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

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

Author

Paul Curtz, Apr 23 2015

Keywords

• nonn

Comments

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) .

Examples

			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.
		

Crossrefs

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

Programs

Mathematica

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 *)

Sage

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

Formula

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.