A191302 -id:A191302 - cp's OEIS frontend

A190339 The denominators of the subdiagonal in the difference table of the Bernoulli numbers.

2, 6, 15, 105, 105, 231, 15015, 2145, 36465, 969969, 4849845, 10140585, 10140585, 22287, 3231615, 7713865005, 7713865005, 90751353, 218257003965, 1641030105, 67282234305, 368217318651, 1841086593255

Offset: 0

Paul Curtz, May 09 2011

Apparently a(n) = A181131(n) for n>=2 (checked numerically up to n=640). - R. J. Mathar, Aug 25 2025
The denominators of the T(n, n+1) with T(0, m) = A164555(m)/A027642(m) and T(n, m) = T(n-1, m+1) - T(n-1, m), n >= 1, m >= 0. For the numerators of the T(n, n+1) see A191972.
The T(n, m) are defined by A164555(n)/A027642(n) and its successive differences, see the formulas.
Reading the array T(n, m), see the examples, by its antidiagonals leads to A085737(n)/A085738(n).
A164555(n)/A027642(n) is an autosequence (eigensequence whose inverse binomial transform is the sequence signed) of the second kind; the main diagonal T(n, n) is twice the first upper diagonal T(n, n+1).
We can get the Bernoulli numbers from the T(n, n+1) in an original way, see A192456/A191302.
Also the denominators of T(n, n+1) of the table defined by A085737(n)/A085738(n), the upper diagonal, called the median Bernoulli numbers by Chen. As such, Chen proved that a(n) is even only for n=0 and n=1 and that a(n) are squarefree numbers. (see Chen link). - Michel Marcus, Feb 01 2013
The sum of the antidiagonals of T(n,m) is 1 in the first antidiagonal, otherwise 0. Paul Curtz, Feb 03 2015

			The first few rows of the T(n, m) array (difference table of the Bernoulli numbers) are:
1,       1/2,     1/6,      0,     -1/30,         0,        1/42,
-1/2,   -1/3,    -1/6,  -1/30,      1/30,      1/42,       -1/42,
1/6,     1/6,    2/15,   1/15,    -1/105,     -1/21,      -1/105,
0,     -1/30,   -1/15, -8/105,    -4/105,     4/105,       8/105,
-1/30, -1/30,  -1/105,  4/105,     8/105,     4/105,   -116/1155,
0,      1/42,    1/21,  4/105,    -4/105,   -32/231,     -16/231,
1/42,   1/42,  -1/105, -8/105, -116/1155,    16/231,  6112/15015,

Ludwig Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.

G. C. Greubel, Table of n, a(n) for n = 0..450
Kwang-Wu Chen, A summation on Bernoulli numbers, Journal of Number Theory, Volume 111, Issue 2, April 2005, Pages 372-391.
Peter Luschny, Computation and asymptotics of the Bernoulli numbers

T := proc(n,m)
    option remember;
    if n < 0 or m < 0 then
        0 ;
    elif n = 0 then
        if m = 1 then
            -bernoulli(m) ;
        else
            bernoulli(m) ;
        end if;
    else
        procname(n-1,m+1)-procname(n-1,m) ;
    end if;
end proc:
A190339 := proc(n)
    denom( T(n+1,n)) ;
end proc: # R. J. Mathar, Apr 25 2013

nmax = 23; b[n_] := BernoulliB[n]; b[1]=1/2; bb = Table[b[n], {n, 0, 2*nmax-1}]; diff = Table[Differences[bb, n], {n, 1, nmax}]; Diagonal[diff] // Denominator (* Jean-François Alcover, Aug 08 2012 *)

def A190339_list(n) :
    T = matrix(QQ, 2*n+1)
    for m in (0..2*n) :
        T[0,m] = bernoulli_polynomial(1,m)
        for k in range(m-1,-1,-1) :
            T[m-k,k] = T[m-k-1,k+1] - T[m-k-1,k]
    for m in (0..n-1) : print([T[m,k] for k in (0..n-1)])
    return [denominator(T[k,k+1]) for k in (0..n-1)]
A190339_list(7) # Also prints the table as displayed in EXAMPLE. Peter Luschny, Jun 21 2012

