A198631 -id:A198631 - cp's OEIS frontend

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

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).

A237425 Denominators of A164555(n)/A027642(n) + A198631(n)/A006519(n+1).

Original entry on oeis.org

1, 1, 6, 4, 30, 2, 42, 8, 30, 2, 66, 4, 2730, 2, 6, 16, 510, 2, 798, 4, 330, 2, 138, 8, 2730, 2, 6, 4, 870, 2, 14322, 32, 510, 2, 6, 4, 1919190, 2, 6, 8, 13530, 2, 1806, 4, 690, 2, 282, 16, 46410, 2, 66, 4, 1590, 2, 798, 8

Offset: 0

Paul Curtz, Feb 07 2014

nonn

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. There are two possibilities. For the first kind, the main diagonal is 0's=A000004, the first two following diagonals being the same (generally not A000004). Integers example: A000045(n).
For the second kind, the main diagonal is the double of the following diagonal. Example: the companion to A000045(n) is A000032(n)=2, 1, 3, ... .
A000032(n)/2 is also a possibility. Here a(n) is the denominator of the sum of two autosequences of second kind involving (fractional) Euler and Bernoulli numbers. The corresponding fractional sequence is also an autosequence of the second kind: 2, 1, 1/6, -1/4, -1/30, 1/2, 1/42, -17/8, -1/30, 31/2, 5/66, -691/4, -691/2730,... . It could be divided by 2.

Cf. 1/n in A003506, A194767, A218289, A168516/A168426, A224270/A046161.

a[n_] := BernoulliB[n] + EulerE[n, 1]/2^IntegerExponent[n, 2]; a[0] = 2; a[1] = 1; Table[a[n] // Denominator, {n, 0, 55}] (* Jean-François Alcover, Feb 11 2014 *)

a(2n) = A002445(n). a(2n+2) = A171977(n+2).

A238398 Numerators of inverse binomial transform of A198631(n)/A006519(n+1) with -1 instead of A198631(1)=1.

Original entry on oeis.org

1, -3, 2, -11, 4, -11, 6, -39, 8, -49, 10, 647, 12, -5487, 14, 929329, 16, -3202325, 18, 221930505, 20, -4722116563, 22, 968383680643, 24, -14717667114201, 26, 2093660879252563, 28, -86125672563201239, 30, 129848163681107300961, 32

Offset: 0

Paul Curtz, Feb 26 2014

sign

From modified fractional Euler numbers.
Inverse binomial transform:
1, -3/2, 2, -11/4, 4, -11/2, 6, -39/8, 8, -49/2, 10, 647/4, 12, -5487/2,... = a(n)/b(n).  b(2n) = A004277(n).
Difference table of c(n) = 1, -1/2, 0, -1/4,... :
1,    -1/2,    0,  -1/4,     0,    1/2,      0,...
-3/2,  1/2, -1/4,   1/4,   1/2,   -1/2,  -17/8,...
2,    -3/4,  1/2,   1/4,    -1,  -13/8,   17/4,...
-11/4, 5/4, -1/4,  -5/4,  -5/8,   47/8,   73/8,...
4,    -3/2,   -1,   5/8,  13/2,   13/4, -107/2,...
-11/2, 1/2, 13/8,  47/8, -13/4, -227/4, -227/8,
6,     9/8, 17/4, -73/8, -107/2, 227/8, 2957/4,...
etc.
c(n) + a(n)/b(n) = 2, -2, 2, -3, 4, -5, 6, -7, 8, -9,... = A233583(n+1) signed. (a(n) discovered in 2013)

Cf. A235774.

max = 40;(* b = A198631 *) b[0] = 1; b[1] = -1; b[n_] := Numerator[EulerE[n, 1]/(2^n-1)]; bb = Table[b[n]/2^IntegerExponent[n+1, 2], {n, 0, max}]; a[n_] := Differences[bb, n] // First // Numerator ; Table[a[n], {n, 0, max}]

A238800 Unreduced numerators in triangle that leads to the Euler numbers A198631(n)/A006519(n+1).

