A176144 -id:A176144 - cp's OEIS frontend

A195240 Numerators of the second differences of the sequence of fractions (-1)^(n+1)*A176618(n)/A172031(n).

0, 1, 1, 7, 8, 11, 10, 7, 8, 19, 14, 337, 1028, 5, -2, -1681, 1936, 22133, -21734, -87223, 87388, 427291, -427222, -118181363, 118182728, 4276553, -4276550, -11874730297, 11874730732, 4307920641583

Offset: 0

Paul Curtz, Sep 13 2011

The array of (-1)^n*A176328(n)/A176591(n) and its first, second, etc. differences in subsequence rows starts as follows:
0,     1,     2,   19/6,   14/3, 199/30,    137/15, ... (-1)^n * A176328(n)/A176591(n),
1,     1,   7/6,    3/2,  59/30,    5/2,    127/42, ... see A176328,
0,   1/6,   1/3,   7/15,   8/15,  11/21,     10/21, ...
1/6, 1/6,  2/15,   1/15, -1/105,  -1/21,    -1/105, ... see A190339
0, -1/30, -1/15, -8/105, -4/105,  4/105, -116/1155, ...
The numerators in the 3rd row, 0, 1/6, 1/3, 7/15, 8/15, 11/21, 10/21, 7/15, 8/15, 19/33, 14/33, 337/1365, 1028/1365, 5/3, -2/3, -1681/255, 1936/255, ... define the current sequence.
The associated denominators are 1, 6 and followed by 3, 15, 15 etc as provided in A172087.
The second column of the array, 1, 1, 1/6, 1/6, -1/30, -1/30, ... contains doubled A000367(n)/A002445(n). These are related to A176150, A176144, and A176184.
In the first subdiagonal of the array we see 1, 1/6, 2/15, -8/150, 8/105, -32/321, 6112/15015, -3712/2145 , ... continued as given by A181130 and A181131.

Vincenzo Librandi, Table of n, a(n) for n = 0..300

read("transforms") ;
evb := [0, 1, 0, seq(bernoulli(n), n=2..30)] ;
ievb := BINOMIALi(evb) ;
[seq((-1)^n*op(n,ievb),n=1..nops(ievb))] ;
DIFF(%) ;
DIFF(%) ;
apply(numer,%) ; # R. J. Mathar, Sep 20 2011

evb = Join[{0, 1, 0}, Table[BernoulliB[n], {n, 2, 32}]]; ievb = Table[ Sum[Binomial[n, k]*evb[[k+1]], {k, 0, n}], {n, 0, Length[evb]-3}]; Differences[ievb, 2] // Numerator (* Jean-François Alcover, Sep 09 2013, after R. J. Mathar *)

a(2*n+1) + a(2*n+2) = A172087(2*n+2) = A172087(2*n+3), n >= 1.

A176184 a(2n) = A027641(n). a(2n+1) = A164555(n).

Original entry on oeis.org

1, 1, -1, 1, 1, 1, 0, 0, -1, -1, 0, 0, 1, 1, 0, 0, -1, -1, 0, 0, 5, 5, 0, 0, -691, -691, 0, 0, 7, 7, 0, 0, -3617, -3617, 0, 0

Offset: 0

Paul Curtz, Apr 11 2010

Essentially the same as A176144. (The signs of the third and fourth entry are swapped.)
This refers to a shuffling of the "original" Bernoulli numbers and the Bernoulli numbers in opposite order compared to the composition discussed in A176150.
The inverse binomial transform of the shuffle in A176150 was 1,0, -1/2, 0, 13/6, -20/3. The shuffling here would yield an inverse binomial transform 1, 0, -3/2, 4, -47/6, 40/3, -21, 95/3 etc.
The difference between the corresponding elements of these two binomial transforms element by element is 0, 0, 1, -4, 10, -20, 35, -56, 84, -120, 165, -220,..., a signed variant of A000292.

A176511 From Bernoulli twin numbers to Catalan numbers arrays (*).First part.We consider array, from Bernoulli twin numbers A051716/A051717 mixed with their companion A172083/A051717 BTC(n)=1,1,-1/2,-3/2,-1/3,2/3,-1/6,-1/6, and successive differences ,named BTC1. a(n) are numerators of BTC(n).Denominators are (double A051717)=1,1,2,2,3,3,6,6,30,30,30,30,.

Original entry on oeis.org

1, 1, -1, -3, -1, 2, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 5, 5

Offset: 0

Paul Curtz, Apr 19 2010

(*) Even case:ECT(n) in A176239. BTC(n) are not Bi-Bernoulli numbers (absolute values of mixed sequences are not the same like BB1(n) in A176144 or BB2(n) in A176184). Rows of array BTC1: 1) 1,1,-1/2,-3/2,-1/3,2/3,-1/6,-1/6,-1/30,-1/30,1/30,1/30,1/42,1/42,; 2) 0,-3/2,-1,7/6,1,-5/6,0,2/15,0,2/15,0; 3) -3/2,1/2,13/6,-1/6,-11/6,5/6,2/15,-2/15,2/15,-2/15; 4) 2,5/3,-7/3,-5/3,8/3,-7/10,-4/15,4/15,-4/15; 5) -1/3,-4,2/3,13/3,-101/30,13/30,8/15,-8/15; 6) -11/3,14/3,11/3,-77/10,19/5,1/10,-16/15; 7) 25/3,-1, -341/30,23/2,-37/10,-29/30; 8) -28/3,-311/30,343/15,. Correction:in A176150 last term (-517) is false.

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A195240 Numerators of the second differences of the sequence of fractions (-1)^(n+1)*A176618(n)/A172031(n).

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A176184 a(2n) = A027641(n). a(2n+1) = A164555(n).

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