A164558 -id:A164558 - cp's OEIS frontend

A219976 Denominators of the Inverse bi-binomial transform of A164558(n)/A027642(n) read downwards antidiagonals.

1, 2, 2, 6, 6, 6, 1, 3, 3, 1, 30, 30, 30, 30, 30, 1, 15, 15, 15, 15, 1, 42, 42, 210, 210, 210, 42, 42, 1, 21, 21, 105, 105, 21, 21, 1, 30, 30, 210, 210, 210, 210, 210, 30, 30, 1, 15, 15, 105, 105, 105, 105, 15, 15, 1

Offset: 0

Paul Curtz, Dec 02 2012

Starting from any sequence a(k) in the first row, we define the array T(n,k) of the inverse bi-binomial transform by T(0,k) = a(k), T(n,k) = T(n-1,k+1) -2*T(n-1,k) n>0. Hence A164558(n)/A027642(n) and successive "bi-differences":
1,       3/2,    13/6,       3,  119/30,        5,   253/42,       7,  239/30, 9;
-1/2,   -5/6,    -4/3,  -61/30,  -44/15,  -167/42,  -106/21, -181/30, -104/15;
1/6,     1/3,   19/30,   17/15, 397/210,    61/21 , 853/210,   77/15;
0,     -1/30,   -2/15, -79/210, -92/105, -367/210, -314/105;
-1/30, -1/15, -23/210, -13/105,   1/210,   53/105;
0,      1/42,    2/21,  53/210,  52/105;
1/42,   1/21,  13/210,  -1/105;
0,     -1/30,   -2/15;
-1/30, -1/15;
0.
The first column is A027641(n)/A027642(n).

			Partial array of denominators:
1,   2,   6,   1,  30,   1,  42,  1, 30,  1;
2,   6,   3,  30,  15,  42,  21, 30, 15;
6,   3,  30,  15, 210,  21, 210, 15;
1,  30,  15, 210, 105, 210, 105;
30, 15, 210, 105, 210, 105;
1,  42,  21, 210, 105;
42, 21, 210, 105;
1,  30,  15;
30, 15;
1.
a(n):
1;
2,   2;
6,   6,  6,;
1,   3,  3,  1;
30, 30, 30, 30, 30;

Cf. A213268.

A164558[n_] := Sum[(-1)^k*Binomial[n, k]*BernoulliB[k], {k, 0, n}] // Numerator; t[0, k_?Positive] := A164558[k] / Denominator[ BernoulliB[k]]; t[n_?Positive, k_] := t[n, k] = t[n-1, k+1] - 2*t[n-1, k]; t[0, 0] = 1; t[, ] = 0; Flatten[ Table[t[n-k , k] // Denominator, {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, Dec 04 2012 *)

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

Original entry on oeis.org

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

Paul Curtz, Feb 03 2015

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.

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).

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 *)

A196838 Numerators of coefficients of Bernoulli polynomials with rising powers of the variable.

Original entry on oeis.org

1, -1, 1, 1, -1, 1, 0, 1, -3, 1, -1, 0, 1, -2, 1, 0, -1, 0, 5, -5, 1, 1, 0, -1, 0, 5, -3, 1, 0, 1, 0, -7, 0, 7, -7, 1, -1, 0, 2, 0, -7, 0, 14, -4, 1, 0, -3, 0, 2, 0, -21, 0, 6, -9, 1, 5, 0, -3, 0, 5, 0, -7, 0, 15, -5, 1, 0, 5, 0, -11, 0, 11, 0, -11, 0, 55, -11, 1

Offset: 0

Wolfdieter Lang, Oct 23 2011

The denominator triangle is found under A196839.
This is the row reversed triangle A053382.
From Wolfdieter Lang, Oct 25 2011: (Start)
This is the Sheffer triangle (z/(exp(z)-1),z), meaning that the column e.g.f.'s are as given below in the formula section. In Roman's book `The Umbral Calculus`, Ch. 2, 5., p. 26ff this is called Appell for (exp(t)-1)/t (see A048854 for the reference).
The e.g.f. for the a- and z-sequence for this Sheffer triangle is 1 and (x-exp(x)+1)/x^2, respectively. See the link under A006232 for the definition. The z-sequence is z(n) = -1/(2*A000217(n+1)). This leads to the recurrence relations given below.
The e.g.f. for the row sums is x/(1-exp(-x)), leading to the rational sequence A164555(n)/A027664(n). The e.g.f. of the alternating row sums is
  x/(exp(x)*(exp(x)-1)), leading to the rational sequence
  (-1)^n*A164558(n)/A027664(n).
(End)

