A176289 -id:A176289 - cp's OEIS frontend

A256003 a(n) = 0 followed by numerators of 2*A176327(n)/A176289(n).

0, 2, 0, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0

Offset: 0

Paul Curtz, May 06 2015

Offset 0 is chosen instead of -1. (The offset 0 corresponds to A176327(n), -1 to 0 followed by A176327(n).)
Denominators: b(n) = 1 followed by A141459(n).
Difference table of a(n)/b(n):
0,          2,      0,   1/3,     0, -1/15, 0, ...
2,         -2,    1/3,  -1/3, -1/15,  1/15, ...
-4,       7/3,   -2/3,  4/15,  2/15, ...
19/3,      -3,  14/15, -2/15, ...
-28/3,  59/15, -16/15, ...
199/15,    -5, ...
-274/15, ...
etc.
Without the first column, the antidiagonal sums are (-1)^n * A254667(n+1).
The Bernoulli numbers A027641(n)/A027642(n) or A164555(n)/A027642(n) come from A000027. 0 followed by the Bernoulli numbers comes from A001477. a(0)=0 is a choice.

Cf. A176327/A176289, A172086/A172087, A141459, A176618, A195240, A254667.

a(2n) = 0. a(2n+1) = A172086(2n), from the main pure Bernoulli twin numbers.

A104712 Pascal's triangle, with the first two columns removed.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715

Offset: 2

Gary W. Adamson, Mar 19 2005

A000295 (Eulerian numbers) gives the row sums.
Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle.
From Peter Luschny, Apr 10 2011: (Start)
A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:
(a) Each partition has at least one singleton block;
(c) k is the size of the largest block of the partition;
(b) m = n - k + 1 is the number of parts of the partition.
Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)).
For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11.
This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts
  1;
  1,  3;
  1,  6,  4;
  1, 10, 10,  5;
  1, 15, 20, 15,  6;
  ...
(End)
Diagonal sums are A001924(n+1). - Philippe Deléham, Jan 11 2014
Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014

			The triangle a(n, k) begins:
  n\k  2   3   4    5    6    7    8   9  10 11 12 13
  2:   1
  3:   3   1
  4:   6   4   1
  5:  10  10   5    1
  6:  15  20  15    6    1
  7:  21  35  35   21    7    1
  8:  28  56  70   56   28    8    1
  9:  36  84 126  126   84   36    9   1
  10: 45 120 210  252  210  120   45  10   1
  11: 55 165 330  462  462  330  165  55  11  1
  12: 66 220 495  792  924  792  495 220  66 12  1
  13: 78 286 715 1287 1716 1716 1287 715 286 78 13  1
... reformatted. - _Wolfdieter Lang_, Mar 20 2015

G. C. Greubel, Rows n=2..100 of triangle, flattened
D. Dugger, A Geometric Introduction to K-Theory
Candice A. Marshall, Construction of Pseudo-Involutions in the Riordan Group, Dissertation, Morgan State University, 2017.
T. Saito, The discriminant and the determinant of a hypersurface of even dimension (p. 4), arXiv:1110.1717 [math.AG], 2011-2012.

Cf. A000295, A007318, A008292, A104713, A027641/A027642 (first Bernoulli numbers B-), A164555/A027642 (second Bernoulli numbers B+), A176327/A176289.

/* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018

t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)

for(n=2, 10, for(k=2,n, print1(binomial(n,k), ", "))) \\ G. C. Greubel, May 15 2018

T(n,k) = binomial(n,k), for 2 <= k <= n.
From Peter Bala, Jul 16 2013: (Start)
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^3, x/(1 - x)).
O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t)) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)
As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
From Tom Copeland, Apr 11 2014: (Start)
A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!.
B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }
C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....
D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)
O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014
Reverse T = [St2]*dP*[St1]- dP = [St2]*(exp(x*M)-I)*[St1]-(exp(x*M)-I) with top two rows of zeros removed, [St1]=padded A008275 just as [St2]=A048993=padded A008277, dP= A132440, M=A238385-I, and I=identity matrix. Cf. A238363. - Tom Copeland, Apr 26 2014
O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015

Edited and extended by David Wasserman, Jul 03 2007

A176327 Numerators of the rational sequence with e.g.f. (x/2)*(1+exp(-x))/(1-exp(-x)).

Original entry on oeis.org

1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0, 2929993913841559, 0, -261082718496449122051

Offset: 0

Paul Curtz, Apr 15 2010

sign

Numerator of the Bernoulli number B_n, except B(1)=0.
A027641 is the main entry for this sequence, which is only a minor variation. - N. J. A. Sloane, Nov 29 2010.
This could formally be defined by building the arithmetic mean of the numerators in A164555(n) and A027641(n).

