A226158 -id:A226158 - cp's OEIS frontend

A347599 Irregular table read by rows, T(n, k) is the rank of the k-th Genocchi permutation of {1,...,n}, permutations sorted in lexicographical order. If no Genocchi permutation of {1,...,n} exists, then T(n, 1) = 0 by convention.

1, 0, 5, 0, 67, 91, 92, 0, 1897, 2017, 2018, 2617, 2619, 2737, 2738, 2739, 2740, 3457, 3458, 3459, 3460, 4177, 4178, 4179, 4180, 0, 99241, 99961, 99962, 104281, 104283, 105001, 105002, 105003, 105004, 110041, 110042, 110043, 110044, 115081, 115082, 115083

Offset: 1

Peter Luschny, Sep 08 2021

Let M be the n X n matrix with M(j, k) = floor((2*j - k ) / n). A Genocchi permutation of order n is a permutation sigma of {1,...,n} if Product_{k=1..n} M(k, sigma(k)) does not vanish.
Let P(n) denote the number of Genocchi permutations of order n. Zhi-Wei Sun conjectured, using permanents, that P(n - 1) = G(n), where G(n) are the Genocchi numbers A036968. From the well-known relation between Genocchi and Bernoulli numbers this implies, assuming the conjecture:
  Bernoulli(n) = P(n - 1) / ((-1)^floor(n/2)*(2^(n + 2) - 2)) for n >= 2.
The related sequence A347600 lists Seidel permutations.

			Table starts:
[1] 1;
[2] 0;
[3] 5;
[4] 0;
[5] 67, 91, 92;
[6] 0;
[7] 1897, 2017, 2018, 2617, 2619, 2737, 2738, 2739, 2740, 3457, 3458, 3459, 3460, 4177, 4178, 4179, 4180;
.
The 17 permutations corresponding to the ranks are for n = 7:
1897 -> [3571246]; 2017 -> [3671245]; 2018 -> [3671254]; 2617 -> [4571236];
2619 -> [4571326]; 2737 -> [4671235]; 2738 -> [4671253]; 2739 -> [4671325];
2740 -> [4671352]; 3457 -> [5671234]; 3458 -> [5671243]; 3459 -> [5671324];
3460 -> [5671342]; 4177 -> [6571234]; 4178 -> [6571243]; 4179 -> [6571324];
4180 -> [6571342].
.
17 / (-510) = -1/30 = Bernoulli(8).

Zhi-Wei Sun, A novel identity connecting permanents to Bernoulli numbers, MathOverflow 2021-09-07.

Cf. A036968, A226158, A027641/A027642, A347600.

using Combinatorics
function GenocchiPermutations(n)
    f(m) = m >= n ? 1 : m < 0 ? -1 : 0
    Mat(n) = [[f(2*j - k) for k in 1:n] for j in 1:n]
    M = Mat(n); P = permutations(1:n); R = Int64[]
    S, rank = 0, 1
    for p in P
        m = prod(M[k][p[k]] for k in 1:n)
        if m != 0
            S += m
            push!(R, rank)
        end
        rank += 1
    end
    # println(n, "  ", S, "  ", S // (2^(n + 2) - 2)) # Bernoulli number
    return R
end
for n in 1:11 println(GenocchiPermutations(n)) end

A239275 a(n) = numerator(2^n * Bernoulli(n, 1)).

Original entry on oeis.org

1, 1, 2, 0, -8, 0, 32, 0, -128, 0, 2560, 0, -1415168, 0, 57344, 0, -118521856, 0, 5749735424, 0, -91546451968, 0, 1792043646976, 0, -1982765704675328, 0, 286994513002496, 0, -3187598700536922112, 0, 4625594563496048066560, 0, -16555640873195841519616, 0, 22142170101965089931264, 0