			The triangle starts with
n\m 0  1  2  3  4  5  6  7  8 ...
0:  1
1: -1  1
2:  1 -1  1
3:  0  1 -3  1
4: -1  0  1 -2  1
5:  0 -1  0  5 -5  1
6:  1  0 -1  0  5 -3  1
7:  0  1  0 -7  0  7 -7  1
8: -1  0  2  0 -7  0 14 -4  1
...
The rational triangle a(n,m)/A196839(n,m) starts with:
n\m   0     1     2    3    4    5     6    7   8 ...
0:    1
1:  -1/2    1
2:   1/6   -1     1
3:    0    1/2  -3/2   1
4:  -1/30   0     1   -2    1
5:    0   -1/6    0   5/3 -5/2   1
6:   1/42   0   -1/2   0   5/2  -3     1
7:    0    1/6    0  -7/6   0   7/2  -7/2   1
8:  -1/30   0    2/3   0  -7/3   0   14/3  -4   1
...
E.g., Bernoulli(2,x) = (1/6)*x^0 - 1*x^1 + 1*x^2.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1991 (Seventh printing).Second ed. 1994.

Naho Kawasaki and Yasuo Ohno, The triangle algorithm for Bernoulli polynomials, Integers, vol. 23 (2023). (See figure 4.)
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.
D. H. Lehmer, A new approach to Bernoulli polynomials, The American mathematical monthly 95.10 (1988): 905-911.

Three versions of coefficients of Bernoulli polynomials: A053382/A053383; for reflected version see A196838/A196839; see also A048998 and A048999.

# Without using Maple's Bernoulli polynomials (Kawasaki and Ohno call it
# the 'triangle algorithm for B(n, x)'):
b := proc(n, m, x) option remember; if n = 0 then 1/(m + 1) else
normal((m + 1)*b(n-1, m + 1, x) - (m + 1 - x)*b(n-1, m, x)) fi end:
Bcoeffs := n -> local k; [seq(coeff(b(n, 0, x), x, k), k = 0..n)]:
for n from 0 to 8 do numer(Bcoeffs(n)) od; # Peter Luschny, Jun 16 2023

row[n_] := CoefficientList[BernoulliB[n, x], x] // Numerator;
Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 15 2018 *)

row(n) = apply(x->numerator(x), Vecrev(bernpol(n)));
tabl(nn) = for (n=0, nn, print(row(n))); \\ Michel Marcus, Jun 15 2018

T(n,m) = numerator([x^m]Bernoulli(n,x)), n>=0, m=0..n.
E.g.f. of Bernoulli(n,x): z*exp(x*z)/(exp(z)-1).
See the Graham et al. reference, eq. (7.80), p. 354.
From Wolfdieter Lang, Oct 25 2011: (Start)
The e.g.f. for column no. m>=0 of the rational triangle B(n,m):=a(n,m)/A096839(n,m) is x^(m+1)/(m!*(exp(x)-1)).
(see the Sheffer-Appell comment above).
The Sheffer a-sequence, given as comment above, leads to the recurrence r(n,m)=(n/m)*r(n-1,m-1), n>=1, m>=1. E.g., -1/6 = B(5,1) = (5/1)*B(4,0)= -5/30 = -1/6.
The Sheffer z-sequence, given as comment above, leads to the recurrence
B(n,0) = n*sum(z(j)*B(n-1,j),j=0..n-1), n>=1. B(0,0)=1.
E.g., -1/30 = B(4,0) = 4*((-1/2)*0 + (-1/6)*(1/2) + (-1/12)*(-3/2) + (-1/20)*1) = -1/30.
(End)
T(n,m) = numerator(binomial(n,m)*Bernoulli(n-m)). - Fabián Pereyra, Mar 04 2020

A215149 a(n) = n * (1 + 2^(n-1)).

Original entry on oeis.org

0, 2, 6, 15, 36, 85, 198, 455, 1032, 2313, 5130, 11275, 24588, 53261, 114702, 245775, 524304, 1114129, 2359314, 4980755, 10485780, 22020117, 46137366, 96469015, 201326616, 419430425, 872415258, 1811939355, 3758096412, 7784628253, 16106127390, 33285996575, 68719476768, 141733920801, 292057776162

Offset: 0

Paul Curtz, Aug 04 2012

Related to Bernoulli numbers.

