A257372 - OEIS

%I #11 May 03 2015 02:48:29

%S 1,6,6,15,30,21,42,15,30,33,66,1365,2730,3,6,255,510,399,798,165,330,

%T 69,138,1365,2730,3,6,435,870,7161,14322,255,510,3,6,959595,1919190,3,

%U 6,6765,13530,903,1806,345,690

%N a(n) = denominators of A255935(n) * triangle T(n,k) for Bernoulli(k+2), k=0 to n-1.

%C Generally, A255935(n) multiplied by triangle T(n,k) for s(k), k=0 to n-1 yields an autosequence of the first kind (a sequence whose main diagonal is 0's).

%C Here s(k) = 1/6, 0, -1/30, ... from A164555(n+2)/A027642(n+2). Hence

%C 0                              =  0/1

%C 1/6, 0                         =  1/6

%C 1/6, 0,     0                  =  1/6

%C 1/6, 0, -1/10, 0               = 1/15

%C 1/6, 0,  -1/5, 0, 0            =-1/30

%C ... .

%C a(n) are the row sums denominators.

%C Compare to A051716(n+2)/A051717(n+2).

%C Hence the difference table

%C 0,       1/6,      1/6,  1/15, -1/30, -1/21, 1/42, ...

%C 1/6,       0,    -1/10, -1/10, -1/70,  1/14, ...

%C -1/6,  -1/10,        0,  3/35,  3/35, ...

%C 1/15,   1/10,     3/35,     0, ...

%C 1/30,  -1/70,    -3/35, ...

%C -1/21, -1/14, ...

%C -1/42, ...

%C ... .

%F a(2n) = A002445(n).

%F a(2n+3) = A001897(n+2).

%F a(2n+2) = A040000(n) * a(2n+1).

%Y Cf. A255935, A027641/A027642, A164555/A027642, A001897, A002445, A040000, A051716/A051717.

%K nonn

%O 0,2

%A _Paul Curtz_, Apr 21 2015