Offset: 0

Paul Curtz, Mar 13 2014

Difference table of f(n) = 2^n *A164555(n)/A027642(n) = a(n)/A141459(n):
  1,           1,      2/3,        0,    -8/15,      0,  32/21, 0,...
  0,        -1/3,     -2/3,    -8/15,     8/15,  32/21, -32/21,...
  -1/3,     -1/3,     2/15,    16/15,  104/105, -64/21,...
  0,        7/15,    14/15,   -8/105, -424/105,...
  7/15,     7/15, -106/105, -416/105,...
  0,      -31/21,   -62/31,
  -31/21, -31/21,...
  0,... etc.
Main diagonal: A212196(n)/A181131(n). See A190339(n).
First upper diagonal: A229023(n)/A181131(n).
The inverse binomial transform of f(n) is g(n). Reciprocally, the inverse binomial transform of g(n) is f(n) with -1 instead of f(1)=1, i.e., f(n) signed.
Sum of the antidiagonals: 1,1,0,-1,0,3,0,-17,... = (-1)^n*A036968(n) = -A226158(n+1).
Following A211163(n+2), f(n) is the coefficients of a polynomial in Pi^n.
Bernoulli numbers, twice, and Genocchi numbers, twice, are linked to Pi.
f(n) - g(n) = -A226158(n).
Also the numerators of the centralized Bernoulli polynomials 2^n*Bernoulli(n, x/2+1/2) evaluated at x=1. The denominators are A141459. - Peter Luschny, Nov 22 2015
(-1)^n*a(n) = 2^n*numerator(A027641(n)/A027642(n)) (that is the present sequence with a(1) = -1 instead of +1). - Wolfdieter Lang, Jul 05 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017. See B(2;n), eq. (53).

Cf. A141459 (denominators), A001896/A001897, A027641/A027642.

seq(numer(2^n*bernoulli(n, 1)), n=0..35); # Peter Luschny, Jul 17 2017

Table[Numerator[2^n*BernoulliB[n, 1]], {n, 0, 100}] (* Indranil Ghosh, Jul 18 2017 *)

from sympy import bernoulli
def a(n): return (2**n * bernoulli(n, 1)).numerator
print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 18 2017

a(n) = numerators of 2^n * A164555(n)/A027642(n).

Numerators of the binomial transform of A157779(n)/(interleave A001897(n), 1)(conjectured).

A297703 The Genocchi triangle read by rows, T(n,k) for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 2, 3, 3, 8, 14, 17, 17, 56, 104, 138, 155, 155, 608, 1160, 1608, 1918, 2073, 2073, 9440, 18272, 25944, 32008, 36154, 38227, 38227, 198272, 387104, 557664, 702280, 814888, 891342, 929569, 929569, 5410688, 10623104, 15448416, 19716064, 23281432, 26031912

Offset: 0

Peter Luschny, Jan 03 2018

			The triangle starts:
0: [     1]
1: [     1,      1]
2: [     2,      3,      3]
3: [     8,     14,     17,     17]
4: [    56,    104,    138,    155,    155]
5: [   608,   1160,   1608,   1918,   2073,   2073]
6: [  9440,  18272,  25944,  32008,  36154,  38227,  38227]
7: [198272, 387104, 557664, 702280, 814888, 891342, 929569, 929569]

Row sums are A005439 with offset 0.
T(n,0) = A005439 with A005439(0) = 1.
T(n,n) = A110501 with offset 0.
Cf. A001469, A014781, A099959, A226158.

function A297703Triangle(len::Int)
    A = fill(BigInt(0), len+2); A[2] = 1
    for n in 2:len+1
        for k in n:-1:2 A[k] += A[k+1] end
        for k in 2: 1:n A[k] += A[k-1] end
        println(A[2:n])
    end
end
println(A297703Triangle(9))

from functools import cache
@cache
def T(n):  # returns row n
    if n == 0: return [1]
    row = [0] + T(n - 1) + [0]
    for k in range(n, 0, -1): row[k] += row[k + 1]
    for k in range(2, n + 2): row[k] += row[k - 1]
    return row[1:]
for n in range(9): print(T(n))  # Peter Luschny, Jun 03 2022

A227577 Square array read by antidiagonals, A(n,k) the numerators of the elements of the difference table of the Euler polynomials evaluated at x=1, for n>=0, k>=0.