Essentially the same as A135854.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).

Cf. A001477, A001787, A001792, A027641, A027642, A094373, A132045, A135854, A157809, A164555, A164558.

[n*(1 + 2^(n-1)): n in [0..40]]; // G. C. Greubel, Apr 19 2018

Table[n(1+2^(n-1)),{n,0,40}] (* or *) LinearRecurrence[{6,-13,12,-4},{0,2,6,15}, 40] (* Harvey P. Dale, Oct 18 2013 *)

a(n) = n*(1+2^(n-1)) \\ Michel Marcus, Mar 10 2013

def A215149(n): return n*(pow(2,n)+2)//2
print([A215149(n) for n in range(41)]) # G. C. Greubel, Jan 18 2025

a(n) = (A157809(n) - A164555(n)) / A027642(n).
a(n) = n (the nonnegative integers A001477(n)) + n*2^(n-1) (their binomial transform A001787(n)).
a(n+1) - a(n) = 2,4,9,21,... = A001792(n) + 1.
a(n+1) - 2*a(n) = 2 before A132045(n+1).
a(n) is the binomial transform of b(n) = 0,2,2,3,4,5,... = A001477(n) with 2 instead of 1. b(n) = (A164558(n) - A027641(n))/A027642(n)?
G.f.: x*(2-6*x+5*x^2) / ( (1-x)^2*(1-2*x)^2 ). - R. J. Mathar, Aug 06 2012
E.g.f.: x*exp(x)*(1 + exp(x)). - G. C. Greubel, Jan 18 2025
a(n) = n * A094373(n). - Alois P. Heinz, Jan 18 2025

A254667 The nonnegative numbers with 2 instead of 1.

Original entry on oeis.org

0, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62

Offset: 0

Paul Curtz, Feb 04 2015

An autosequence of the first kind is a sequence whose main diagonal in the difference table is A000004 = 0's.
This is the case for a(n).
Difference table of a(n):
   0,  2,  2,  3,  4,  5, ...
   2,  0,  1,  1,  1,  1, ...
  -2,  1,  0,  0,  0,  0, ...
   3, -1,  0,  0,  0,  0, ...
  -4,  1,  0,  0,  0,  0, ...
   5, -1,  0,  0,  0,  0, ...
  etc.
The inverse binomial transform of a(n) is (-1)^(n+1)*a(n).
0 followed by A000012(n) is not in the OEIS. See A054977.
What is the meaning of a(n)?
Among many others, A015441 is an autosequence of the first kind.
General form for such autosequence.
Starting from the first upper diagonal s0, s1, s2, s3, s4, ...,
the autosequence is
0, s0, s0, s0 + s1, s0 + 2*s1, s0 + 3*s1 + s2, s0 + 4*s1 + 3*s2, ... .
After 0, the corresponding coefficients are A011973(n).

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

G. C. Greubel, Table of n, a(n) for n = 0..2500
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000004, A001477, A011973, A015441, A027641, A027642, A054977, A063524, A164555, A164558, A233583.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x-2*x^2+x^3)/(1-x)^2)); // G. C. Greubel, Aug 03 2018

CoefficientList[Series[(2*x-2*x^2+x^3)/(1-x)^2, {x, 0, 60}], x] (* G. C. Greubel, Aug 03 2018 *)
a[ n_] := n + Boole[n == 1]; (* Michael Somos, Aug 19 2018 *)
Join[{0,2},Range[2,70]] (* Harvey P. Dale, Oct 10 2024 *)

{a(n) = n + (n==1)}; /* Michael Somos, Feb 09 2015 */

a(n) = (A164558(n) + (-1)^(n+1)*A164555(n))/A027642(n).
a(n) = A063524(n) + A001477(n). - David A. Corneth, Aug 03 2018
G.f.: (2*x - 2*x^2 + x^3) / (1 - x)^2. - Michael Somos, Feb 09 2015

A157809 Numerator of Bernoulli(n,3).