Original entry on oeis.org

1, 1, 1, -2, 1, -3, 1, -4, 2, 1, -5, 5, 1, -6, 9, -10, 1, -7, 14, -35, 1, -8, 20, -80, 26, 1, -9, 27, -150, 117, 1, -10, 35, -250, 325, -454, 1, -11, 44, -385, 715, -2497, 1, -12, 54, -560, 1365, -8172, 5914, 1, -13

Offset: 0

Paul Curtz, Mar 05 2014

We use the array ASPEC mentioned in A191302:
2, 1,  1,  1,  1,  1,   1,   1,...
2, 3,  4,  5,  6,  7,   8,   9,...
2, 5,  9, 14, 20, 27,  35,  44,...
2, 7, 16, 30, 50, 77, 112, 156,...
with the first upper diagonal of the difference table of the autosequence A198631(n)/A006519(n+1), i.e., 1/2, -1/4, 1/4, -5/8, 13/4, -227/8, 2957/8,...
written by columns:
1/2
1/2,
1/2, -1/4,
1/2, -1/4,
1/2, -1/4, 1/4,
1/2, -1/4, 1/4,
1/2, -1/4, 1/4, -5/8,
1/2, -1/4, 1/4, -5/8,
etc.
Hence, by multiplication of this double triangle by ASPEC, the beginning of the double triangle ESPEC is obtained:
E(0) =     1 =   1
E(1) =   1/2 = 1/2
E(2) =     0 = 1/2 -2/4
E(3) =  -1/4 = 1/2 -3/4
E(4) =     0 = 1/2 -4/4  +2/4
E(5) =   1/2 = 1/2 -5/4  +5/4
E(6) =     0 = 1/2 -6/4  +9/4 -10/8
E(7) = -17/8 = 1/2 -7/4 +14/4 -35/8
E(8) =     0 = 1/2 -8/4 +20/4 -80/8 +26/4.
The terms of the sequence are the reduced numerators. Like A192456(n) for Bernoulli numbers A164555(n)/A027642(n).

			a(n) by triangle
1,
1,
1, -2,
1, -3,
1, -4,  2,
1, -5,  5,
1, -6,  9, -10,
1, -7, 14, -35,
1, -8, 20, -80, 26,
etc.

A240980 Numerators of f(n) with 2*f(n+1) = f(n) + A198631(n)/A006519(n+1), f(0)=0.

Original entry on oeis.org

0, 1, 1, 1, 0, 0, 1, 1, -1, -1, 15, 15, -169, -169, 10753, 10753, -28713, -28713, 1586789, 1586789, -27542974, -13771487, 4694573547, 4694573547, -60230569205, -60230569205, 7328718272473, 7328718272473, -1043166080490099, -1043166080490099, 343459524172314625, 343459524172314625

Offset: 0

Paul Curtz, Aug 06 2014

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. (Examples: 1) A000045(n) is of the first kind. 2) 1/(n+1) is of the second kind).
f(n), companion to A198631(n)/A006519(n+1), is an autosequence of the first kind.
The difference table of f(n) is:
     0,  1/2,  1/2,  1/4,    0,    0, ...
   1/2,    0, -1/4, -1/4,    0,  1/4, ...
  -1/2, -1/4,    0,  1/4,  1/4, -3/8, ...
   1/4,  1/4,  1/4,    0, -5/8, -5/8, ...
etc.
The main diagonal is 0's=A000004. The first two upper diagonal are equal.
a(n) are the numerators of f(n).
f(n) is the first sequence of the family of alternated autosequences of the first and of the second kind
   0,  1/2,  1/2,  1/4,  0,    0, ...
   1,  1/2,    0, -1/4,  0,  1/2, ... = A198631(n)/A006519(n+1),
   0, -1/2, -1/2,  1/4,  1, -1/2, ...
  -1, -1/2,    1,  7/4, -2,   -8, ...
etc.
Like A164555(n)/A027642(n), A198631(n)/A006519(n+1) is an autosequence which has its main diagonal equal to the first upper diagonal multiplied by 2. See A190339(n).
The first column is 0 followed by A122045(n).
For the numerators of the second column see A241209(n).

			2*f(1) = 0 + 1, f(1) = 1/2;
2*f(2) = 1/2 + 1/2, f(2) = 1/2;
2*f(3) = 1/2 + 0, f(3) = 1/4.

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

Cf. A122045, A190339, A233808.

Clear[f]; f[0] = 0; f[1] = 1/2; f[n_] := f[n] = (1/2)*(EulerE[n-1, 1]/2^IntegerExponent[n-1, 2] + f[n-1]); Table[f[n] // Numerator, {n, 0, 31}] (* Jean-François Alcover, Aug 06 2014 *)

A281825 Numerators of the binomial transform of A198631(n)/A006519(n+1) with A198631(1) = -1 instead of 1.