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 1, -1, -1, 0, 1, 1, 1, 0, -1, -1, -1, 1, 1, 1, 0, -1, -1, -5, -1, -1, 0, 17, 17, 13, 5, -5, -13, -17, -17, 0, 17, 17, 47, 13, 47, 17, 17, 0, -31, -31, -107, -73, -13, 13, 73, 107, 31, 31, 0, -31, -31, -355

Offset: 0

Paul Curtz, Jul 16 2013

sign

The difference table of the Euler polynomials evaluated at x=1:
    1,   1/2,     0,    -1/4,     0,     1/2,      0,      -17/8, ...
  -1/2, -1/2,   -1/4,    1/4,    1/2,   -1/2,   -17/8,      17/8, ...
    0,   1/4,    1/2,    1/4;    -1,   -13/8,    17/4,     107/8, ...
   1/4,  1/4,   -1/4,   -5/4,   -5/8,   47/8,    73/8,    -355/8, ...
    0,  -1/2,    -1,     5/8    13/2,   13/4,  -107/2,    -655/8, ...
  -1/2, -1/2,   13/8,   47/8,  -13/4, -227/4,  -227/8,    5687/8, ...
    0,  17/8,   17/4,  -73/8, -107/2,  227/8,  2957/4,    2957/8, ...
  17/8, 17/8, -107/8, -355/8,  655/8, 5687/8, -2957/8, -107125/8, ...
To compute the difference table, take
    1,   1/2;
  -1/2;
The next term is always half of the sum of the antidiagonals. Hence (-1/2 + 1/2 = 0)
    1,   1/2,   0;
  -1/2, -1/2;
    0;
The first column (inverse binomial transform) lists the numbers (1, -1/2, 0, 1/4, ..., not in the OEIS; corresponds to A027641/A027642). See A209308 and A060096.
A198631(n)/A006519(n+1) is an autosequence. See A181722.
Note the main diagonal: 1, -1/2, 1/2, -5/4, 13/2, -227/4, 2957/4, -107125/8, .... (See A212196/A181131.)
This twice the first upper diagonal. The autosequence is of the second kind.
From 0, -1, the algorithm gives A226158(n), full Genocchi numbers, autosequence of the first kind.
The difference table of the Bernoulli polynomials evaluated at x=1 is (apart from signs) A085737/A085738 and its analysis by Ludwig Seidel was discussed in the Luschny link. - Peter Luschny, Jul 18 2013

			Read by antidiagonals:
    1;
  -1/2,  1/2;
    0,  -1/2,   0;
   1/4,  1/4, -1/4, -1/4;
    0,   1/4,  1/2,  1/4,   0;
  -1/2, -1/2, -1/4,  1/4,  1/2,  1/2;
    0,  -1/2, - 1,  -5/4,  -1,  -1/2,   0;
  ...
Row sums: 1, 0, -1/2, 0, 1, 0, -17/4, 0, ... = 2*A198631(n+1)/A006519(n+2).
Denominators: 1, 1, 2, 1, 1, 1, 4, 1, ... = A160467(n+2)?

Peter Luschny, The computation and asymptotics of the Bernoulli numbers.
OEIS Wiki, Autosequence

Cf. A164555/A027642 in A190339.

DifferenceTableEulerPolynomials := proc(n) local A,m,k,x;
A := array(0..n,0..n); x := 1;
for m from 0 to n do for k from 0 to n do A[m,k]:= 0 od od;
for m from 0 to n do A[m,0] := euler(m,x);
   for k from m-1 by -1 to 0 do
      A[k,m-k] := A[k+1,m-k-1] - A[k,m-k-1] od od;
LinearAlgebra[Transpose](convert(A, Matrix)) end:
DifferenceTableEulerPolynomials(7);  # Peter Luschny, Jul 18 2013

t[0, 0] = 1; t[0, k_] := EulerE[k, 1]; t[n_, 0] := -t[0, n]; t[n_, k_] := t[n, k] = t[n-1, k+1] - t[n-1, k]; Table[t[n-k, k] // Numerator, {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 18 2013 *)

def DifferenceTableEulerPolynomialsEvaluatedAt1(n) :
    @CachedFunction
    def ep1(n):          # Euler polynomial at x=1
        if n < 2: return 1 - n/2
        s = add(binomial(n,k)*ep1(k) for k in (0..n-1))
        return 1 - s/2
    T = matrix(QQ, n)
    for m in range(n) :  # Compute difference table
        T[m,0] = ep1(m)
        for k in range(m-1,-1,-1) :
            T[k,m-k] = T[k+1,m-k-1] - T[k,m-k-1]
    return T
def A227577_list(m):
    D = DifferenceTableEulerPolynomialsEvaluatedAt1(m)
    return [D[k,n-k].numerator() for n in range(m) for k in (0..n)]
A227577_list(12)  # Peter Luschny, Jul 18 2013

Corrected by Jean-François Alcover, Jul 17 2013

A227608 Denominators of A225825(n) difference table written by antidiagonals.

