A051717 -id:A051717 - cp's OEIS frontend

A141424 Numerators of second column of the inverse of the triangle of polynomial coefficients P(0,x)=1, 2P(n,x)=(1+x)*[(1+x)^(n-1)+x^(n-1)].

1, -3, 3, -5, 5, -7, 7, 3, -3, -121, 121, 1261, -1261, -20583, 20583, 888403, -888403, -24729925, 24729925, 862992399, -862992399, -36913939769, 36913939769, 1899853421885, -1899853421885, -115841483491323, 115841483491323, 8258802033519361

Offset: 0

Paul Curtz, Aug 06 2008

sign

For the denominators see A053644.

The P(n,x) polynomials are based on the Euler polynomials and the inverse matrix of their coefficients is described in Example section of A133135. First column is A033999, third column is A133135.

Cf. A051717.

max = 27; p[0] = 1; p[n_] := (1+x)*((1+x)^(n-1)+x^(n-1))/2; t = Table[Coefficient[p[n], x, k], {n, 0, max+2}, {k, 0, max+2}]; a[n_] := Inverse[t][[All, 2]][[n+2]] // Numerator; Table[a[n], {n, 0, max}] (* Jean-François Alcover, Dec 16 2013 *)

lista(n) = {m = matrix(n, n); m[1, 1] = 1; for (i=2, n, pol = (1+x)*((1+x)^(i-2)+x^(i-2))/2; for (j=1, n, m[i, j] = polcoeff(pol, j-1, x););); m = 1/m; for (i=2, n, print1(numerator(m[i, 2]), ", ");); print();} \\ Michel Marcus, Aug 16 2013

Edited by Michel Marcus, Aug 16 2013

A165161 Numerator of the n-th term in the first differences of the binomial transform of the "original" Bernoulli numbers.

Original entry on oeis.org

1, 2, 5, 29, 31, 43, 41, 29, 31, 71, 61, 2039, 3421, 13, -1, -3107, 4127, 44665, -43069, -174281, 174941, 854651, -854375, -236361361, 236366821, 8553109, -8553097, -23749460159, 23749461899, 8615841290327

Offset: 0

Paul Curtz, Sep 06 2009

The binomial transform of the "original" Bernoulli numbers is 1, 3/2, 13/6, ... as mentioned in A164558.
The first differences of that sequence are 3/2 - 1 = 1/2, 13/6 - 3/2 = 2/3, 5/6, 29/30, 31/30, ... and the numerators of these differences are listed here.
The bisection a(2n) reappears (up to signs) as A162173(n+1).

Cf. A051717 (denominators), A164555, A027642.

read("transforms") :
A164555 := proc(n) if n <= 2 then 1; else numer(bernoulli(n)) ; end if; end proc:
A027642 := proc(n) denom(bernoulli(n)) ; end proc:
nmax := 40:
BINOMIAL([seq(A164555(n)/A027642(n), n=0..nmax)]) :
map(numer,DIFF(%)) ; # R. J. Mathar, Jul 07 2011

a(2n) + A000367(n) = A006954(n+1) = A051717(2n+1).

a(2n+1) + a(2n+2) = A051717(2n+2) + A051717(2n+3), n > 0.

A233808 Autosequence preceding A198631(n)/A006519(n+1). Numerators.

Original entry on oeis.org

0, 0, 1, 3, 3, 5, 5, 7, 7, -3, -3, 121, 121, -1261, -1261, 20583, 20583, -888403, -888403, 24729925, 24729925, -862992399, -862992399, 36913939769, 36913939769, -1899853421885, -1899853421885