Original entry on oeis.org

1, 1, 0, -3, -2, -7, -4, -23, -6, -45, -8, 655, -10, -5483, -12, 929361, -14, -3202321, -16, 221930513, -18, -4722116559, -20, 968383680659, -22, -14717667114197, -24, 2093660879252571, -26, -86125672563201235, -28, 129848163681107301025, -30

Offset: 0

Paul Curtz, Jan 31 2017

sign

What is the correct name for the rational sequence c(n) = 1, -1/2, 0, -1/4, 0, 1/2, 0, -17/8, 0, 31/2, 0, ... (a variant of the second fractional Euler numbers)?

Its binomial transform is f(n) = 1, 1/2, 0, -3/4, -2, -7/2, -4, -23/8, -6, -45/2, -8, 655/4, -10, ... = a(n)/A006519(n+1).

Cf. A001477, A006519, A198631, A199969, A209308, A238398.

A198631 := proc(n)
    1/(1+exp(-x)) ;
    coeftayl(%,x=0,n) ;
    numer(%*n!) ;
end proc:
A006519 := proc(n)
    2^padic[ordp](n,2) ;
end proc:
L := [seq( A198631(n)/A006519(n+1),n=0..40)] ;
subsop(2=-1/2,L) ;
b := BINOMIAL(%) ;
for i from 1 to nops(b) do
    printf("%d,",numer(b[i])) ;
end do: # R. J. Mathar, Feb 21 2017

By definition f(0) - c(0), f(1) + c(1), f(2) - c(2), f(3) + c(3), ... is an autosequence of the first kind, here 1 - 1 = 0, 1/2 - 1/2 = 0, 0 - 0 = 0, -3/4 - 1/4 = -1, -2 - 0 = -2, -7/2 + 1/2 = -3, ... i.e., t(n) = 0, 0, followed by -A001477(n), not in the OEIS, but the corresponding autosequence of the second kind is: A199969 = 0, 0, -2, -3, -4, ... Hence f(n) from c(n) and t(n).

A060096 Numerator of coefficients of Euler polynomials (rising powers).

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 1, 0, -3, 1, 0, 1, 0, -2, 1, -1, 0, 5, 0, -5, 1, 0, -3, 0, 5, 0, -3, 1, 17, 0, -21, 0, 35, 0, -7, 1, 0, 17, 0, -28, 0, 14, 0, -4, 1, -31, 0, 153, 0, -63, 0, 21, 0, -9, 1, 0, -155, 0, 255, 0, -126, 0, 30, 0, -5, 1, 691, 0, -1705, 0, 2805, 0, -231, 0, 165, 0, -11, 1, 0, 2073, 0, -3410, 0, 1683, 0, -396

Offset: 0

Wolfdieter Lang, Mar 29 2001

From S. Roman, The Umbral Calculus (see the reference in A048854), p. 101, (4.2.10) (corrected): E(n,x)= sum(sum(binomial(n,m)*((-1/2)^j)*j!*S2(n-m,j),j=0..k)*x^m,m=0..n), with S2(n,m)=A008277(n,m) and S2(n,0)=1 if n=0 else 0 (Stirling2).
From Wolfdieter Lang, Oct 31 2011: (Start)
This is the Sheffer triangle (2/(exp(x)+1),x) (which would be called in the above mentioned S. Roman reference Appell for (exp(t)+1)/2) (see p. 27).
  The e.g.f. for the row sums is 2/(1+exp(-x)). The row sums look like A198631(n)/A006519(n+1), n>=0.
  The e.g.f. for the alternating row sums is 2/(exp(x)*(exp(x)+1)). These sums look like (-1)^n*A143074(n)/ A006519(n+1).
  The e.g.f. for the a-sequence of this Sheffer array is 1. The z-sequence has e.g.f. (1-exp(x))/(2*x). This z-sequence is -1/(2*A000027(n))=-1/(2*(n+1)) (see the link under A006232 for the definition of a- and z-sequences). This leads to the recurrences given below.
The alternating power sums for the first n positive integers are given by sum((-1)^(n-j)*j^k,j=1..n) = (E(k, x=n+1)+(-1)^n*E(k, x=0))/2, k>=1, n>=1,with the row polynomials E(n, x)(see the Abramowitz-Stegun reference, p. 804, 23.1.4, and an addendum in the W. Lang link under A196837).
(End)

			n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1:  -1    1
2:   0   -1    1
3:   1    0   -3    1
4:   0    1    0   -2    1
5:  -1    0    5    0   -5    1
6:   0   -3    0    5    0   -3    1
7:  17    0  -21    0   35    0   -7    1
8:   0   17    0  -28    0   14    0   -4  1
...
The rational triangle a(n,m)/A060097(n,m) starts
n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1: -1/2   1
2:   0   -1    1
3:  1/4   0  -3/2   1
4:   0    1    0   -2    1
5: -1/2   0   5/2   0  -5/2   1
6:   0   -3    0    5    0   -3    1
7: 17/8   0 -21/2   0  35/4   0  -7/2   1
8:   0   17    0  -28    0   14    0   -4  1
...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 809.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 20, equations 20:4:1 - 20:4:8 at pages 177-178.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A060097.