T(0, m) = A164555(m)/A027642(m) and T(n, m) = T(n-1, m+1) - T(n-1, m), n >= 1, m >= 0.
T(1, m) = A051716(m+1)/A051717(m+1);
T(n, n) = 2*T(n, n+1).
T(n+1, n+1) = (-1)^(1+n)*A181130(n+1)/A181131(n+1). - R. J. Mathar, Jun 18 2011
a(n) = A141044(n)*A181131(n). - Paul Curtz, Apr 21 2013

Edited and Maple program added by Johannes W. Meijer, Jun 29 2011, Jun 30 2011

New name from Peter Luschny, Jun 21 2012

A192456 Numerators in triangle that leads to the Bernoulli numbers.

Original entry on oeis.org

1, 1, 1, -1, 1, -1, 1, -2, 2, 1, -5, 1, 1, -1, 3, -8, 1, -7, 14, -4, 1, -4, 4, -64, 8, 1, -3, 9, -8, 12, 1, -5, 7, -40, 20, -32, 1, -11, 44, -44, 44, -16, 1, -2, 18, -64, 4, -192, 6112

Offset: 0

Paul Curtz, Jul 01 2011

For the denominators and detailed information see A191302.

Cf. A191302 (denominators).

nmax:=14: mmax:=nmax: A164555:=proc(n): if n=1 then 1 else numer(bernoulli(n)) fi: end: A027642:=proc(n): if n=1 then 2 else denom(bernoulli(n)) fi: end: for m from 0 to 2*mmax do T(0,m):=A164555(m)/A027642(m) od: for n from 1 to nmax do for m from 0 to 2*mmax do T(n,m):=T(n-1,m+1)-T(n-1,m) od: od: seq(T(n,n+1),n=0..nmax): for n from 0 to nmax do ASPEC(n,0):=2: for m from 1 to mmax do ASPEC(n,m):= (2*n+m)*binomial(n+m-1,m-1)/m od: od: for n from 0 to nmax do seq(ASPEC(n,m),m=0..mmax) od: for n from 0 to nmax do for m from 0 to 2*mmax do SBD(n,m):=0 od: od: for m from 0 to mmax do for n from 2*m to nmax do SBD(n,m):= T(m,m+1) od: od: for n from 0 to nmax do seq(SBD(n,m), m= 0..mmax/2) od: for n from 0 to nmax do BSPEC(n,2) := SBD(n,2)*ASPEC(2,n-4) od: for m from 0 to mmax do for n from 0 to nmax do BSPEC(n,m) := SBD(n,m)*ASPEC(m,n-2*m) od: od: for n from 0 to nmax do seq(BSPEC(n,m), m=0..mmax/2) od: seq(add(BSPEC(n, k), k=0..floor(n/2)) ,n=0..nmax): Tx:=0: for n from 0 to nmax do for m from 0 to floor(n/2) do a(Tx):= numer(BSPEC(n,m)): Tx:=Tx+1: od: od: seq(a(n),n=0..Tx-1); # Johannes W. Meijer, Jul 02 2011

(* a=ASPEC, b=BSPEC *) nmax = 13; a[n_, 0] = 2; a[n_, m_] := (2n+m)*Binomial[n+m-1, m-1]/m; b[n_] := BernoulliB[n]; b[1]=1/2; bb = Table[b[n], {n, 0, nmax}]; diff = Table[ Differences[bb, n], {n, 1, nmax}]; dd = Diagonal[diff]; sbd[n_, m_] := If[n >= 2m, -dd[[m+1]], 0]; b[n_, m_] := sbd[n, m]*a[m, n-2m]; Table[b[n, m], {n, 0, nmax}, {m, 0, Floor[n/2]}] // Flatten // Numerator (* Jean-François Alcover, Aug 09 2012 *)

Edited and Maple program added by Johannes W. Meijer, Jul 02 2011

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

A191972 The numerators of T(n, n+1) with T(0, m) = A164555(m)/A027642(m) and T(n, m) = T(n-1, m+1) - T(n-1, m), n >= 1, m >= 0.