Offset: 0

Paul Curtz, Dec 16 2013

The fractions are g(n)=0, 0, 1, 3/2, 3/2, 5/4, 5/4, 7/4, 7/4, -3/8, -3/8, 121/8, 121/8, -1261/8, -1261/8, 20583/8, 20583/8, -888403/16, -888403/16,... . The denominators are 1, 1, followed by A053644(n+1).
g(n+2) - g(n+1) = A198631(n)/A006519(n+1).
The corresponding fractions to g(n) are f(n) in A165142(n).
g(n) differences table:
0,       0,    1,  3/2,  3/2,  5/4,
0,       1,  1/2,    0, -1/4,    0,
1,    -1/2, -1/2, -1/4,  1/4,  1/2,    Euler twin numbers (new),
-3/2,    0,  1/4,  1/2,  1/4,   -1,
3/2,   1/4,  1/4, -1/4, -5/4, -5/8,
-5/4,    0, -1/2,   -1,  5/8, 13/2, etc.
Like A198631(n)/A006519(n+1),g(n) is an autosequence of the second kind.
If we proceed, here for Euler polynomials, like in A233565 for Bernoulli polynomials, we obtain
1) A133138(n)/A007395(n) (unreduced form) or
2) A233508(n)/A232628(n) (reduced form),the first array in A133135.
The Bernoulli's corresponding fractions to 1) are A193815(n)/(A003056(n) with 1 instead of 0).

Cf. A051716/A051717, Bernoulli twin numbers.

max = 27; p[0] = 1; p[n_] := (1 + x)*((1 + x)^(n - 1) + x^(n - 1))/2; t = Table[Coefficient[p[n], x, k], {n, 0, max + 2}, {k, 0, max + 2}]; a[n_] := (-1)^n*Inverse[t][[n, 2]] // Numerator; a[0] = 0; Table[a[n], {n, 0, max}] (* Jean-François Alcover, Jan 11 2016 *)

a(n) = 0, 0, followed by (-1)^n *A141424(n).

A237109 a(n) is the numerator of 2n / ((n+2) (n+3)).

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 7, 8, 3, 5, 11, 4, 13, 7, 5, 16, 17, 3, 19, 20, 7, 11, 23, 8, 25, 13, 9, 28, 29, 5, 31, 32, 11, 17, 35, 12, 37, 19, 13, 40, 41, 7, 43, 44, 15, 23, 47, 16, 49, 25, 17, 52, 53, 9, 55, 56, 19, 29, 59, 20, 61, 31, 21, 64, 65, 11, 67, 68, 23, 35, 71, 24

Offset: 1

Paul Curtz, Feb 03 2014

Previous name was: Numerators of the third row of the Akiyama-Tanigawa algorithm (or transformation) applied to A001008(n+1)/A002805(n+1).
Successive rows:
3/2,  11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ...;
-1/3, -1/2, -3/5, -2/3, -5/7, -3/4, -7/9, -4/5, ... = A026741(n+1)/A026741(n+3);
1/6, 1/5, 1/5, 4/21, 5/28, 1/6, 7/45, 8/55,  3/22, ...;
-1/30, 0, ...;
-1/30.
First column denominators: 2,3,6,30,30,... = A051717(n+1).
A001008(n)/A002805(n) is the inverse Akiyama-Tanigawa transformation applied to A027641(n)/A027642(n). A051716(n)/A051717(n) comes from 0 followed by A164555(n)/A027642(n). Then, from the two Bernoulli numbers.

G. C. Greubel, Table of n, a(n) for n = 1..2500

[Numerator(2*n/((n+2)*(n+3))): n in [1..50]]; // G. C. Greubel, Aug 07 2018

a[1, n_] := HarmonicNumber[n+1]; a[n_, m_] := a[n, m] = m*(a[n-1, m]-a[n-1, m+1]); Table[a[3, m] // Numerator, {m, 1, 72}] (* Jean-François Alcover, Feb 11 2014 *)
a[ n_] := n / {1, 2, 3, 1, 1, 6, 1, 1, 3, 2, 1, 3}[[Mod[n, 12, 1]]]; (* Michael Somos, Aug 01 2017 *)