A060096 := proc(n,m) coeff(euler(n,x),x,m) ; numer(%) ;end proc:
seq(seq(A060096(n,m),m=0..n),n=0..12) ; # R. J. Mathar, Dec 21 2010

Numerator[Flatten[Table[CoefficientList[EulerE[n, x], x], {n, 0, 12}]]] (* Jean-François Alcover, Apr 29 2011 *)

E(n, x)= sum((a(n, m)/b(n, m))*x^m, m=0..n), denominators b(n, m)= A060097(n, m).
From Wolfdieter Lang, Oct 31 2011: (Start)
E.g.f. for E(n, x) is 2*exp(x*z)/(exp(z)+1).
E.g.f. of column no. m, m>=0, is 2*x^{m+1}/(m!*(exp(x)+1)).
Recurrences for E(n,m):=a(n,m)/A060097(n,m) from the Sheffer a-and z-sequence:
E(n,m)=(n/m)*E(n-1,m-1), n>=1,m>=1.
E(n,0)=-n*sum(E(n-1,j)/(2*(j+1)),j=0..n-1), n>=1, E(0,0)=1.
(see the Sheffer comments above).
(End)
E(n,m) = binomial(n,m)*sum(((-1)^j)*j!*S2(n-m,j)/2^j ,j=0..n-m), 0<=m<=n, with S2 given by A008277. From S. Roman, The umbral calculus, reference under A048854, eq. (4.2.10), p. 101, with a=1, and a misprint corrected: replace 1/k! by binomial(n,k) (also in the two preceding formulas). - Wolfdieter Lang, Nov 03 2011
The first (m=0) column of the rational triangle is conjectured to be E(n,0) = ((-1)^n)*A198631(n) / A006519(n+1). See also the first column shown in A209308 (different signs). - Wolfdieter Lang, Jun 15 2015

Table rewritten by Wolfdieter Lang, Oct 31 2011

A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 4, 4, 8, 8, 1, 4, 8, 4, 16, 2, 2, 1, 8, 32, 32, 1, 2, 4, 4, 16, 32, 64, 8, 8, 16, 16, 64, 64, 128, 128, 1, 8, 16, 8, 32, 64, 128, 32, 256, 2, 2, 8, 16, 64, 64, 128, 64, 512, 512, 1, 2, 4, 8, 32, 64, 128, 16, 128, 512, 1024

Offset: 0

Paul Curtz, Jan 18 2013

1/2^n and successive rows are
1,       1/2,   1/4,   1/8,  1/16,  1/32,   1/64, 1/128, 1/256,...
1/2,     1/2,   3/8,   1/4,  5/32,  3/32,  7/128,  1/32,...       = A000265/A075101, the Oresme numbers n/2^n. Paul Curtz, Jan 18 2013 and May 11 2016
0,       1/4,   3/8,   3/8,  5/16, 15/64, 21/128,...              = (0 before A069834)/new,
-1/4,   -1/4,     0,   1/4, 25/64, 27/64,...
0,      -1/2,  -3/4, -9/16, -5/32,...
1/2,     1/2, -9/16, -13/8,...
0,      17/8, 51/16,...
-17/8, -17/8,...
0
The first column is A198631/(A006519?), essentially the fractional Euler numbers 1, -1/2, 0, 1/4, 0,...  in A060096.
Numerators b(n): 1, 1, 1, 0, 1, 1, -1, 1, 3, 1, ... .
Coll(n+1) - 2*Coll(n) = -1/2, -5/8, -1/2, -11/32, -7/32, -17/128, -5/64, -23/512, ... = -A075677/new, from Collatz problem.
There are three different Bernoulli numbers:
The first Bernoulli numbers are  1, -1/2, 1/6, 0,... = A027641(n)/A027642(n).
The second Bernoulli numbers are 1,  1/2, 1/6, 0,... = A164555(n)/A027642(n). These are the binomial transform of the first one.
The third Bernoulli numbers are  1,   0,  1/6, 0,... = A176327(n)/A027642(n), the half sum. Via A177427(n) and A191567(n), they yield the Balmer series A061037/A061038.
There are three different fractional Euler numbers:
1) The first are  1, -1/2, 0, 1/4, 0, -1/2,... in A060096(n).
Also Akiyama-Tanigawa algorithm for ( 1, 3/2, 7/4, 15/8, 31/16, 63/32,... = A000225(n+1)/A000079(n) ).
2) The second are 1, 1/2, 0, -1/4, 0,  1/2,... , mentioned by Wolfdieter Lang in A198631(n).
3) The third are  0, 1/2, 0, -1/4, 0,  1/2,... , half difference of 2) and 1).
Also Akiyama-Tanigawa algorithm for ( 0, -1/2, -3/4, -7/8, -15/16, -31/32,... =  A000225(n)/A000079(n) ). See A097110(n).

