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A256595 Triangle A074909(n) with 0's as second column.

%I A256595 #43 Dec 02 2018 23:04:55
%S A256595 1,1,0,1,0,3,1,0,6,4,1,0,10,10,5,1,0,15,20,15,6,1,0,21,35,35,21,7,1,0,
%T A256595 28,56,70,56,28,8,1,0,36,84,126,126,84,36,9,1,0,45,120,210,252,210,
%U A256595 120,45,10,1,0,55,165,330,462,462,330,165,55,11
%N A256595 Triangle A074909(n) with 0's as second column.
%C A256595 For Bernoulli numbers, B(1) excluded.
%C A256595 B(n) is calculated via
%C A256595 B(0) = 1;
%C A256595 B(0) + 0 = 1;
%C A256595 B(0) + 0 + 3*B(2) = 3/2;
%C A256595 B(0) + 0 + 6*B(2) + 4*B(3) = 2;
%C A256595 etc.
%C A256595 The diagonal is A026741(n+1)/A040001(n).
%C A256595 Row sums: 1, 1, 4, 11, 26, 57, ..., essentially Euler numbers A000295. See A130103, A008292 and A173018.
%C A256595 There is an infinitude of Bernoulli number sequences. They are of the form
%C A256595 B(n,q) = 1, q, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, ... .
%C A256595 Chronologically, the first, and the most regular, is, for q=1/2, A164555(n)/A027642(n), from Jacob Bernoulli (1654-1705), published in Ars Conjectandi in 1713 and(?) Seko Kowa (1642-1708) in 1712. See A159688. The second is, for q=-1/2, B(n,-1/2) = A027641(n)/A027642(n), from B(n,1/2) via Pascal's triangle. We could choose Be(n,q) instead of B(n,q) to avoid confusion with Sloane's B(n,p) for A027641(n)/A027642(n) (p=-1), A164555(n)/A027642(n) (p=1), A164558(n)/A027642(n) (p=2), A157809(n)/A027642(n) (p=3), ..., successive binomial transforms of the previous sequence.
%C A256595 This motivates the proposal of the (independent of q) sequence Bernoulli(n+2):
%C A256595 B(n+2) = 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, ... and its inverse binomial transform. See A190339.
%D A256595 Jacob Bernoulli, Ars Conjectandi (1713).
%H A256595 Wikipedia, <a href="http://en.wikipedia.org/wiki/Seki_Takakazu">Seki Takakazu</a> (also known as Seki Kowa).
%e A256595 1,
%e A256595 1, 0,
%e A256595 1, 0,  3,
%e A256595 1, 0,  6, 4,
%e A256595 1, 0, 10, 10,  5,
%e A256595 1, 0, 15, 20, 15, 6,
%e A256595 1, 0, 21, 35, 35, 21, 7,
%e A256595 etc.
%t A256595 T[_, 0] = 1; T[_, 1] = 0; T[n_, k_] := Binomial[n+1, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jan 11 2016 *)
%Y A256595 Cf. A007318, A074909, A026741, A004001, A000295, A130103, A008292, A173018, A027641/A027642, A164555/A027642, A176327/A176289.
%K A256595 nonn,tabl
%O A256595 0,6
%A A256595 _Paul Curtz_, Apr 03 2015