{a(n) = if( n<0, -a(-n), numerator( 2*n / ((n+2) * (n+3))))}; /* Michael Somos, Aug 01 2017 */

a(n) = -a(-n) for all n in Z. - Michael Somos, Aug 01 2017
From Amiram Eldar, Nov 17 2022: (Start)
Multiplicative with a(2) = 1, a(2^e) = 2^e for e > 1, a(3^e) = 3^(e-1), and a(p^e) = p^e for p >= 5.
Sum_{k=1..n} a(k) ~ (49/144) * n^2. (End)
Dirichlet g.f.: zeta(s-1)*(1-1/2^s+2/4^s)*(1-2/3^s). - Amiram Eldar, Jan 05 2023

New name using Somos's Pari code from Joerg Arndt, May 27 2018

Keyword:mult added by Andrew Howroyd, Jul 31 2018

A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

Original entry on oeis.org

1, 1, -3, 1, -5, 5, 1, -7, 25, -5, 1, -9, 23, -35, 49, 1, -11, 73, -27, 112, -49, 1, -13, 53, -77, 629, -91, 58, 1, -15, 145, -130, 1399, -451, 753, -58, 1, -17, 95, -135, 2699, -2301, 8573, -869, 341, 1, -19, 241

Offset: 0

Paul Curtz, Apr 27 2012

In A190339 we saw that (the second Bernoulli numbers) A164555/A027642 is an eigensequence (its inverse binomial transform is the sequence signed) of the second kind, see A192456/A191302. We consider this array preceded by 1 for the second row, by 1, -3/2, for the third one; 1 is chosen and is followed by the differences of successive rows.
Hence
                    1    1/2   1/6      0
            1    -1/2   -1/3  -1/6  -1/30
      1   -3/2    1/6    1/6  2/15   1/15
  1 -5/2   5/3      0  -1/30 -1/15 -8/105.
The second row is A051716/A051717.
The (reduced) triangle before the square array (T(n,m) in A190339) is a(n)/b(n)=
B(0)=    1 = 1                 Redbernou1li
B(1)= -1/2 = 1  -3/2
B(2)=  1/6 = 1  -5/2  5/3
B(3)=    0 = 1  -7/2 25/6  -5/3
B(4)=-1/30 = 1  -9/2 23/3 -35/6  49/30
B(5)=    0 = 1 -11/2 73/6 -27/2 112/15 -49/30.
For the main diagonal, see A165142.
Denominator b(n) will be submitted.
This transform is valuable for every eigensequence of the second kind. For instance Leibniz's 1/n (A003506).
With increasing exponents for coefficients, polynomials CB(n,x) create Redbernou1li. See the formula.
Triangle Bernou1li for A027641/A027642 with the same denominator A080326 for every column is
1
1  -3/2
1  -5/2 10/6
1  -7/2 25/6 -10/6
1  -9/2 46/6 -35/6  49/30
1 -11/2 73/6 -81/6 224/30 -49/30.
For numerator by columns,see A000012, -A144396, A100536, Q(n)=n*(2*n^2+9*n+9)/2 , new.
Triangle Checkbernou1 with the same denominator A080326 for every row is
1/1
(2    -3)/2
(6   -15  +10)/6
(6   -21  +25  -10)/6
(30 -135 +230 -175  +49)/30
(30 -165 +365 -405 +224 -49)/30;
Hence for numerator: 1, 2-3, 16-15, 31-31, 309-310, 619-619, 8171-8166.
Absolute sum: 1, 5, 31, 62, 619, 1238, 17337. Reduced division by A080326:
1, 5/2, 31/6, 31/3, 619/30, 619/15, 5779/70, = A172030(n+1)/A172031(n+1).

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane), A051714/A051715 (Akiyama-Tanigawa), A192456/A191302 for other triangles that lead to the Bernoulli numbers.

CB(0,x) = 1,
CB(1,x) = 1 - 3*x/2,
CB(n,x) = (1-x)*CB(n-1,x) + B(n)*x^n , n > 1.

