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

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, 28, 56, 70, 56, 28, 8, 1, 0, 36, 84, 126, 126, 84, 36, 9, 1, 0, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 55, 165, 330, 462, 462, 330, 165, 55, 11

Offset: 0

Author

Paul Curtz, Apr 03 2015

Keywords

• nonn
• tabl

Comments

For Bernoulli numbers, B(1) excluded.

B(n) is calculated via

B(0) = 1;

B(0) + 0 = 1;

B(0) + 0 + 3*B(2) = 3/2;

B(0) + 0 + 6*B(2) + 4*B(3) = 2;

etc.

The diagonal is A026741(n+1)/A040001(n).

Row sums: 1, 1, 4, 11, 26, 57, ..., essentially Euler numbers A000295. See A130103, A008292 and A173018.

There is an infinitude of Bernoulli number sequences. They are of the form

B(n,q) = 1, q, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, ... .

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.

This motivates the proposal of the (independent of q) sequence Bernoulli(n+2):

B(n+2) = 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, ... and its inverse binomial transform. See A190339.

Examples

			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,
etc.
		

References

• Jacob Bernoulli, Ars Conjectandi (1713).

Links

• Wikipedia, Seki Takakazu (also known as Seki Kowa).

Crossrefs

Cf. A007318, A074909, A026741, A004001, A000295, A130103, A008292, A173018, A027641/A027642, A164555/A027642, A176327/A176289.

Programs

Mathematica

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 *)