Antti Karttunen, Table of n, a(n) for n = 0..200

Cf. A176289 (denominators), A027642, A141056, A164020, A165823

seq(numer((bernoulli(i,0)+bernoulli(i,1))/2),i=0..40); # Peter Luschny, Jun 17 2012

terms = 41; egf = (x/2)*((1 + Exp[-x])/(1 - Exp[-x])) + O[x]^(terms+1);
CoefficientList[egf, x]*Range[0, terms-1]! // Numerator (* Jean-François Alcover, Jun 13 2017 *)

apply(numerator, Vec(serlaplace((x/2)*(1+exp(-x))/(1-exp(-x))))) \\ Charles R Greathouse IV, Sep 26 2017

a(2n+1) = 0. a(2n ) = A000367(n).

a(n) = A164555(n) = A027641(n) if n <>1.

New name from Peter Luschny, Jun 18 2012

A239315 Array read by antidiagonals: denominators of the core of the classical Bernoulli numbers.

Original entry on oeis.org

15, 15, 15, 105, 105, 105, 21, 105, 105, 21, 105, 105, 105, 105, 105, 15, 105, 105, 105, 105, 15, 165, 165, 1155, 231, 1155, 165, 165, 33, 165, 165, 231, 231, 165, 165, 33, 15015, 15015, 15015, 15015, 15015, 15015, 15015, 15015, 15015

Offset: 0

Paul Curtz, Mar 15 2014

We consider the autosequence A164555(n)/A027642(n) (see A190339(n)) and its difference table without the first two rows and the first two columns:
2/15,     1/15,    -1/105,   -1/21,   -1/105,     1/15,   7/165,   -5/33,...
-1/15,  -8/105,    -4/105,   4/105,    8/105,   -4/165, -32/165,...
-1/105,  4/105,     8/105,   4/105, -116/1155, -28/165,...
1/21,    4/105,    -4/105, -32/231,   -16/231,...
-1/105, -8/105, -116/1155,  16/231,...
-1/15,  -4/165,    28/165,...
7/165,  32/165,...
5/33,... etc.
This is an autosequence of the second kind.
The antidiagonals are palindromes in absolute values.
a(n) are the denominators. Multiples of 3.
Sum of odd antidiagonals: 2/15, -2/21, 2/15, -10/33, 1382/1365,... = -2*A000367(n+2)/A001897(n+2).
The sum of the even antidiagonals is A000004.
2/15, 0, -2/21,... = -4*A027641(n+4)/A027642(n+4) = -4*A164555(n)/A027642(n+4) and others.

			As a triangle:
15,
15,   15,
105, 105, 105,
21,  105, 105, 21,
105, 105, 105, 105, 105,
etc.

Cf. A085737/A085738, A168516/A168426 (autosequence), A027641, A176327/A176289, A235774, A165161/A051717(n+1).

max = 12; tb = Table[BernoulliB[n], {n, 0, max}]; td = Table[Differences[tb, n][[3 ;; -1]], {n, 2, max - 1}]; Table[td[[n - k + 1, k]] // Denominator, {n, 1, max - 3}, {k, 1, n}] // Flatten (* Jean-François Alcover, Apr 11 2014 *)

A176447 a(2n) = -n, a(2n+1) = 2n+1.

Original entry on oeis.org

0, 1, -1, 3, -2, 5, -3, 7, -4, 9, -5, 11, -6, 13, -7, 15, -8, 17, -9, 19, -10, 21, -11, 23, -12, 25, -13, 27, -14, 29, -15, 31, -16, 33, -17, 35, -18, 37, -19, 39, -20, 41, -21, 43, -22, 45, -23, 47, -24, 49, -25, 51, -26, 53, -27, 55, -28, 57, -29, 59, -30, 61, -31, 63, -32, 65, -33, 67, -34, 69, -35

Offset: 0

Paul Curtz, Apr 18 2010

There is more complicated way of defining the sequence: consider the sequence of modified Bernoulli numbers EVB(n) = A176327(n)/A176289(n) and its inverse binomial transform IEVB(n) = A176328(n)/A176591(n). Then a(n) is the numerator of the difference EVB(n)-IEVB(n). The denominator of the difference is 1 if n=0, else A040001(n-1).
A particularity of EVB(n) is: its (forward) binomial transform is 1, 1, 7/6, 3/2, 59/30,.. = (-1)^n*IEVB(n).
Note that A026741 is related to the Rydberg-Ritz spectrum of the hydrogen atom.

			G.f. = x - x^2 + 3*x^3 - 2*x^4 + 5*x^5 - 3*x^6 + 7*x^7 - 4*x^8 + 9*x^9 - 5*x^10 + ...