A224964 Irregular triangle of the denominators of the unreduced fractions that lead to the second Bernoulli numbers.

Original entry on oeis.org

2, 2, 2, 6, 2, 6, 2, 6, 15, 2, 6, 15, 2, 6, 15, 105, 2, 6, 15, 105, 2, 6, 15, 105, 105, 2, 6, 15, 105, 105, 2, 6, 15, 105, 105, 231, 2, 6, 15, 105, 105, 231, 2, 6, 15, 105, 105, 231, 15015, 2, 6, 15, 105, 105, 231, 15015

Offset: 0

Paul Curtz, Apr 21 2013

The triangle of fractions A192456(n)/A191302(n) leading to the second Bernoulli numbers written in A191302(n) is the reduced case. The unreduced case is
B(0) =   1   = 2/2         (1 or 2/2 chosen arbitrarily)
B(1)         = 1/2
B(2) =  1/6  = 1/2 - 2/6
B(3) =   0   = 1/2 - 3/6
B(4) = -1/30 = 1/2 - 4/6 +  2/15
B(5) =   0   = 1/2 - 5/6 +  5/15
B(6) =  1/42 = 1/2 - 6/6 +  9/15 -  8/105
B(7) =   0   = 1/2 - 7/6 + 14/15 - 28/105
B(8) = -1/30 = 1/2 - 8/6 + 20/15 - 64/105 + 8/105.
The constant values along the columns of denominators are A190339(n).
With B(0)=1, B(2) = 1/2 -1/3, (reduced case), the last fraction of the B(2*n) is
1, -1/3, 2/15, -8/105, 8/105, ... = A212196(n)/A181131(n).
We can continue this method of sum of fractions yielding Bernoulli numbers.
Starting from 1/6 for B(2*n+2), we have:
B(2) = 1/6
B(4) = 1/6 - 3/15
B(6) = 1/6 - 5/15 + 20/105
B(8) = 1/6 - 7/15 + 56/105 - 28/105.
With the odd indices from 3, all these B(n) are the Bernoulli twin numbers -A051716(n+3)/A051717(n+3).

			Triangle begins
  2;
  2;
  2, 6;
  2, 6;
  2, 6, 15;
  2, 6, 15;
  2, 6, 15, 105;
  2, 6, 15, 105;
  2, 6, 15, 105, 105;
  2, 6, 15, 105, 105;
  2, 6, 15, 105, 105, 231;
  2, 6, 15, 105, 105, 231;
  2, 6, 15, 105, 105, 231, 15015;
  2, 6, 15, 105, 105, 231, 15015;

Cf. A051716, A051717, A141044, A181131, A190339, A191302, A192456, A212196.

nmax = 7; b[n_] := BernoulliB[n]; b[1] = 1/2; bb = Table[b[n], {n, 0, 2*nmax-1}]; diff = Table[ Differences[bb, n], {n, 1, nmax}]; A190339 = diff // Diagonal // Denominator; Table[ Table[ Take[ A190339, n], {2}], {n, 1, nmax}] // Flatten (* Jean-François Alcover, Apr 25 2013 *)

T(n,k) = A190339(k).

A238235 Numerators of Euler twin numbers Et(n).

Original entry on oeis.org

1, -1, -1, -1, 1, 1, -1, -17, 17, 31, -31, -691, 691, 5461, -5461, -929569, 929569, 3202291, -3202291, -221930581, 221930581, 4722116521, -4722116521, -968383680827, 968383680827, 14717667114151, -14717667114151