Original entry on oeis.org

1, 2, 2, 6, 3, 6, 2, 3, 3, 2, 30, 15, 15, 15, 30, 2, 15, 15, 15, 15, 2, 42, 21, 105, 105, 105, 21, 42, 2, 21, 21, 105, 105, 21, 21, 2, 30, 15, 105, 105, 105, 105, 105, 15, 30, 2, 15, 15, 105, 105, 105, 105, 15, 15, 2, 66, 33, 165, 165, 1155, 231, 1155, 165, 165, 33, 66, 2, 33, 33, 165, 165, 231, 231, 165, 165, 33, 33, 2

Offset: 0

Paul Curtz, Aug 10 2013

			1,
-1/2,      1/2,
-1/6,     -2/3,     -1/6,
1/2,       1/3,     -1/3,     -1/2,
7/30,    11/15,    16/15,    11/15,     7/30,
-3/2,   -19/15,    -8/15,     8/15,    19/15,    3/2,
-31/42, -47/21, -368/105, -424/105, -368/105, -47/21, -31/42.
Row sums: 1, 0/2, -6/6, 0/6, 90/30, 0/30, -3570/210, 0/210, 32550/210,... .
Are the denominators A034386(n+1)?
Reduced row sums: 1, 0, -1, 0, 3, 0, -17, 0, 155,... = -A036968(n+1)? See A226158(n+2). First 100 terms checked by Jean-François Alcover.

Cf. A085738

max = 12; b[0] = 1; b[n_] := Numerator[ BernoulliB[n, 1/2] - (n+1)*EulerE[n, 0]]; t = Table[b[n], {n, 0, max}] / Table[ Sum[ Boole[ PrimeQ[d+1]]/(d+1), {d, Divisors[n]}] // Denominator, {n, 0, max}]; dt = Table[ Differences[t, n], {n, 0, max}]; Table[ dt[[n-k+1, k]] // Denominator, {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 12 2013 *)

More terms from Jean-François Alcover, Aug 12 2013

A240485 a(n) = -Zeta(1-n)n(2^(n+1) - 4) - Zeta(-n)(n+1)(2^(n+2) - 2), for n = 0 the limit is understood.

Original entry on oeis.org

1, 3, 2, -1, -2, 3, 6, -17, -34, 155, 310, -2073, -4146, 38227, 76454, -929569, -1859138, 28820619, 57641238, -1109652905, -2219305810, 51943281731, 103886563462, -2905151042481, -5810302084962, 191329672483963, 382659344967926, -14655626154768697

Offset: 0

Paul Curtz, Apr 06 2014

sign

Let G(m, n) denote the difference table of a(n):
1,    3,   2,  -1,  -2,   3,   6, -17, -34,...
2,   -1,  -3,  -1,   5,   3, -23, -17,...
-3,  -2,   2,   6,  -2, -26,   6,...
1,    4,   4,  -8, -24,  32,...
3,    0, -12, -16,  56,...
-3, -12,  -4,  72,...
-9,   8,  76,...
17,  68,...
51,...
a(n) = G(0, n).
The main diagonal G(n, n) = 1, -1, 2, -8, 56, -608,... is essentially a signed version of A005439.
The first upper diagonal is the main diagonal multiplied by 3. G(n, n+1) = 3*G(n, n).
G(m, n) = G(m, n-1) + G(m+1,n-1).
Inverse binomial transform: after 1, 2, -3, A110501(n+1) is interleaved with 3*A110501(n+1), signed two by two. I. e. b(n) = 1, 2, -3, 1, 3, -3, -9, 17, 51,... . a(n+2) + b(n+2) = -1, 0, 1, 0, -3, 0, 17,... = A226158(n+2).
This is particular to the Genocchi numbers. If the first upper diagonal is proportional to the main diagonal (1, -1, 2, -8,...), the sequence and the inverse binomial transform are simply connected to the Genocchi numbers.

Cf. A001469, A005439, A110501, A226158, A230324.