Original entry on oeis.org

1, -1, 1, -4, 4, -16, 3056, -1856, 181312, -35853056, 1670556928, -39832634368, 545273832448, -19385421824, 53026545299456, -2753673793480966144, 68423881271489019904, -22654998127210332160

Offset: 0

Paul Curtz, Jun 20 2011

For the denominators of T(n, n+1) see A190339, where detailed information can be found.

			T(n,n+1) = [1/2, -1/6, 1/15 , -4/105, 4/105, -16/231, 3056/15015, -1856/2145, 181312/36465, ...]

Cf. A191302, A191754, A181131.

nmax:=20: mmax:=nmax: A164555:=proc(n): if n=1 then 1 else numer(bernoulli(n)) fi: end: A027642:=proc(n): if n=1 then 2 else denom(bernoulli(n)) fi: end: for m from 0 to 2*mmax do T(0,m):=A164555(m)/A027642(m) od: for n from 1 to nmax do for m from 0 to 2*mmax do T(n,m):=T(n-1,m+1)-T(n-1,m) od: od: for n from 0 to nmax do seq(T(n,m),m=0..mmax) od: seq(numer(T(n,n+1)),n=0..nmax-1); # Johannes W. Meijer, Jun 30 2011

nmax = 17; b[n_] := BernoulliB[n]; b[1] = 1/2; bb = Table[b[n], {n, 0, 2*nmax+1}]; dd = Table[Differences[bb, n], {n, 1, nmax }]; a[0] = 1; a[n_] := dd[[n, n+2]] // Numerator; Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Oct 02 2012 *)

T(n, n+1) = T(n, n)/2.

a(n+2) = (-1)^n*A181130(n+2)/2.

Thanks to R. J. Mathar by Paul Curtz, Jun 20 2011

Edited by Johannes W. Meijer, Jun 30 2011

A238800 Unreduced numerators in triangle that leads to the Euler numbers A198631(n)/A006519(n+1).

Original entry on oeis.org

1, 1, 1, -2, 1, -3, 1, -4, 2, 1, -5, 5, 1, -6, 9, -10, 1, -7, 14, -35, 1, -8, 20, -80, 26, 1, -9, 27, -150, 117, 1, -10, 35, -250, 325, -454, 1, -11, 44, -385, 715, -2497, 1, -12, 54, -560, 1365, -8172, 5914, 1, -13

Offset: 0

Paul Curtz, Mar 05 2014

We use the array ASPEC mentioned in A191302:
2, 1,  1,  1,  1,  1,   1,   1,...
2, 3,  4,  5,  6,  7,   8,   9,...
2, 5,  9, 14, 20, 27,  35,  44,...
2, 7, 16, 30, 50, 77, 112, 156,...
with the first upper diagonal of the difference table of the autosequence A198631(n)/A006519(n+1), i.e., 1/2, -1/4, 1/4, -5/8, 13/4, -227/8, 2957/8,...
written by columns:
1/2
1/2,
1/2, -1/4,
1/2, -1/4,
1/2, -1/4, 1/4,
1/2, -1/4, 1/4,
1/2, -1/4, 1/4, -5/8,
1/2, -1/4, 1/4, -5/8,
etc.
Hence, by multiplication of this double triangle by ASPEC, the beginning of the double triangle ESPEC is obtained:
E(0) =     1 =   1
E(1) =   1/2 = 1/2
E(2) =     0 = 1/2 -2/4
E(3) =  -1/4 = 1/2 -3/4
E(4) =     0 = 1/2 -4/4  +2/4
E(5) =   1/2 = 1/2 -5/4  +5/4
E(6) =     0 = 1/2 -6/4  +9/4 -10/8
E(7) = -17/8 = 1/2 -7/4 +14/4 -35/8
E(8) =     0 = 1/2 -8/4 +20/4 -80/8 +26/4.
The terms of the sequence are the reduced numerators. Like A192456(n) for Bernoulli numbers A164555(n)/A027642(n).

			a(n) by triangle
1,
1,
1, -2,
1, -3,
1, -4,  2,
1, -5,  5,
1, -6,  9, -10,
1, -7, 14, -35,
1, -8, 20, -80, 26,
etc.

