A110501 -id:A110501 - cp's OEIS frontend

A166353 Exponential Riordan array [1+x*tan(x/2),x].

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 0, 6, 0, 1, 0, 5, 0, 10, 0, 1, 3, 0, 15, 0, 15, 0, 1, 0, 21, 0, 35, 0, 21, 0, 1, 17, 0, 84, 0, 70, 0, 28, 0, 1, 0, 153, 0, 252, 0, 126, 0, 36, 0, 1, 155, 0, 765, 0, 630, 0, 210, 0, 45, 0, 1

Offset: 0

Paul Barry, Oct 12 2009

First column is aerated Genocchi number variant with e.g.f. 1+x*tan(x/2).

Row sums are A166354. Diagonal sums are A166355.

			Triangle begins
1,
0, 1,
1, 0, 1,
0, 3, 0, 1,
1, 0, 6, 0, 1,
0, 5, 0, 10, 0, 1,
3, 0, 15, 0, 15, 0, 1,
0, 21, 0, 35, 0, 21, 0, 1,
17, 0, 84, 0, 70, 0, 28, 0, 1,
0, 153, 0, 252, 0, 126, 0, 36, 0, 1,
155, 0, 765, 0, 630, 0, 210, 0, 45, 0, 1

Cf. A110501, A166354, A166355.

(* The function RiordanArray is defined in A256893. *)
RiordanArray[1 + # Tan[#/2]&, #&, 11, True] // Flatten (* Jean-François Alcover, Jul 19 2019 *)

T(n,k) = [k<=n]*G((n-k)/2)*C(n,k)*(1+(-1)^(n-k))/2 where

G(n)=0^n+2(-1)^n*(1-4^n)*sum{k=0..2n, sum{j=0..k, (-1)^j*C(k,j)*j^(2n)/(k+1)}}.

A166354 Row sums of exponential Riordan array [1+x*tan(x/2),x], A166353.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 34, 78, 200, 568, 1806, 6282, 24052, 99100, 443178, 2107966, 10775664, 58092112, 334087750, 2012990930, 12863046636, 85662585604, 602124105122, 4391793687974, 33676375206568, 266989039507576

Offset: 0

Paul Barry, Oct 12 2009

Binomial transform of aerated Genocchi number variant with e.g.f. 1+x*tan(x/2).

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A110501.

CoefficientList[Series[E^x*(1+x*Tan[x/2]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 02 2013 *)

E.g.f.: exp(x)*(1+x*tan(x/2)).
a(n)=sum{k=0..n, C(n,k)*G(k/2)(1+(-1)^k)/2} where
G(n)=0^n+2(-1)^n*(1-4^n)*sum{k=0..2n, sum{j=0..k, (-1)^j*C(k,j)*j^(2n)/(k+1)}}.
a(n) ~ n! * 2*(exp(Pi)+(-1)^n*exp(-Pi))/Pi^n. - Vaclav Kotesovec, Oct 02 2013

A221971 G.f.: A(x,y) = Sum_{n>=0} n! * x^ny^n Product_{k=1..n} (1 + kx) / (1 + kx*y + k^2x^2y).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 4, 1, 0, 0, 3, 11, 1, 0, 0, 0, 27, 26, 1, 0, 0, 0, 17, 148, 57, 1, 0, 0, 0, 0, 278, 646, 120, 1, 0, 0, 0, 0, 155, 2590, 2481, 247, 1, 0, 0, 0, 0, 0, 4073, 18304, 8805, 502, 1, 0, 0, 0, 0, 0, 2073, 58427, 109699, 29682, 1013, 1, 0, 0, 0

Offset: 0

Paul D. Hanna, Feb 01 2013

			Triangle begins:
1;
0, 1;
0, 1, 1;
0, 0, 4, 1;
0, 0, 3, 11, 1;
0, 0, 0, 27, 26, 1;
0, 0, 0, 17, 148, 57, 1;
0, 0, 0, 0, 278, 646, 120, 1;
0, 0, 0, 0, 155, 2590, 2481, 247, 1;
0, 0, 0, 0, 0, 4073, 18304, 8805, 502, 1;
0, 0, 0, 0, 0, 2073, 58427, 109699, 29682, 1013, 1;
0, 0, 0, 0, 0, 0, 80712, 614819, 590254, 96648, 2036, 1;
0, 0, 0, 0, 0, 0, 38227, 1665829, 5340996, 2948040, 307255, 4083, 1; ...

Paul D. Hanna, Rows n = 0..32, flattened.

Cf. A208237 (row sums), A110501 (central terms), A005439 (column sums), A136126.