Original entry on oeis.org

1, 5, 37, 15, 1079, 85, 8317, 455, 30959, 2313, 338585, 11275, 67124549, 53261, 688219, 245775, 267391423, 1114129, 1882776439, 4980755, 3460132789, 22020117, 6367811021, 96469015, 549385297589, 419430425, 5243044651, 1811939355, 3245794417411, 7784628253

Offset: 0

N. J. A. Sloane, Nov 10 2009

From Paul Curtz, Feb 18 2015 (Start)
The fractions 1, 5/2, 37/6, 15, 1079/30, 85, 8317/42, 455, 30959/30 etc are the binomial transform of the sequence of fractions Bernoulli(n,2) = 1, 3/2, 13/6, 3, 119/30, 5, 253/42 specified in A164558.
Their table of repeated differences starts
1,         5/2,   37/6,     15, 1079/30, ...
3/2,      11/3,   53/6, 629/30, ...
13/6,     31/6, 182/15, ...
3,      209/30, ...
119/30, ...
etc.
The sums of the antidiagonals in this table of differences are n*2^(n-1)
1                       =  1
3/2  +  5/2             =  4
13/6 + 11/3 + 37/6      = 12
3    + 31/6 + 53/6 + 15 = 32
etc, see A001787.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..250

For denominators see A027642.

seq(numer(bernoulli(n,3)),n=0..50); # Robert Israel, Jul 03 2016

Table[Numerator[BernoulliB[n, 3]], {n, 0, 50}] (* Vincenzo Librandi, Mar 16 2014 *)

A164869 n*A027642(n).

Original entry on oeis.org

0, 2, 12, 3, 120, 5, 252, 7, 240, 9, 660, 11, 32760, 13, 84, 15, 8160, 17, 14364, 19, 6600, 21, 3036, 23, 65520, 25, 156, 27, 24360, 29, 429660, 31, 16320, 33, 204, 35, 69090840, 37, 228, 39, 541200, 41, 75852, 43, 30360, 45, 12972, 47, 2227680, 49, 3300, 51, 82680

Offset: 0

Paul Curtz, Aug 29 2009

nonn

Cf. A164877 (bisection).

a(n) = A164558(n) - A164555(n).

a(n) = A027642(n) * A001477(n).

Extended by R. J. Mathar, Sep 03 2009

A235774 Let b(k) = A164555(k)/A027642(k), the sequence of "original" Bernoulli numbers with -1 instead of A164555(0)=1; then a(n) = numerator of the n-th term of the binomial transform of the b(k) sequence.

Original entry on oeis.org

-1, -1, 1, 1, 59, 3, 169, 5, 179, 7, 533, 9, 26609, 11, 79, 13, 3523, 15, 56635, 17, -168671, 19, 857273, 21, -236304031, 23, 8553247, 25, -23749438409, 27, 8615841677021, 29, -7709321025917, 31, 2577687858559, 33, -26315271552988224913

Offset: 0

Paul Curtz, Jan 15 2014

(a(n)/A027642(n)) = -1, -1/2, 1/6, 1, 59/30, 3, 169/42, 5, 179/30, 7, 533/66, 9,.. .
Difference table for a(n)/A027642(n):
-1, -1/2,   1/6,      1,  59/30,     3, 169/42, ...
1/2, 2/3,   5/6,  29/30,  31/30, 43/42,  41/42, ... = A165161(n)/A051717(n+1)
1/6, 1/6,  2/15,   1/15, -1/105, -1/21, -1/105, ... not in the OEIS
0, -1/30, -1/15, -8/105, -4/105, 4/105,  8/105, ... etc.
Compare with the array in A190339.

Cf. A164558/A027642, A165161(n)/A051717(n+1).

b[0] = -1; b[1] = 1/2; b[n_] := BernoulliB[n]; a[n_] := Sum[Binomial[n, k]*b[k], {k, 0, n}] // Numerator; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Jan 30 2014 *)

(a(n+1) - a(n))/A027642(n) = A165161(n)/A051717(n+1).
(A164558(n) - a(n))/A027642(n) = 2's = A007395.
(a(n) - A164555(n))/A027642(n) = n - 2 = A023444(n).

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.

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 *)

cp's OEIS Frontend

A219976 Denominators of the Inverse bi-binomial transform of A164558(n)/A027642(n) read downwards antidiagonals.

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

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A196838 Numerators of coefficients of Bernoulli polynomials with rising powers of the variable.

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A215149 a(n) = n * (1 + 2^(n-1)).

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A254667 The nonnegative numbers with 2 instead of 1.

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A157809 Numerator of Bernoulli(n,3).

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A164869 n*A027642(n).

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A235774 Let b(k) = A164555(k)/A027642(k), the sequence of "original" Bernoulli numbers with -1 instead of A164555(0)=1; then a(n) = numerator of the n-th term of the binomial transform of the b(k) sequence.

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