A226158 -id:A226158 - cp's OEIS frontend

A232800 Denominators written by antidiagonals of interleaved A063524(n) and A002427(n)/A006955(n).

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 1, 2, 1, 6, 6, 3, 3, 6, 6, 1, 6, 3, 1, 3, 6, 1, 6, 6, 1, 3, 3, 1, 6, 6, 1, 6, 3, 3, 1, 3, 3, 6, 1, 10, 10, 15, 5, 15, 15, 5, 15, 10, 10, 1, 10, 5, 15, 15, 1, 15, 15, 5, 10, 1

Offset: 0

Paul Curtz, Nov 30 2013

The numerators of Br(n)=0, 1, 1, 1/2, 0, -1/6, 0, 1/6, 0, -3/10, 0, 5/6, 0,... are in A229979(n).
(Corresponding complementary Euler numbers Er(n):
A198631(2n+1)/A006519(2n+2)=1/2, -1/4, 1/2, -17/8, 31/2, -691/4,... =Ef2(n). (2*n+2)*Ef2(n)=1, -1, 3, -17, 155, 2073,...=-A001469(n+1),Genocchi numbers. Er(n)=interleaved (A063524(n) and -A001469(n+1)) =0, 1, 1, -1, 0, 3, 0, -17, 0, 155,... =-A226158(n).)

			1,
1, 1,
1, 1, 1,
2, 2, 2, 2,
1, 2, 1, 2, 1,
6, 6, 3, 3, 6, 6, etc.
Triangle of denominators of Br(n),complementary Bernoulli numbers,written by antidiagonals (see A229979).

A242246 Numerators of n*A164555(n-1)/A027642(n-1).

Original entry on oeis.org

0, 1, 1, 1, 0, -1, 0, 1, 0, -3, 0, 5, 0, -691, 0, 35, 0, -3617, 0, 43867, 0, -1222277, 0, 854513, 0, -1181820455, 0, 76977927, 0, -23749461029, 0, 8615641276005, 0, -84802531453387, 0, 90219075042845, 0

Offset: 0

Paul Curtz, May 09 2014

sign

First multiplied shifted (second) Bernoulli numbers.
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,... .
f(n+1) - f(n) = A051716(n)/A051717(n).
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).
Here we apply the transform to the Bernoulli numbers A164555(n)/A027642(n).
We take n*(0 followed by A164555(n)/A027642(n)).
Hence the autosequence of first kind
TB1(n) = 0, 1, 1, 1/2, 0, -1/6, 0, 1/6, 0, -3/10, 0, 5/6, O, -691/210,.. .
a(n) are the numerators.
The first seven rows of the differencece table of TB1(n) are
0,       1,    1,  1/2,   0, - 1/6,    0,      1/6,...
1,       0, -1/2, -1/2,  -1/6,  1/6,  1/6,    -1/6,... =A140351(n+1)/b(n+1)
-1,   -1/2,    0,  1/3,   1/3,    0, -1/3,   -2/15,...
1/2,   1/2,  1/3,    0,  -1/3, -1/3,  1/5,   11/15,...
0,    -1/6, -1/3, -1/3,     0, 8/15, 8/15,    -4/5,...
-1/6, -1/6,    0,  1/3,  8/15,    0, -4/3,    -4/3,...
0,     1/6,  1/3,  1/5, -8/15, -4/3,    0, 512/105,... .
First and second upper diagonals: 1, -1/2, 1/3, -1/3, 8/15, -4/3, 512/105,... .
Sum of the antidiagonals:
0, 1, 1, 0, -1/2, 0, 1/2, 0, -5/6, 0, 13/6, 0, -49/6, 0,... .
(Note that the same transform applied to the second fractional Euler numbers A198631(n)/A006519(n+1) yields the Genocchi numbers -A226158(n)).
This transform can be continued:
TB2(n) = n*(0 followed by TB1(n)) =
0, 0, 2, 3, 2, 0, -1, 0, 4/3, 0, -3, 0, 10, 0, -691/15, 0, 280, 0,...
is an autosequence of second kind.
TB3(n) = 0, 0, 0, 6, 12, 10, 0, -7, 0, 12, 0, -33, 0, 130, 0, 691, 0,...
is apparently an integer autosequence of the first kind.

Cf. A199969 (autosequence).

a(n) = 0 followed by (A050925(n) = 1, -1, 1, 0,... ) with 1 instead of -1.

a(2n) = A063524(n). a(2n+1) = A002427(n).

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A232800 Denominators written by antidiagonals of interleaved A063524(n) and A002427(n)/A006955(n).

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A242246 Numerators of n*A164555(n-1)/A027642(n-1).

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