{T(n,k)=polcoeff(polcoeff(sum(m=0, n,m!*x^m*y^m*prod(k=1, m, (1+k*x)/(1+k*x*y+k^2*x^2*y +x*O(x^n)))),n,x),k,y)}
for(n=0, 12, for(k=0, n, print1(T(n, k), ", ")); print(""))

Row sums equal A208237.
Central terms equal A110501, the Genocchi numbers of first kind (unsigned).
Columns sums equal A005439, the Genocchi numbers of second kind.

A223925 a(2n+1) = 2*n-1; a(2n)= 4^n.

Original entry on oeis.org

1, 4, 3, 16, 5, 64, 7, 256, 9, 1024, 11, 4096, 13, 16384, 15, 65536, 17, 262144, 19, 1048576, 21, 4194304, 23, 16777216, 25, 67108864, 27, 268435456, 29, 1073741824, 31

Offset: 1

Paul Curtz, Mar 29 2013

If A132050(n) has offset 1 (proposed),
A132049(n)/A132050(n) = 2, 4, 3, 16/5, 25/8, 192/61,... leads to Pi (Euler, 1735)
A132049(n)/a(n) = (2/1=2, 4/4=1, 3/3=1, 16/16=1, 25/5=5, 192/64=3,... ). The second bisection  1, 1, 3, 17, 155, begins like A110501.
Conjecture: a(2n) is always a divisor of A132049(2n).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-9,0,4).

Cf. A005408, A000302, A132050, A100102.

Table[ If[ OddQ[n], n, 4^(n/2)], {n, 1, 31}] (* Jean-François Alcover, Apr 02 2013 *)
CoefficientList[Series[(1 + 4 x - 3 x^2 - 8 x^3 - 4 x^4 + 4 x^5) / ((1 - x)^2 (1 + x)^2 (1 - 2 x) (1 + 2 x)), {x, 0, 35}], x] (* Vincenzo Librandi, Jul 20 2013 *)
LinearRecurrence[{0,6,0,-9,0,4},{1,4,3,16,5,64},40] (* Harvey P. Dale, Jul 30 2018 *)

G.f.: x*(1+4*x-3*x^2-8*x^3-4*x^4+4*x^5)/((1-x)^2*(1+x)^2*(1-2x)*(1+2x)). - Philippe Deléham, Apr 01 2013

a(n) = 6*a(n-2) -9*a(n-4) + 4*a(n-6) with a(1) = 1, a(2) = 4, a(3) = 3, a(4) = 16, a(5) = 5, a(6) = 64. - Philippe Deléham, Apr 01 2013

Conjecture about A132049(n)/a(n) modified by Jean-François Alcover, Apr 12 2013

A236935 The infinite Seidel matrix H read by antidiagonals upwards; H = (H(n,k): n,k >= 0).

Original entry on oeis.org

1, 0, -1, -1, -1, 0, 0, 1, 2, 2, 5, 5, 4, 2, 0, 0, -5, -10, -14, -16, -16, -61, -61, -56, -46, -32, -16, 0, 0, 61, 122, 178, 224, 256, 272, 272, 1385, 1385, 1324, 1202, 1024, 800, 544, 272, 0, 0, -1385, -2770, -4094, -5296, -6320, -7120, -7664, -7936, -7936, -50521, -50521, -49136, -46366, -42272, -36976, -30656, -23536, -15872, -7936, 0

Offset: 0

N. J. A. Sloane, Feb 17 2014

This is, in essence, a signed version of the triangle in A008280.

			Array begins:
     1   -1    0    2    0 -16   0 272 0 ...
     0   -1    2    2  -16 -16 272 272 ...
    -1    1    4  -14  -32 256 544 ...
     0    5  -10  -46  224 800 ...
     5   -5  -56  178 1024 ...
     0  -61  122 1202 ...
   -61   61 1324 ...
     0 1385 ...
  1385 ...
  ...

D. Dumont and G. Viennot, A combinatorial interpretation of the Seidel generation of Genocchi numbers, Preprint, Annotated scanned copy.
D. Dumont and G. Viennot, A combinatorial interpretation of the Seidel generation of Genocchi numbers, Annals of Discrete Mathematics, 6 (1980), 77-87.
Dominique Foata and Guo-Niu Han, Seidel Triangle Sequences and Bi-Entringer Numbers, November 20, 2013.
Dominique Foata and Guo-Niu Han, Seidel Triangle Sequences and Bi-Entringer Numbers, European Journal of Combinatorics, 42 (2014), 243-260.
L. Seidel, Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, Vol. 7 (1877), pp. 157-187; see Beilage 4 (p. 187).