A240485 := proc(n) if n = 0 then 1 elif n = 1 then 3 else
m := 2*iquo(n-1, 2) + 2; -2^irem(n-1, 2)*m*euler(m-1, 0) fi end:
seq(A240485(n), n=0..27); # Peter Luschny, Apr 09 2014

a[n_] := Which[n == 0, 1, n == 1, 3, True, m = 2*Quotient[n-1, 2]+2; -2^Mod[n-1, 2]*m*EulerE[m-1, 0]]; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Apr 09 2014, after Peter Luschny *)

def A240485(n):
    if n < 3: return [1,3,2][n]
    m = 2*((n+1)//2)
    b = 2*(1-2^m)*bernoulli(m)
    if is_even(n): b = 2*b
    return (-1)^ceil((n^2+1)/2)*b
[A240485(n) for n in (0..24)]  # Peter Luschny, Apr 08 2014

a(2*n+1) = a(2*n+2)/2 for n > 0.
-a(2*n+2)/2 = A226158(2*n+2) = A001469(n+1) = (2*n+2)*E(2*n+1, 0) where E(n, x) are the Euler polynomials.
a(n) = -2*A226158(n) - A226158(n+1).
E.g.f.: (2*exp(x)*(3*x+exp(x)*(2*x+1)+1))/(exp(x)+1)^2. - Peter Luschny, Apr 10 2014

A277627 Square array read by antidiagonals downwards: T(n,k), n>=0, k>=0, in which column 0 is equal to A057427: 0, 1, 1, 1, ..., and for k > 0 column k lists two zeros followed by the partial sums of column k-1.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 3, 1, 0, 0, 0, 0, 0, 1, 4, 1, 0, 0, 0, 0, 0, 0, 3, 5, 1, 0, 0, 0, 0, 0, 0, 0, 6, 6, 1, 0, 0, 0, 0, 0, 0, 0, 1, 10, 7, 1, 0, 0, 0, 0, 0, 0, 0, 0, 4, 15, 8, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 21, 9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 20, 28, 10, 1

Offset: 0

Paul Curtz, Oct 24 2016

In other words, for n > 0 the column k lists 2*k+1 zeros together with the partial sums of the positive terms of column k-1. - Omar E. Pol, Oct 25 2016
Comments from the author:
1) ZSPEC =
  0, 0, 0, 0, 0, 0, 0, 0, ...
  1, 0, 0, 0, 0, 0, 0, 0, ...
  1, 0, 0, 0, 0, 0, 0, 0, ...
  1, 1, 0, 0, 0, 0, 0, 0, ...
  1, 2, 0, 0, 0, 0, 0, 0, ...
  1, 3, 1, 0, 0, 0, 0, 0, ...
  1, 4, 3, 0, 0, 0, 0, 0, ...
  1, 5, 6, 1, 0, 0, 0, 0, ...
etc.
The columns are the autosequences of the first kind of the title (column 1: 0, 0, followed by A001477(n); column 2: 0, 0, 0, 0, followed by A000217(n), etc) .
The positive terms are the Pascal triangle written by diagonals (A011973).
First column: A060576(n+1). Or A057427(n), n>-1, thanks to Omar E. Pol.
Row sums: A000045(n), autosequence of the first kind.
Alternated row sums and subtractions: 0, 1, 1, 0, -1, -1, 0 = A128834(n), autosequence of the first kind.
Antidiagonal sums: 0, 1, 1, 1, 2, 3, 4, 6, ... = A078012(n+2).
Application.
Numbers in triangle leading to the Genocchi numbers -A226158(n).
We multiply the columns of ZSPEC by d(n) = 1, -1, 2, -8, 56, -608, ... from A005439.
Hence, with only the first 0,
  0,
  1,
  1,
  1, -1,
  1, -2,
  1, -3,  2,
  1, -4,  6,
  1, -5, 12,   -8,
  1, -6, 20,  -32,
  1, -7, 30,  -80,  56,
  1, -8, 42, -160, 280,
  etc.
The row sums is -A226158(n).
2) Now consider the case of the autosequences of the second kind.
First step.
            2, 1, 1,  1,  1,  1, ...        = A054977(n)
      0, 0, 2, 3, 4,  5,  6,  7, ...        = A199969(n) with offset 0
0, 0, 0, 0, 2, 5, 9, 14, 20, 27, ...        see A000096
etc.
The positive terms are ASPEC in A191302. By triangle, they are either A029653(n) with A029653(0) = 2 instead of 1 or A029635(n).
Second step. YSPEC =
  2, 0,  0, 0, 0, 0, ...
  1, 0,  0, 0, 0, 0, ...
  1, 2,  0, 0, 0, 0, ...
  1, 3,  0, 0, 0, 0, ...
  1, 4,  2, 0, 0, 0, ...
  1, 5,  5, 0, 0, 0, ...
  1, 6,  9, 2, 0, 0, ...
  1, 7, 14, 7, 0, 0, ...
  etc.
Diagonals by triangle: A029635(n).
This is the companion to ZSPEC.
Row sums: A000032(n), autosequence of the second kind.
Alternated row sums and subtractions: period 6 repeat 2, 1, -1, -2, -1, 1 = A087204(n), autosequence of the second kind.
Application.
Numbers in triangle leading to A230324(n), a companion to -A226158(n).
We multiply the columns of YSPEC by d(n) 1, -1, 2, -8, 56, ... (see above).
Hence, without zeros:
  2,
  1,
  1, -2,
  1, -3,
  1, -4,  4,
  1, -5, 10,
  1, -6, 18,  -16,
  1, -7, 28,  -56,
  1, -8, 40, -128, 112,
  1, -9, 54, -240, 504,
  etc.
The row sum is A230324(n).