A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

Original entry on oeis.org

1, 1, -3, 1, -5, 5, 1, -7, 25, -5, 1, -9, 23, -35, 49, 1, -11, 73, -27, 112, -49, 1, -13, 53, -77, 629, -91, 58, 1, -15, 145, -130, 1399, -451, 753, -58, 1, -17, 95, -135, 2699, -2301, 8573, -869, 341, 1, -19, 241

Offset: 0

Paul Curtz, Apr 27 2012

In A190339 we saw that (the second Bernoulli numbers) A164555/A027642 is an eigensequence (its inverse binomial transform is the sequence signed) of the second kind, see A192456/A191302. We consider this array preceded by 1 for the second row, by 1, -3/2, for the third one; 1 is chosen and is followed by the differences of successive rows.
Hence
                    1    1/2   1/6      0
            1    -1/2   -1/3  -1/6  -1/30
      1   -3/2    1/6    1/6  2/15   1/15
  1 -5/2   5/3      0  -1/30 -1/15 -8/105.
The second row is A051716/A051717.
The (reduced) triangle before the square array (T(n,m) in A190339) is a(n)/b(n)=
B(0)=    1 = 1                 Redbernou1li
B(1)= -1/2 = 1  -3/2
B(2)=  1/6 = 1  -5/2  5/3
B(3)=    0 = 1  -7/2 25/6  -5/3
B(4)=-1/30 = 1  -9/2 23/3 -35/6  49/30
B(5)=    0 = 1 -11/2 73/6 -27/2 112/15 -49/30.
For the main diagonal, see A165142.
Denominator b(n) will be submitted.
This transform is valuable for every eigensequence of the second kind. For instance Leibniz's 1/n (A003506).
With increasing exponents for coefficients, polynomials CB(n,x) create Redbernou1li. See the formula.
Triangle Bernou1li for A027641/A027642 with the same denominator A080326 for every column is
1
1  -3/2
1  -5/2 10/6
1  -7/2 25/6 -10/6
1  -9/2 46/6 -35/6  49/30
1 -11/2 73/6 -81/6 224/30 -49/30.
For numerator by columns,see A000012, -A144396, A100536, Q(n)=n*(2*n^2+9*n+9)/2 , new.
Triangle Checkbernou1 with the same denominator A080326 for every row is
1/1
(2    -3)/2
(6   -15  +10)/6
(6   -21  +25  -10)/6
(30 -135 +230 -175  +49)/30
(30 -165 +365 -405 +224 -49)/30;
Hence for numerator: 1, 2-3, 16-15, 31-31, 309-310, 619-619, 8171-8166.
Absolute sum: 1, 5, 31, 62, 619, 1238, 17337. Reduced division by A080326:
1, 5/2, 31/6, 31/3, 619/30, 619/15, 5779/70, = A172030(n+1)/A172031(n+1).

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane), A051714/A051715 (Akiyama-Tanigawa), A192456/A191302 for other triangles that lead to the Bernoulli numbers.

CB(0,x) = 1,
CB(1,x) = 1 - 3*x/2,
CB(n,x) = (1-x)*CB(n-1,x) + B(n)*x^n , n > 1.

A191532 Triangle T(n,k) read by rows: T(n,n) = 2n+1, T(n,k)=k for k

Original entry on oeis.org

1, 0, 3, 0, 1, 5, 0, 1, 2, 7, 0, 1, 2, 3, 9, 0, 1, 2, 3, 4, 11, 0, 1, 2, 3, 4, 5, 13, 0, 1, 2, 3, 4, 5, 6, 15, 0, 1, 2, 3, 4, 5, 6, 7, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 19, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 23, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 25, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 27

Offset: 0

Paul Curtz, Jun 05 2011

We can build products of linear polynomials with these T(n,k) defining the absolute terms:
1+n = A000027(1+n)                               =2,  3,  4,   5,   6,   7,
n*(3+n)/2 = A000096(1+n)                         =2,  5,  9,  14,  20,  27,
n*(1+n)*(5+n)/6 = A005581(2+n)                   =2,  7, 16,  30,  50,  77,
n*(1+n)*(2+n)*(7+n)/24 = A005582(1+n)            =2,  9, 25,  55, 105, 182,
n*(1+n)*(2+n)*(3+n)*(9+n)/120 = A005583(n)       =2, 11, 36,  91, 196, 378,
n*(1+n)*(2+n)*(3+n)*(4+n)*(11+n)/720 = A005584(n)=2, 13, 49, 140, 336, 714,

			1;
0,3;
0,1,5;
0,1,2,7;
0,1,2,3,9;
0,1,2,3,4,11;

Cf. A191302.

