A256595 -id:A256595 - cp's OEIS frontend

A256675 Denominators of the inverse binomial transform of Bernoulli(n+2).

6, 6, 15, 15, 105, 21, 105, 15, 165, 33, 15015, 1365, 1365, 3, 255, 255, 33915, 399, 21945, 165, 3795, 69, 31395, 1365, 1365, 3, 435, 435, 1038345, 7161, 608685, 255, 255, 3, 959595, 959595, 959595, 3, 6765, 6765, 2036265, 903, 103845, 345, 16215, 141, 1090635

Offset: 0

Paul Curtz, Apr 07 2015

nonn

Difference table of B(n+2):
1/6,        0,  -1/30,     0,   1/42,     0, -1/30, ...
-1/6,   -1/30,   1/30,  1/42,  -1/42, -1/30,   ...
2/15,    1/15, -1/105, -1/21, -1/105,   ...
-1/15, -8/105, -4/105, 4/105,    ...
-1/105, 4/105,  8/105,   ...
1/21,   4/105,    ...
-1/105,   ...
...
a(n) is the denominator of the n-th term of the first column.
a(n+2) is the denominator of the n-th term of the third row.
See A239315(n), which is the table without the first two rows.
Inverse binomial transform: 1/6, -1/6, 2/15, -1/15, -1/105, 1/21, -1/105, -1/15, 7/165, 5/33, -2663/15015, ... .

Cf. A190339, A239315, A029765, A001897, A141459, A172087, A168516, A177735, A256595.

max = 42; bb = Table[BernoulliB[n+2], {n, 0, max}]; dd = Table[Differences[bb, n], {n, 0, max}]; dd[[All, 1]] // Denominator (* Jean-François Alcover, Apr 09 2015 *)

lista(nn) = {A = vector(nn, n, bernfrac(n+1)); for (i=1, #A-1, for(j=0,i-1,A[i+1]-=binomial(i,j)*A[j+1])); for (i=1, #A, print1(denominator(A[i]), ", "));} \\ Michel Marcus, Apr 08 2015

a(2n) = A029765(n).
a(2n+3) = A001897(n+2).
a(2n)/a(2n+1) = A177735(n).
a(2n+4)/a(2n+3) = A177735(n+3).

A257106 Denominators of the inverse binomial transform of the Bernoulli numbers with B(1)=2/3.

Original entry on oeis.org

1, 3, 6, 2, 10, 6, 42, 6, 30, 2, 22, 6, 2730, 6, 6, 2, 170, 6, 798, 6, 330, 2, 46, 6, 2730, 6, 6, 2, 290, 6, 14322, 6, 510, 2, 2, 6, 1919190, 6, 6, 2, 4510, 6, 1806, 6, 690, 2, 94, 6, 46410, 6, 66, 2, 530, 6, 798, 6, 870, 2, 118, 6, 56786730, 6, 6, 2, 170, 6

Offset: 0

Paul Curtz, Apr 23 2015

nonn

Difference table of Bernoulli numbers with B(1)=2/3:
1,       2/3,    1/6,      0, -1/30,     0,  1/42, 0, ...
-1/3,   -1/2,   -1/6,  -1/30,  1/30,  1/42, -1/42, ...
-1/6,    1/3,   2/15,   1/15, -1/105, -1/21, ...
1/2,    -1/5,  -1/15, -8/105, -4/105, ...
-7/10,  2/15, -1/105,  4/105, ...
5/6,    -1/7,   1/21, ...
-41/42, 2/15, ...
7/6, ...
...
First column: 1, -1/3, -1/6, 1/2, -7/10, 5/6, -41/42, 7/6, -41/30, 3/2, -35/22, 11/6, ... . a(n) is the n-th term of the denominators.
Antidiagonal sums: 1, 1/3, -1/2, 2/3, -5/6, 1, -7/6, 4/3, -3/2, 5/3, -11/6, 2, ... . See A060789(n).
a(2n+2)/a(2n+1) = 2, 5, 7, 5, 11, 455, ... .
By definition, for B(1) = b, the inverse binomial transform is
Bi(b) =       1,  -1 + b, 7/6 - 2*b, -3/2 + 3*b, 59/30 + 4*b, ...
      = A176328(n)/A176591(n) - (-1)^n *n*b.
With Bic(b) = 0, -1/2 + b, 1 - 2*b,  -3/2 + 3*b,   2 + 4*b, ...
            = (-1)^n *(A001477(n)/2 - n*b),
Bi(b) = (-1)^n *(A164555(n)/A027642(n) + A001477(n)/2 - n*b) =
      = A027641(n)/A027642(n) + Bic(b) .

			a(0) = 1-0, a(1) = -1/2 +1/6 = -1/3, a(2) = 1/6 -1/3 = -1/6, a(3) = 0 +1/2.

Cf. A256595, A027641/A027642(n), A164555(n)/A027642(n), A060789, A176328/A176591, A001477, A109007.

max = 66; B[1] = 2/3; B[n_] := BernoulliB[n]; BB = Array[B, max, 0]; a[n_] := Differences[BB, n] // First // Denominator; Table[a[n], {n, 0, max-1}] (* Jean-François Alcover, May 11 2015 *)

def A257106_list(len, B1) :
    T = matrix(QQ, 2*len+1)
    for m in (0..2*len) :
        T[0, m] = bernoulli_polynomial(1, m) if m <> 1 else B1
        for k in range(m-1, -1, -1) :
            T[m-k, k] = T[m-k-1, k+1] - T[m-k-1, k]
    return [denominator(T[k, 0]) for k in (0..len-1)]
A257106_list(66, 2/3) # Peter Luschny, May 09 2015

Conjecture: a(2n+1) = 3 followed by period 3: repeat 2, 6, 6.
Conjecture: a(2n) = A002445(n)/(period 3: repeat 1, 1, 3).
a(n) = A027641(n)/A027642(n) - (-1)^n *n/6.

cp's OEIS Frontend

A256675 Denominators of the inverse binomial transform of Bernoulli(n+2).

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