Cf. A008280, A099023, A122045, A155585, A239005.

a(n) = 2^n*2^(n+1)*(subst(bernpol(n+1, x), x, 3/4) - subst(bernpol(n+1, x), x, 1/4))/(n+1) /* A122045 */
H(n,k) = sum(i=0, k, (-1)^i*binomial(k,i)*a(n+k-i)) /* Petros Hadjicostas, Feb 21 2021 */
/* Second PARI program (same a(n) for A122045 as above) */
H(n,k) = (-1)^(n+k)*sum(i=0, k, binomial(k,i)*a(n+i)) /* Petros Hadjicostas, Feb 21 2021 */

From Petros Hadjicostas, Feb 20 2021: (Start)
H(n,0) = A122045(n).
H(0,k) = (-1)^n*A155585(n).
H(n,k) = Sum_{i=0..n} binomial(n,i)*H(0,k+i).
H(n,k) = Sum_{i=0..k} (-1)^i*binomial(k,i)*H(n+k-i,0).
H(n,n) = A099023(n).
Bivariate e.g.f.: Sum_{n,k>=0} H(n,k)*(x^n/n!)*(y^k/k!) = 2*exp(x)/(1 + exp(2*(x+y))).
H(n,k) = (-1)^(n+k)*A239005(n+k,k), where the latter is a triangle.
H(n,k) = -A008280(n+k,k) if ((n+k) mod 4) == 1 or 2, and H(n,k) = A008280(n+k,k) if ((n+k) mod 4) == 3 or 0, provided A008280 is read as a triangle. (End)

More terms from Petros Hadjicostas, Feb 21 2021

A240485 a(n) = -Zeta(1-n)n(2^(n+1) - 4) - Zeta(-n)(n+1)(2^(n+2) - 2), for n = 0 the limit is understood.

Original entry on oeis.org

1, 3, 2, -1, -2, 3, 6, -17, -34, 155, 310, -2073, -4146, 38227, 76454, -929569, -1859138, 28820619, 57641238, -1109652905, -2219305810, 51943281731, 103886563462, -2905151042481, -5810302084962, 191329672483963, 382659344967926, -14655626154768697

Offset: 0

Paul Curtz, Apr 06 2014

sign

Let G(m, n) denote the difference table of a(n):
1,    3,   2,  -1,  -2,   3,   6, -17, -34,...
2,   -1,  -3,  -1,   5,   3, -23, -17,...
-3,  -2,   2,   6,  -2, -26,   6,...
1,    4,   4,  -8, -24,  32,...
3,    0, -12, -16,  56,...
-3, -12,  -4,  72,...
-9,   8,  76,...
17,  68,...
51,...
a(n) = G(0, n).
The main diagonal G(n, n) = 1, -1, 2, -8, 56, -608,... is essentially a signed version of A005439.
The first upper diagonal is the main diagonal multiplied by 3. G(n, n+1) = 3*G(n, n).
G(m, n) = G(m, n-1) + G(m+1,n-1).
Inverse binomial transform: after 1, 2, -3, A110501(n+1) is interleaved with 3*A110501(n+1), signed two by two. I. e. b(n) = 1, 2, -3, 1, 3, -3, -9, 17, 51,... . a(n+2) + b(n+2) = -1, 0, 1, 0, -3, 0, 17,... = A226158(n+2).
This is particular to the Genocchi numbers. If the first upper diagonal is proportional to the main diagonal (1, -1, 2, -8,...), the sequence and the inverse binomial transform are simply connected to the Genocchi numbers.

Cf. A001469, A005439, A110501, A226158, A230324.

A240485 := proc(n) if n = 0 then 1 elif n = 1 then 3 else
m := 2*iquo(n-1, 2) + 2; -2^irem(n-1, 2)*m*euler(m-1, 0) fi end:
seq(A240485(n), n=0..27); # Peter Luschny, Apr 09 2014

a[n_] := Which[n == 0, 1, n == 1, 3, True, m = 2*Quotient[n-1, 2]+2; -2^Mod[n-1, 2]*m*EulerE[m-1, 0]]; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Apr 09 2014, after Peter Luschny *)

def A240485(n):
    if n < 3: return [1,3,2][n]
    m = 2*((n+1)//2)
    b = 2*(1-2^m)*bernoulli(m)
    if is_even(n): b = 2*b
    return (-1)^ceil((n^2+1)/2)*b
[A240485(n) for n in (0..24)]  # Peter Luschny, Apr 08 2014

a(2*n+1) = a(2*n+2)/2 for n > 0.
-a(2*n+2)/2 = A226158(2*n+2) = A001469(n+1) = (2*n+2)*E(2*n+1, 0) where E(n, x) are the Euler polynomials.
a(n) = -2*A226158(n) - A226158(n+1).
E.g.f.: (2*exp(x)*(3*x+exp(x)*(2*x+1)+1))/(exp(x)+1)^2. - Peter Luschny, Apr 10 2014