T(n,k) = A002262(n-1,k).

sum_{k=0..n} T(n,k) = A000217(1+n).

A224964 Irregular triangle of the denominators of the unreduced fractions that lead to the second Bernoulli numbers.

Original entry on oeis.org

2, 2, 2, 6, 2, 6, 2, 6, 15, 2, 6, 15, 2, 6, 15, 105, 2, 6, 15, 105, 2, 6, 15, 105, 105, 2, 6, 15, 105, 105, 2, 6, 15, 105, 105, 231, 2, 6, 15, 105, 105, 231, 2, 6, 15, 105, 105, 231, 15015, 2, 6, 15, 105, 105, 231, 15015

Offset: 0

Paul Curtz, Apr 21 2013

The triangle of fractions A192456(n)/A191302(n) leading to the second Bernoulli numbers written in A191302(n) is the reduced case. The unreduced case is
B(0) =   1   = 2/2         (1 or 2/2 chosen arbitrarily)
B(1)         = 1/2
B(2) =  1/6  = 1/2 - 2/6
B(3) =   0   = 1/2 - 3/6
B(4) = -1/30 = 1/2 - 4/6 +  2/15
B(5) =   0   = 1/2 - 5/6 +  5/15
B(6) =  1/42 = 1/2 - 6/6 +  9/15 -  8/105
B(7) =   0   = 1/2 - 7/6 + 14/15 - 28/105
B(8) = -1/30 = 1/2 - 8/6 + 20/15 - 64/105 + 8/105.
The constant values along the columns of denominators are A190339(n).
With B(0)=1, B(2) = 1/2 -1/3, (reduced case), the last fraction of the B(2*n) is
1, -1/3, 2/15, -8/105, 8/105, ... = A212196(n)/A181131(n).
We can continue this method of sum of fractions yielding Bernoulli numbers.
Starting from 1/6 for B(2*n+2), we have:
B(2) = 1/6
B(4) = 1/6 - 3/15
B(6) = 1/6 - 5/15 + 20/105
B(8) = 1/6 - 7/15 + 56/105 - 28/105.
With the odd indices from 3, all these B(n) are the Bernoulli twin numbers -A051716(n+3)/A051717(n+3).

			Triangle begins
  2;
  2;
  2, 6;
  2, 6;
  2, 6, 15;
  2, 6, 15;
  2, 6, 15, 105;
  2, 6, 15, 105;
  2, 6, 15, 105, 105;
  2, 6, 15, 105, 105;
  2, 6, 15, 105, 105, 231;
  2, 6, 15, 105, 105, 231;
  2, 6, 15, 105, 105, 231, 15015;
  2, 6, 15, 105, 105, 231, 15015;

Cf. A051716, A051717, A141044, A181131, A190339, A191302, A192456, A212196.

nmax = 7; b[n_] := BernoulliB[n]; b[1] = 1/2; bb = Table[b[n], {n, 0, 2*nmax-1}]; diff = Table[ Differences[bb, n], {n, 1, nmax}]; A190339 = diff // Diagonal // Denominator; Table[ Table[ Take[ A190339, n], {2}], {n, 1, nmax}] // Flatten (* Jean-François Alcover, Apr 25 2013 *)

T(n,k) = A190339(k).

cp's OEIS Frontend

A190339 The denominators of the subdiagonal in the difference table of the Bernoulli numbers.

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A192456 Numerators in triangle that leads to the Bernoulli numbers.

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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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A191972 The numerators of T(n, n+1) with T(0, m) = A164555(m)/A027642(m) and T(n, m) = T(n-1, m+1) - T(n-1, m), n >= 1, m >= 0.

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A238800 Unreduced numerators in triangle that leads to the Euler numbers A198631(n)/A006519(n+1).

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A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

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A191532 Triangle T(n,k) read by rows: T(n,n) = 2n+1, T(n,k)=k for k

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A224964 Irregular triangle of the denominators of the unreduced fractions that lead to the second Bernoulli numbers.

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