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Third fractional autosequence after (1) (Br0(n)=) ) = ) A164555/A027642 and (2) Br(n)=) = A229979/(c(n)=) = 1,1,1,2,1,6,...=,... = 1 interleaved with A006955 or 1 followed by A050932.Thanks; thanks to _Jean-François Alcover). _). Hence
Br2(n)=) = 0, 0, 1, 3/2, 1, 0, -1/2, 0, 2/3, 0, -3/2, 0, 5, 0, -691/30,...,, ...,
0, , 0, , 1, , 3/2, 1, , 0, -, -1/2,..., ...
0, , 1, , 1/2, -, -1/2, -1, -, -1/2, , 1/2,..., ...
1, -, -1/2, -1, -, -1/2, , 1/2, 1, , 1/6,..., ...
- -3/2, -, -1/2, , 1/2, 1, , 1/2, -, -5/6, -, -3/2,..., ...
1, , 1, , 1/2, -, -1/2, -, -4/3, -, -2/3, 2,..., ...
0, -, -1/2, -1, -, -5/6, , 2/3, , 8/3, , 4/3,..., ...
- -1/2, -, -1/2, , 1/6, , 3/2, 2, -, -4/3, -8,... ., ... .
First Bernoulli polynomials, i.e., for B(1)=-) = -1/2,, A196838/A196839,, with 0's instead of the spaces:
1, , 0, 0, 0, 0, , 0, , 0, , 0, 0,..., ...
- -1/2, , 1, 0, 0, 0, , 0, , 0, , 0, 0,..., ...
1/6, -, -1, 1, 0, 0, , 0, , 0, , 0, 0,..., ...
0, , 1/2, -3/2, , 1, 0, , 0, , 0, , 0, 0,..., ...
- -1/30, , 0, 1, -2, 1, , 0, , 0, , 0, 0,..., ...
0, -, -1/6, , 0, , 5/3, -5/2, 1, , 0, , 0, 0,..., ...
1/42, , 0, -, -1/2, , 0, , 5/2, -3, , 1, , 0, 0,..., ...
0, , 1/6, , 0, -, -7/6, , 0, , 0, -7/2, 1, 0,..., ...
- -1/30, , 0, , 2/3, , 0. -, -7/3, 0, 14/3, -4, 10,... ., ... .
Second column: A229979/c(n) with -1 instead of 1, first column in A229979.
Third column: Br2(n) with -3/2 instead of 3/2, first column of the first array ..
1, 1, 1, 1, 1, 1, 1, 1, 1, 1,..., ...
0, 1, 2, 3, 4, 5, 6, 7, 8, 9,... =, ... = A001477,
0, 0, 1, 3, 6, 10, 15, 21, 28, 36,..., ...
0, 0, 0, 1, 4, 10, 20, 35, 56, 84,... . , ... . See A052553.
(Br0(n),), Br(n),), Br2(n),), Br3(n),... ), ... lead to A193815)..)
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