G. C. Greubel, Table of n, a(n) for the first 25 antidiagonals, flattened
OEIS Wiki, Autosequence

Cf. A011973 (without 0's), A007318 (Pascal's triangle).
Cf. A000045 (row sums), A078012 (antidiagonal sums).
Columns: A060576 or A057427 (k=0), A001477 (k=1), A000217 (k=2).
Cf. A000004, A000012, A000027, A000032, A005439, A029635, A029653, A054977, A087204, A128834, A191302, A199969, A226158, A230324.

kMax = 13; col[0] = Join[{0}, Array[1&, kMax]]; col[k_] := col[k] = Join[{0, 0}, col[k-1][[1 ;; -3]] // Accumulate]; T[n_, k_] := col[k][[n+1]]; Table[T[n-k, k], {n, 0, kMax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Nov 15 2016 *)

Better definition from Omar E. Pol, Oct 25 2016

A240677 a(n) = 6Zeta(1-n)n*(2^n-1) - Zeta(-n)(n+1)(2^(n+2)-2), for n = 0 the limit is understood.

Original entry on oeis.org

1, -2, -3, -1, 3, 3, -9, -17, 51, 155, -465, -2073, 6219, 38227, -114681, -929569, 2788707, 28820619, -86461857, -1109652905, 3328958715, 51943281731, -155829845193, -2905151042481, 8715453127443

Offset: 0

Paul Curtz, Apr 10 2014

sign

G2(m, n), difference table of a(n):
1,   -2, -3, -1,   3,   3, -9, -17, 51,...
-3,  -1,  2,  4,   0, -12, -8,  68,...
2,    3,  2, -4, -12,   4, 76,...
1,   -1, -6, -8,  16,  72,...
-2,  -5, -2, 24,  56,...
-3,   3, 26, 32,...
6,   23,  6,...
17, -17,...
-34,...
etc.
The main diagonal G2(n,n) = 1, -1, 2, -8,... is essentially a signed version of A005439.
The first upper diagonal is the main diagonal multiplied by -2. G2(n, n+1) = -2*G2(n, n).
G2(m, n) = G2(m, n-1) + G2(m+1, n-1).
a(n) = (-1)^n*b(n) of A240485(n).
Inverse binomial transform: (-1)^n*A240485(n).
a(n) and A240485(n) are reciprocal. Like for instance (-1)^n and 2^n.

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A240485.

A240677 := n -> `if`(n=0, 1, 6*Zeta(1-n)*n*(2^n-1) - Zeta(-n)*(n+1)*(2^(n+2)-2)); seq(A240677(n), n=0..24); # Peter Luschny, Apr 11 2014

g[0] = 0; g[1] = -1; g[n_] := n*EulerE[n - 1, 0]; a[n_] := 3*g[n] - g[n + 1]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 10 2014 *)

x = 'x+O('x^66);
A = -2*exp(x)*(2*x+exp(x)*(3*x-1)-1)/(exp(x)+1)^2;
Vec( serlaplace(A) )  /* Peter Luschny, Apr 10 2014 */

a(n) = 3*A226158(n) - A226158(n+1).
a(n+3) = -A001469(n+1).
a(2n+4) = -3*a(2n+3).
a(n) = A240485(n) + 5*A226158(n).
E.g.f.: -2*exp(x)*(2*x+exp(x)*(3*x-1)-1)/(exp(x)+1)^2. - Peter Luschny, Apr 10 2014

A278331 Shifted sequence of second differences of Genocchi numbers.