			Triangle begins:
  1,
  2, 2,
  1, 2,  4,
  4, 4,  8,  8,
  1, 4,  8,  4, 16,
  2, 2,  1,  8, 32, 32,
  1, 2,  4,  4, 16, 32,  64,
  8, 8, 16, 16, 64, 64, 128, 128,
  ...

G. C. Greubel, Table of n, a(n) for n = 0..5049
A. F. Horadam, Oresme Numbers, Fibonacci Quarterly, 12, #3, 1974, pp. 267-271.

Cf. Second Bernoulli numbers A164555(n)/A027642(n) via Akiyama-Tanigawa algorithm for 1/(n+1), A272263.

max = 10; t[0, k_] := 1/2^k; t[n_, k_] := t[n, k] = (k + 1)*(t[n - 1, k] - t[n - 1, k + 1]); denoms = Table[t[n, k] // Denominator, {n, 0, max}, {k, 0, max - n}]; Table[denoms[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 05 2013 *)

A089171 Numerators of series coefficients of 1/(1 + cosh(sqrt(x))).

Original entry on oeis.org

1, -1, 1, -17, 31, -691, 5461, -929569, 3202291, -221930581, 4722116521, -56963745931, 14717667114151, -2093660879252671, 86125672563201181, -129848163681107301953, 868320396104950823611, -209390615747646519456961, 14129659550745551130667441

Offset: 0

Wouter Meeussen, Dec 07 2003

Unsigned version is equal to A002425 up to n=11, but differs beyond that point.

Unsigned version: numerators of series coefficients of 1/(1 + cos(sqrt(x))); see Mathematica. - Clark Kimberling, Dec 06 2016

N. J. A. Sloane, Table of n, a(n) for n = 0..299

Cf. A002425, A050970, A002430, A046990.
Cf. A000182, A198631, A279009, A279010.
Cf. A276592, A279370.

with(numtheory): c := n->(2^(2*n)-1)*bernoulli(2*n)/(2*n)!; seq(numer(c(n)),n=1..20); # C. Ronaldo

Numerator[CoefficientList[Series[1/(1+Cosh[Sqrt[x]]), {x, 0, 24}], x]]
Numerator[CoefficientList[Series[1/(1+Cos[Sqrt[x]]), {x, 0, 30}], x]]
(* unsigned version, Clark Kimberling, Dec 06 2016 *)

a(n) = numerator(c(n+1)) where c(n)=(2^(2*n)-1)*B(2*n)/(2*n)!, B(k) denotes the k-th Bernoulli number. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 19 2004
Numerators of expansion of cosec(x)-cot(x) = 1/2*x+1/4*x^3/3!+1/2*x^5/5!+17/8*x^7/7!+31/2*x^9/9!+... - Ralf Stephan, Dec 21 2004 (Comment was applied to wrong entry, corrected by Alessandro Musesti (musesti(AT)gmail.com), Nov 02 2007)
E.g.f.: 1/sin(x)-cot(x). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: x/G(0); G(k) = 4*k+2-x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: (1+x/(x-2*Q(0)))/2; Q(k) = 8*k+2+x/(1+(2*k+1)*(2*k+2)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: x/(x+Q(0)); Q(k) = x+(x^2)/((4*k+1)*(4*k+2)-(4*k+1)*(4*k+2)/(1+(4*k+3)*(4*k+4)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 22 2011
E.g.f.: T(0)/2, where T(k) = 1 - x^2/(x^2 - (4*k+2)*(4*k+6)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 12 2013
Aerated, these are the numerators of the Taylor series coefficients of 2 * tanh(x/2) (cf. A000182 and A198631). - Tom Copeland, Oct 19 2016

A255935 Triangle read by rows: a(n) = Pascal's triangle A007318(n) + A197870(n+1).

