A254630 Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).

1, 1, 3, 1, 2, 13, 0, 1, 5, 3, -1, -1, 2, 29, 119, 0, -1, -1, 1, 31, 5, 1, 1, -1, -8, -1, 43, 253, 0, 1, 1, 4, -4, -1, 41, 7, -1, -1, -1, 4, 8, 4, -1, 29, 239, 0, -1, -1, -8, -4, 4, 8, 1, 31, 9, 5, 5, 7, -4, -116, -32, -116, -4, 7, 71, 665, 0

Offset

0,3

Comments

The difference table of Bernoulli(n,2) or B(n,2) = A164558(n)/A027642(n) is defined by placing the fractions in the upper row and calculating further rows as the differences of their preceding row:

1, 3/2, 13/6, 3, 119/30, ...

1/2, 2/3, 5/6, 29/30, ...

1/6, 1/6, 2/15, ...

0, -1/30, ...

-1/30, ...

etc.

The first column is A164555(n)/A027642(n).

In particular, the sums of the antidiagonals

1 = 1

1/2 + 3/2 = 2

1/6 + 2/3 + 13/6 = 3

0 + 1/6 + 5/6 + 3 = 4

etc. are the positive natural numbers. (This is rewritten for Bernoulli(n,3) in A157809).

We also have for Bernoulli(.,2)

B(0,2) = 1

B(0,2) + 2*B(1,2) = 4

B(0,2) + 3*B(1,2) + 3*B(2,2) = 12

B(0,2) + 4*B(1,2) + 6*B(2,2) + 4*B(3,2) = 32

etc. with right hand sides provided by A001787.

More generally sum_{s=0..t-1} binomial(t,s)*Bernoulli(s,q) gives A027471(t) for q=3, A002697 for q=4 etc, by reading A104002 downwards the q-th column.

Links

Table of n, a(n) for n=0..66.

Mathematica

nmax = 11; A164558 = Table[BernoulliB[n, 2], {n, 0, nmax}]; D164558 = Table[ Differences[A164558, n], {n, 0, nmax}]; Table[ D164558[[n-k+1, k+1]] // Numerator, {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 04 2015 *)

Crossrefs

Cf. A027641, A027642, A074909, A085737, A085738, A104002, A157809, A157920, A157930, A157945, A157946, A157965, A164555, A164558, A190339, A158302, A181131 (numerators and denominators of the main diagonal).

Sequence in context: A136125 A092580 A004468 * A145463 A144107 A199647

Adjacent sequences: A254627 A254628 A254629 * A254631 A254632 A254633

Keyword

sign,tabl,frac,easy

Author

Paul Curtz, Feb 03 2015

Status

approved