Original entry on oeis.org

0, -2, -2, 6, 14, -34, -138, 310, 1918, -4146, -36154, 76454, 891342, -1859138, -27891050, 57641238, 1080832286, -2219305810, -50833628826, 103886563462, 2853207760750, -5810302084962, -188424521441482, 382659344967926, 14464296482284734, -29311252309537394, -1277229462293249018

Offset: 0

Jean-François Alcover and Paul Curtz, Nov 18 2016

sign

This is an autosequence of the first kind (array of successive differences shows typical zero diagonal).
Last digits are apparently of period 20.
From A226158(n) for the continuity of autosequences of the first kind.
b(n) = 0, 1, -1, 0, 1, 0, -3, 0, 17, ... = A226158(n) with 1 as second term instead of -1.
c(n) = 0, 0, -1, 0, 1, 0, -3, 0, 17, ... = A226158(n) with 0 as second term instead of -1.
Respective difference tables:
0, -1, -1,  0,  1,  0, -3,   0,   17, ...
-1, 0,  1,  1, -1, -3 , 3,  17,  -17, ...
1,  1,  0, -2, -2,  6, 14, -34, -138, ...
etc,
0,  1, -1,  0,  1,  0, -3,   0,   17, ... = 0 followed by A036968(n+1)
1, -2,  1,  1, -1, -3,  3,  17,  -17, ...
-3, 3,  0, -2, -2,  6, 14, -34, -138, ...
etc,
0,  0, -1,  0,  1,  0, -3,   0,   17, ...
0, -1,  1,  1, -1, -3,  3,  17,  -17, ...
-1, 2,  0, -2, -2,  6, 14, -34, -138, ...
etc.
Since it is in the three tables, a(n) is the core of the Genocchi numbers.

Eric Weisstein's MathWorld, Genocchi Number.
Wikipedia, Genocchi number

Cf. A001469, A014781, A036968, A005439 (a(n) second and third diagonals), A164555/A027642, A209308, A226158, A240581(n)/A239315(n) (core of Bernoulli numbers).

g[0] = 0; g[1] = -1; g[n_] := n*EulerE[n-1, 0]; G = Table[g[n], {n, 0, 30}]; Drop[Differences[G, 2], 2]
(* or, from Seidel's triangle A014781: *)
max = 26; T[1, 1] = 1; T[n_, k_] /; 1 <= k <= (n + 1)/2 := T[n, k] = If[EvenQ[n], Sum[T[n - 1, i], {i, k, max}], Sum[T[n - 1, i], {i, 1, k}]]; T[, ] = 0; a[n_] := With[{k = Floor[(n - 1)/2] + 1}, (-1)^k*T[n + 3, k]]; Table[a[n], {n, 0, max}]

a(n) = (n+2)*E(n+1, 0) - 2*(n+3)*E(n+2, 0) + (n+4)*E(n+3, 0), where E(n,x) is the n-th Euler polynomial.

a(n) = -2*(2^(n+2)-1)*B(n+2) + 4*(2^(n+3)-1)*B(n+3) - 2*(2^(n+4)-1)*B(n+4), where B(n) is the n-th Bernoulli number.

A333303 T(n, k) = [x^k] (-2)^n*(B(n, x/2) - B(n, (x+1)/2)) where B(n, x) are the Bernoulli polynomials. Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

0, 1, 1, -2, 0, -3, 3, -1, 0, 6, -4, 0, 5, 0, -10, 5, 3, 0, -15, 0, 15, -6, 0, -21, 0, 35, 0, -21, 7, -17, 0, 84, 0, -70, 0, 28, -8, 0, 153, 0, -252, 0, 126, 0, -36, 9, 155, 0, -765, 0, 630, 0, -210, 0, 45, -10, 0, -1705, 0, 2805, 0, -1386, 0, 330, 0, -55, 11

Offset: 0

Peter Luschny, May 07 2020

Can be seen as the Bernoulli counterpart of the Euler triangles A247453 and A109449.