R. J. Mathar, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A008795, A026741, A040001, A176289, A176327, A176328, A176591.

[n*(1-3*(-1)^n)/4: n in [0..60]]; // Vincenzo Librandi, Aug 04 2011

a[n_?EvenQ]:=-(n/2); a[n_?OddQ]:=n; Table[a[n], {n, 100}] (* Alonso del Arte, Dec 01 2010 *)
a[ n_] := n / If[ Mod[ n, 2] == 1, 1, -2]; (* Michael Somos, Jun 11 2013 *)
CoefficientList[Series[x (1 - x + x^2)/((x - 1)^2*(1 + x)^2), {x, 0, 70}], x]  (* Michael De Vlieger, Dec 10 2016 *)
LinearRecurrence[{0,2,0,-1},{0,1,-1,3},80] (* Harvey P. Dale, Nov 01 2017 *)

{a(n) = n / if( n%2, 1, -2)}; /* Michael Somos, Jun 11 2013 */

From R. J. Mathar, Dec 01 2010:  (Start)
a(n) = (-1)^n*A026741(n) = n*(1-3*(-1)^n)/4.
G.f.: x*(1-x+x^2) / ( (x-1)^2*(1+x)^2 ).
a(n) = +2*a(n-2) -a(n-4).  (End)
a(n) = -a(-n) for all n in Z. - Michael Somos, Jun 11 2013
From Michael Somos, Aug 30 2014: (Start)
Euler transform of length 6 sequence [ -1, 3, 1, 0, 0, -1].
0 = - 1 - a(n) - a(n+1) + a(n+2) + a(n+3) for all n in Z.
0 = 1 + a(n)*(-2 -a(n) + a(n+2)) - 2*a(n+1) - a(n+2) for all n in Z. (End)
From Michael Somos, May 04 2015: (Start)
a(n) is multiplicative with a(2^e) = -(2^(e-1)) if e>0, a(p^e) = p^e otherwise.
G.f.: (f(x) - 3 * f(-x)) / 4 where f(x) := x / (1 - x)^2.
G.f.: x * (1 - x) * (1 - x^6) / ((1 - x^2)^3 * (1 - x^3)). (End)
From Amiram Eldar, Sep 21 2023: (Start)
Dirichlet g.f.: zeta(s-1) * (1 - 3/2^s).
Sum_{k=0..n} a(k) = A008795(n-1), for n > 0.
Sum_{k=0..n} a(k) ~ n^2/8. (End)

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

Original entry on oeis.org

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

Paul Curtz, Apr 03 2015

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.

			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.

Jacob Bernoulli, Ars Conjectandi (1713).

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

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

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 *)

A340243 a(n) = denominator((2n-1)zeta(2n)/Pi^(2n)).

Original entry on oeis.org

2, 6, 30, 189, 1350, 10395, 58046625, 1403325, 21709437750, 2292899734125, 80596287646875, 640374140030625, 8779111824511153125, 443779279041223125, 20913098524817639765625, 195202717402382161174828125, 2015813566807172297008593750, 367589532770719654160390625

Offset: 0

Artur Jasinski, Jan 01 2021

For numerators a(n+1) see A046988.

			1/2, 1/6, 1/30, 1/189, 1/1350, 1/10395, 691/58046625, 2/1403325, 3617/21709437750, 43867/2292899734125, ...

Cf. A002432, A027642, A046988, A176289, A164555.

a := n -> denom((2*n-1)*Zeta(2*n)/Pi^(2*n));
seq(a(n), n=0..17); # Peter Luschny, Jan 12 2021

Denominator[Table[(2 n - 1)*Zeta[2 n]/Pi^(2 n), {n, 0, 16}]]

a(n) = denominator((2*n-1)*2^(2*n-1)*bernfrac(2*n)/(2*n)!); \\ Michel Marcus, Jun 15 2022

a(n) = denominator((2*n-1)*2^(2*n-1)*Bernoulli(2*n)/(2*n)!). - Peter Luschny, Jan 12 2021

cp's OEIS Frontend

A256003 a(n) = 0 followed by numerators of 2*A176327(n)/A176289(n).

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A104712 Pascal's triangle, with the first two columns removed.

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A176327 Numerators of the rational sequence with e.g.f. (x/2)*(1+exp(-x))/(1-exp(-x)).

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A239315 Array read by antidiagonals: denominators of the core of the classical Bernoulli numbers.

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A176447 a(2n) = -n, a(2n+1) = 2n+1.

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

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A340243 a(n) = denominator((2*n-1)*zeta(2*n)/Pi^(2*n)).

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A340243 a(n) = denominator((2n-1)zeta(2n)/Pi^(2n)).