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%I #11 May 17 2014 11:27:45
%S 0,1,1,1,0,-1,0,1,0,-3,0,5,0,-691,0,35,0,-3617,0,43867,0,-1222277,0,
%T 854513,0,-1181820455,0,76977927,0,-23749461029,0,8615641276005,0,
%U -84802531453387,0,90219075042845,0
%N Numerators of n*A164555(n-1)/A027642(n-1).
%C First multiplied shifted (second) Bernoulli numbers.
%C A164555(n-1)/A027642(n-1) = 0 followed by (A164555(n)/A027642(n)=1, 1/2, 1/6,...) = f(n) = 0, 1, 1/2, 1/6, 0,... .
%C f(n+1) - f(n) = A051716(n)/A051717(n).
%C Generally we consider a transform applied to the autosequences of first or second kind. An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. It is of the first kind if the main diagonal is A000004=0's. It is of the second kind if the main diagonal is the first upper diagonal multiplied by 2. A000045(n) is an autosequence of the first kind. A164555(n)/A027642(n) is an autosequence of the second kind. See A190339 (and A241269).
%C Here we apply the transform to the Bernoulli numbers A164555(n)/A027642(n).
%C We take n*(0 followed by A164555(n)/A027642(n)).
%C Hence the autosequence of first kind
%C TB1(n) = 0, 1, 1, 1/2, 0, -1/6, 0, 1/6, 0, -3/10, 0, 5/6, O, -691/210,.. .
%C a(n) are the numerators.
%C The first seven rows of the differencece table of TB1(n) are
%C 0, 1, 1, 1/2, 0, - 1/6, 0, 1/6,...
%C 1, 0, -1/2, -1/2, -1/6, 1/6, 1/6, -1/6,... =A140351(n+1)/b(n+1)
%C -1, -1/2, 0, 1/3, 1/3, 0, -1/3, -2/15,...
%C 1/2, 1/2, 1/3, 0, -1/3, -1/3, 1/5, 11/15,...
%C 0, -1/6, -1/3, -1/3, 0, 8/15, 8/15, -4/5,...
%C -1/6, -1/6, 0, 1/3, 8/15, 0, -4/3, -4/3,...
%C 0, 1/6, 1/3, 1/5, -8/15, -4/3, 0, 512/105,... .
%C First and second upper diagonals: 1, -1/2, 1/3, -1/3, 8/15, -4/3, 512/105,... .
%C Sum of the antidiagonals:
%C 0, 1, 1, 0, -1/2, 0, 1/2, 0, -5/6, 0, 13/6, 0, -49/6, 0,... .
%C (Note that the same transform applied to the second fractional Euler numbers A198631(n)/A006519(n+1) yields the Genocchi numbers -A226158(n)).
%C This transform can be continued:
%C TB2(n) = n*(0 followed by TB1(n)) =
%C 0, 0, 2, 3, 2, 0, -1, 0, 4/3, 0, -3, 0, 10, 0, -691/15, 0, 280, 0,...
%C is an autosequence of second kind.
%C TB3(n) = 0, 0, 0, 6, 12, 10, 0, -7, 0, 12, 0, -33, 0, 130, 0, 691, 0,...
%C is apparently an integer autosequence of the first kind.
%F a(n) = 0 followed by (A050925(n) = 1, -1, 1, 0,... ) with 1 instead of -1.
%F a(2n) = A063524(n). a(2n+1) = A002427(n).
%Y Cf. A199969 (autosequence).
%K sign
%O 0,10
%A _Paul Curtz_, May 09 2014
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