			B*(8, z) = 1024*(Zeta(-7, (z+1)/2) - Zeta(-7, z/2))
         = -17 + 84*z^2 - 70*z^4 + 28*z^6 - 8*z^7.
Triangle starts:
[ 0] [  0]
[ 1] [  1]
[ 2] [  1,    -2]
[ 3] [  0,    -3,    3]
[ 4] [ -1,     0,    6,   -4]
[ 5] [  0,     5,    0,  -10,   5]
[ 6] [  3,     0,  -15,    0,  15,    -6]
[ 7] [  0,   -21,    0,   35,   0,   -21,    7]
[ 8] [-17,     0,   84,    0, -70,     0,   28,  -8]
[ 9] [  0,   153,    0, -252,   0,   126,    0, -36,  9]
[10] [155,     0, -765,    0, 630,     0, -210,   0, 45, -10]
[11] [  0, -1705,    0, 2805,   0, -1386,    0, 330,  0, -55, 11]

Michael De Vlieger, Table of n, a(n) for n = 0..11325 (rows 0 <= n <= 150, flattened)
Digital Library of Mathematical Functions, Lerch’s Transcendent.
Peter H. N. Luschny, An introduction to the Bernoulli function, arXiv:2009.06743 [math.HO], 2020.
Eric Weisstein's World of Mathematics, Lerch Transcendent.

Row sums are (-1)^n*A226158(n). Alternating row sums are A239977(n).

Cf. A181983, A247453, A109449, (A053382/A053383) Bernoulli polynomials.

B[n_, x_] := (-2)^n (BernoulliB[n, x/2] - BernoulliB[n, (x + 1)/2]);
Prepend[Table[CoefficientList[B[n, x], x], {n, 1, 11}], 0] // Flatten

def Bstar(n,x):
    return (-2)^n*(bernoulli_polynomial(x/2,n) - bernoulli_polynomial((x+1)/2,n))
print(flatten([expand(Bstar(n, x)).list() for n in (0..11)]))

Let B*(n, x) denote the alternating Bernoulli rational polynomial functions defined by Z*(s, x) = Phi(-1, s, x) and B*(s, x) = -s Z*(1 - s, x). Here Phi(z, s, x) is the Hurwitz-Lerch transcendent defined as an analytic continuation of Sum_{k>=0} z^k/(k+x)^s. Then T(n, k) = (-1)^n [x^k] 2 B*(n, x).
T(n, 0) = 2*(2^n - 1)*Bernoulli(n, 1) = n*Euler(n - 1, 1) = -A226158(n).
Main diagonal is (-1)^(n+1)*n = A181983(n).

cp's OEIS Frontend

A347599 Irregular table read by rows, T(n, k) is the rank of the k-th Genocchi permutation of {1,...,n}, permutations sorted in lexicographical order. If no Genocchi permutation of {1,...,n} exists, then T(n, 1) = 0 by convention.

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A239275 a(n) = numerator(2^n * Bernoulli(n, 1)).

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A297703 The Genocchi triangle read by rows, T(n,k) for n>=0 and 0<=k<=n.

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A227577 Square array read by antidiagonals, A(n,k) the numerators of the elements of the difference table of the Euler polynomials evaluated at x=1, for n>=0, k>=0.

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A227608 Denominators of A225825(n) difference table written by antidiagonals.

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A240485 a(n) = -Zeta(1-n)*n*(2^(n+1) - 4) - Zeta(-n)*(n+1)*(2^(n+2) - 2), for n = 0 the limit is understood.

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A277627 Square array read by antidiagonals downwards: T(n,k), n>=0, k>=0, in which column 0 is equal to A057427: 0, 1, 1, 1, ..., and for k > 0 column k lists two zeros followed by the partial sums of column k-1.

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A240677 a(n) = 6*Zeta(1-n)*n*(2^n-1) - Zeta(-n)*(n+1)*(2^(n+2)-2), for n = 0 the limit is understood.

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A240485 a(n) = -Zeta(1-n)n(2^(n+1) - 4) - Zeta(-n)(n+1)(2^(n+2) - 2), for n = 0 the limit is understood.

A240677 a(n) = 6Zeta(1-n)n*(2^n-1) - Zeta(-n)(n+1)(2^(n+2)-2), for n = 0 the limit is understood.