%I #11 Apr 11 2014 06:44:08
%S 15,15,15,105,105,105,21,105,105,21,105,105,105,105,105,15,105,105,
%T 105,105,15,165,165,1155,231,1155,165,165,33,165,165,231,231,165,165,
%U 33,15015,15015,15015,15015,15015,15015,15015,15015,15015
%N Array read by antidiagonals: denominators of the core of the classical Bernoulli numbers.
%C We consider the autosequence A164555(n)/A027642(n) (see A190339(n)) and its difference table without the first two rows and the first two columns:
%C 2/15, 1/15, -1/105, -1/21, -1/105, 1/15, 7/165, -5/33,...
%C -1/15, -8/105, -4/105, 4/105, 8/105, -4/165, -32/165,...
%C -1/105, 4/105, 8/105, 4/105, -116/1155, -28/165,...
%C 1/21, 4/105, -4/105, -32/231, -16/231,...
%C -1/105, -8/105, -116/1155, 16/231,...
%C -1/15, -4/165, 28/165,...
%C 7/165, 32/165,...
%C 5/33,... etc.
%C This is an autosequence of the second kind.
%C The antidiagonals are palindromes in absolute values.
%C a(n) are the denominators. Multiples of 3.
%C Sum of odd antidiagonals: 2/15, -2/21, 2/15, -10/33, 1382/1365,... = -2*A000367(n+2)/A001897(n+2).
%C The sum of the even antidiagonals is A000004.
%C 2/15, 0, -2/21,... = -4*A027641(n+4)/A027642(n+4) = -4*A164555(n)/A027642(n+4) and others.
%e As a triangle:
%e 15,
%e 15, 15,
%e 105, 105, 105,
%e 21, 105, 105, 21,
%e 105, 105, 105, 105, 105,
%e etc.
%t max = 12; tb = Table[BernoulliB[n], {n, 0, max}]; td = Table[Differences[tb, n][[3 ;; -1]], {n, 2, max - 1}]; Table[td[[n - k + 1, k]] // Denominator, {n, 1, max - 3}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Apr 11 2014 *)
%Y Cf. A085737/A085738, A168516/A168426 (autosequence), A027641, A176327/A176289, A235774, A165161/A051717(n+1).
%K nonn,tabl,frac
%O 0,1
%A _Paul Curtz_, Mar 15 2014
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