Offset: 0

Paul Curtz, Feb 20 2014

sign

Et(n) = 1, -1/2, -1/2, -1/4, 1/4, 1/2, -1/2, -17/8, 17/8, 31/2, -31/2, -691/4, 691/4, 5461/2, -5461/2,... =a(n)/b(n) is mentioned in A233808.
Denominators: b(n)= 1, 2, 2, 4, 4, 2, 2, 8, 8,... = A065176(n) with 1 instead of 0.
Et(n) is the first difference of 0, followed by A198631(n)/A006519(n+1).
Et(n+2) = -1/2, -1/4, 1/4, 1/2,... is an autosequence of the second kind. Its main diagonal is the double of the following diagonal, the inverse binomial transform of Et(n+2) being Et(n+2) signed.
The denominators of the difference table of Et(n+2) are even numbers of the form 2^p. For the Bernoulli twin numbers A051716(n+1)/A051717(n+2), the denominators of the difference table, A168426(n), are multiples of 3.

Cf. A051716/A051717 (Bernoulli twin numbers).

Join[{1, -1, -1}, Table[{nu = Numerator[EulerE[2*n+1, 1]], -nu}, {n, 1, 12}]] // Flatten (* Jean-François Alcover, Feb 24 2014 *)

Binomial transform of A141424(n)/(A053644(n) with 1 instead of 0).

a(2n+3) = (-1)^n*A002425(n+2) = -a(2n+4).

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).

A257372 a(n) = denominators of A255935(n) * triangle T(n,k) for Bernoulli(k+2), k=0 to n-1.

Original entry on oeis.org

1, 6, 6, 15, 30, 21, 42, 15, 30, 33, 66, 1365, 2730, 3, 6, 255, 510, 399, 798, 165, 330, 69, 138, 1365, 2730, 3, 6, 435, 870, 7161, 14322, 255, 510, 3, 6, 959595, 1919190, 3, 6, 6765, 13530, 903, 1806, 345, 690

Offset: 0

Paul Curtz, Apr 21 2015

nonn

Generally, A255935(n) multiplied by triangle T(n,k) for s(k), k=0 to n-1 yields an autosequence of the first kind (a sequence whose main diagonal is 0's).
Here s(k) = 1/6, 0, -1/30, ... from A164555(n+2)/A027642(n+2). Hence
0                              =  0/1
1/6, 0                         =  1/6
1/6, 0,     0                  =  1/6
1/6, 0, -1/10, 0               = 1/15
1/6, 0,  -1/5, 0, 0            =-1/30
... .
a(n) are the row sums denominators.
Compare to A051716(n+2)/A051717(n+2).
Hence the difference table
0,       1/6,      1/6,  1/15, -1/30, -1/21, 1/42, ...
1/6,       0,    -1/10, -1/10, -1/70,  1/14, ...
-1/6,  -1/10,        0,  3/35,  3/35, ...
1/15,   1/10,     3/35,     0, ...
1/30,  -1/70,    -3/35, ...
-1/21, -1/14, ...
-1/42, ...
... .

Cf. A255935, A027641/A027642, A164555/A027642, A001897, A002445, A040000, A051716/A051717.

a(2n) = A002445(n).
a(2n+3) = A001897(n+2).
a(2n+2) = A040000(n) * a(2n+1).

cp's OEIS Frontend

A141424 Numerators of second column of the inverse of the triangle of polynomial coefficients P(0,x)=1, 2P(n,x)=(1+x)*[(1+x)^(n-1)+x^(n-1)].

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A165161 Numerator of the n-th term in the first differences of the binomial transform of the "original" Bernoulli numbers.

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Maple

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A233808 Autosequence preceding A198631(n)/A006519(n+1). Numerators.

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A237109 a(n) is the numerator of 2*n / ((n+2) * (n+3)).

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A182397 Numerators in triangle that leads to the (first) Bernoulli numbers A027641/A027642.

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A224964 Irregular triangle of the denominators of the unreduced fractions that lead to the second Bernoulli numbers.

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A238235 Numerators of Euler twin numbers Et(n).

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

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A257372 a(n) = denominators of A255935(n) * triangle T(n,k) for Bernoulli(k+2), k=0 to n-1.

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A237109 a(n) is the numerator of 2n / ((n+2) (n+3)).