Original entry on oeis.org

0, 1, 2, 1, 2, 0, 1, 3, 3, 2, 1, 4, 6, 4, 0, 1, 5, 10, 10, 5, 2, 1, 6, 15, 20, 15, 6, 0, 1, 7, 21, 35, 35, 21, 7, 2, 1, 8, 28, 56, 70, 56, 28, 8, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 2, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0

Offset: 0

Paul Curtz, Mar 11 2015

Consider the difference table of a sequence with A000004(n)=0's as main diagonal. (Example: A000045(n).) We call this sequence an autosequence of the first kind.
Based on Pascal's triangle, a(n) =
0,                   T1
1, 2,
1, 2, 0,
1, 3, 3, 2,
etc.
transforms every sequence s(n) in an autosequence of the first kind via the multiplication by the triangle
s0,                  T2
s0, s1,
s0, s1, s2,
s0, s1, s2, s3,
etc.
Examples.
1) s(n) = A198631(n)/A006519(n+1), the second fractional Euler numbers (see A209308). This yields 0*1, 1*1+2*1/2=2, 1*1+2*1/2+0*0=2, 1*1+3*1/2++3*0+2*(-1/4)=2, ... .
The autosequence is 0 followed by 2's or 2*(0,1,1,1,1,1,1,1,... = b(n)).
b(n), the basic autosequence of the first kind, is not in the OEIS (see A140575 and A054977).
2) s(n) = A164555(n)/A027642(n), the second Bernoulli numbers, yields 0,2,2,3,4,5,6,7,... = A254667(n).
Row sums of T1: A062510(n) = 3*A001045(n).
Antidiagonal sums of T1: A111573(n).
With 0's instead of the spaces, every column, i.e.,
0, 1, 1, 1, 1,  1,  1,  1,  1,  1,   1, ...
0, 2, 2, 3, 4,  5,  6,  7,  8,  9,  10, ... = A001477(n) with 0 instead of 1 = A254667(n)
0, 0, 0, 3, 6, 10, 15, 21, 28, 36,  45, ... = A161680(n) with 0 instead of 1
0, 0, 0, 2, 4, 10, 20, 35, 56, 84, 120, ...
etc., is an autosequence of the first kind.
With T(0,0) = 1, it is (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 24 2015

			Triangle starts:
0;
1, 2;
1, 2, 0;
1, 3, 3, 2;
1, 4, 6, 4, 0;
1, 5, 10, 10, 5, 2;
1, 6, 15, 20, 15, 6, 0;
...

Cf. A001045, A001477, A010673, A011973, A054977, A062510, A007318, A197870, A006519, A198631, A140575, A209308, A164555, A027642, A111573, A161680, A254667.

a[n_, k_] := If[k == n, 2*Mod[n, 2], Binomial[n, k]]; Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 23 2015 *)

a(n) = Pascal's triangle A007318(n) with main diagonal A010673(n) (= period 2: repeat 0, 2) instead of 1's=A000012(n).
a(n) = reversal abs(A140575(n)).
a(n) = A007318(n) + A197870(n+1).
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 0, T(1,0) = 1, T(1,1) = 2, T(n,k) = 0 if k>n or if k<0 . - Philippe Deléham, May 24 2015
G.f.: (-1-2*x*y+x^2*y+x^2*y^2)/((x*y+1)*(x*y+x-1)) - 1. - R. J. Mathar, Aug 12 2015

cp's OEIS Frontend

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

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A237425 Denominators of A164555(n)/A027642(n) + A198631(n)/A006519(n+1).

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A238398 Numerators of inverse binomial transform of A198631(n)/A006519(n+1) with -1 instead of A198631(1)=1.

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A238800 Unreduced numerators in triangle that leads to the Euler numbers A198631(n)/A006519(n+1).

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A240980 Numerators of f(n) with 2*f(n+1) = f(n) + A198631(n)/A006519(n+1), f(0)=0.

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A281825 Numerators of the binomial transform of A198631(n)/A006519(n+1) with A198631(1) = -1 instead of 1.

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A060096 Numerator of coefficients of Euler polynomials (rising powers).

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A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

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A089171 Numerators of series coefficients of 1/(1 + cosh(sqrt(x))).

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