.

T := proc(n, k) coeff(bernoulli(n, x) / n!, x, k); ifelse(% = 0, 0, 1/%) end:
seq(seq(T(n, k), k = 0..n), n = 0..9);

T[n_, k_] := With[{c = Coefficient[BernoulliB[n, x]/n!, x, k]}, If[c == 0, 0, 1/c]];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 07 2024, after Maple code *)

T(n, k) = ([x^k] b(n, x))^(-1), provided [x^k] b(n, x) != 0, otherwise 0.
Integral_{x=0..1} b(n, x)*b(m, x) = (-1)^(m + 1)*b(n + m, 1). [N. E. Nørlund]
Note that n!*b(n, 1) are the Bernoulli numbers (with B_1 = 1/2).

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A120082 Numerators of expansion for Debye function for n=1: D(1,x).

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A060054 Numerators of numbers appearing in the Euler-Maclaurin summation formula.

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A321331 Triangle read by rows: T(n, k) = (k+1)*S2(n+1, k+1), for n >= k >= 0, and S2 = A048993 (Stirling2).

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A358111 The multiplicative inverse of the coefficients of the factorially normalized Bernoulli polynomials (provided they do not vanish, otherwise by convention 0).

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