%I #40 Nov 22 2021 10:25:46
%S 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,
%T 6,2,290,6,14322,6,510,2,2,6,1919190,6,6,2,4510,6,1806,6,690,2,94,6,
%U 46410,6,66,2,530,6,798,6,870,2,118,6,56786730,6,6,2,170,6
%N Denominators of the inverse binomial transform of the Bernoulli numbers with B(1)=2/3.
%C Difference table of Bernoulli numbers with B(1)=2/3:
%C 1, 2/3, 1/6, 0, -1/30, 0, 1/42, 0, ...
%C -1/3, -1/2, -1/6, -1/30, 1/30, 1/42, -1/42, ...
%C -1/6, 1/3, 2/15, 1/15, -1/105, -1/21, ...
%C 1/2, -1/5, -1/15, -8/105, -4/105, ...
%C -7/10, 2/15, -1/105, 4/105, ...
%C 5/6, -1/7, 1/21, ...
%C -41/42, 2/15, ...
%C 7/6, ...
%C ...
%C 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.
%C 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).
%C a(2n+2)/a(2n+1) = 2, 5, 7, 5, 11, 455, ... .
%C By definition, for B(1) = b, the inverse binomial transform is
%C Bi(b) = 1, -1 + b, 7/6 - 2*b, -3/2 + 3*b, 59/30 + 4*b, ...
%C = A176328(n)/A176591(n) - (-1)^n *n*b.
%C With Bic(b) = 0, -1/2 + b, 1 - 2*b, -3/2 + 3*b, 2 + 4*b, ...
%C = (-1)^n *(A001477(n)/2 - n*b),
%C Bi(b) = (-1)^n *(A164555(n)/A027642(n) + A001477(n)/2 - n*b) =
%C = A027641(n)/A027642(n) + Bic(b) .
%F Conjecture: a(2n+1) = 3 followed by period 3: repeat 2, 6, 6.
%F Conjecture: a(2n) = A002445(n)/(period 3: repeat 1, 1, 3).
%F a(n) = A027641(n)/A027642(n) - (-1)^n *n/6.
%e 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.
%t 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 *)
%o (Sage)
%o def A257106_list(len, B1) :
%o T = matrix(QQ, 2*len+1)
%o for m in (0..2*len) :
%o T[0, m] = bernoulli_polynomial(1, m) if m <> 1 else B1
%o for k in range(m-1, -1, -1) :
%o T[m-k, k] = T[m-k-1, k+1] - T[m-k-1, k]
%o return [denominator(T[k, 0]) for k in (0..len-1)]
%o A257106_list(66, 2/3) # _Peter Luschny_, May 09 2015
%Y Cf. A256595, A027641/A027642(n), A164555(n)/A027642(n), A060789, A176328/A176591, A001477, A109007.
%K nonn
%O 0,2
%A _Paul Curtz_, Apr 23 2015