A272481 E.g.f. A(x,y) = cos((x - xy)/2) / cos((x + xy)/2) represented as a triangle, read by rows, where row n lists of coefficients T(n,k) of x^(2n)y^k/n! in A(x,y), for k=0..2*n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 3, 1, 0, 0, 3, 15, 25, 15, 3, 0, 0, 17, 119, 329, 455, 329, 119, 17, 0, 0, 155, 1395, 5325, 11235, 14301, 11235, 5325, 1395, 155, 0, 0, 2073, 22803, 110605, 311355, 563013, 683067, 563013, 311355, 110605, 22803, 2073, 0, 0, 38227, 496951, 2918825, 10241231, 23904881, 39102063, 45956625, 39102063, 23904881, 10241231, 2918825, 496951, 38227, 0, 0, 929569, 13943535, 96075665, 403075855, 1150348017, 2362479119, 3600524785, 4136759055, 3600524785, 2362479119, 1150348017, 403075855, 96075665, 13943535, 929569, 0

Offset: 0

Paul D. Hanna, May 01 2016

Row sums equal the Euler numbers, A000364.
Column 1 equals A110501, the unsigned Genocchi numbers of first kind.
Main diagonal equals A272482, where A272482(n) = A005799(n)/2^n * (2*n)!/(n!)^2.
Sum_{k=0..2*n} (-1)^k*T(n,k) = (-1)^n.
Sum_{k=0..2*n} (-2)^k*T(n,k) = 2*(-1)^n for n>0.
Sum_{k=0..2*n} 2^k*T(n,k) = (-1)^n*A210657(n).
Sum_{k=0..2*n} 3^k*T(n,k) = A000281(n).
Sum_{k=0..2*n} 4^k*T(n,k) = A272158(n).
Sum_{k=0..2*n} 2^k*3^(2*n-k)*T(n,k) = A272467(n).

			E.g.f.: A(x,y) = 1 + x^2*(y)/2! + x^4*(y + 3*y^2 + y^3)/4! +
x^6*(3*y + 15*y^2 + 25*y^3 + 15*y^4 + 3*y^5)/6! +
x^8*(17*y + 119*y^2 + 329*y^3 + 455*y^4 + 329*y^5 + 119*y^6 + 17*y^7)/8! +
x^10*(155*y + 1395*y^2 + 5325*y^3 + 11235*y^4 + 14301*y^5 + 11235*y^6 + 5325*y^7 + 1395*y^8 + 155*y^9)/10! +
x^12*(2073*y + 22803*y^2 + 110605*y^3 + 311355*y^4 + 563013*y^5 + 683067*y^6 + 563013*y^7 + 311355*y^8 + 110605*y^9 + 22803*y^10 + 2073*y^11)/12! +...
where A(x,y) = cos((x - x*y)/2) / cos((x + x*y)/2).
This triangle of coefficients of x^(2*n)*y^k/(2*n)!, k=0..2*n, begins:
[1];
[0, 1, 0];
[0, 1, 3, 1, 0];
[0, 3, 15, 25, 15, 3, 0];
[0, 17, 119, 329, 455, 329, 119, 17, 0];
[0, 155, 1395, 5325, 11235, 14301, 11235, 5325, 1395, 155, 0];
[0, 2073, 22803, 110605, 311355, 563013, 683067, 563013, 311355, 110605, 22803, 2073, 0];
[0, 38227, 496951, 2918825, 10241231, 23904881, 39102063, 45956625, 39102063, 23904881, 10241231, 2918825, 496951, 38227, 0];
[0, 929569, 13943535, 96075665, 403075855, 1150348017, 2362479119, 3600524785, 4136759055, 3600524785, 2362479119, 1150348017, 403075855, 96075665, 13943535, 929569, 0]; ...

Paul D. Hanna, Table of n, a(n) for n = 0..1088, of flattened triangle for rows 0..32.

Cf. A000364, A110501, A272482, A210657, A000281, A272158, A272467.

{T(n,k) = my(X=x+x*O(x^(2*n))); (2*n)!*polcoeff(polcoeff( cos((X-x*y)/2)/cos((X+x*y)/2), 2*n,x), k,y)}
for(n=0,10, for(k=0,2*n, print1(T(n,k),", "));print(""))

E.g.f.: A(x,y) = (cos(x) + cos(x*y)) / (1 + cos(x + x*y)).
E.g.f.: A(x,y) = (sin(x) + sin(x*y)) / sin(x + x*y).
E.g.f.: A(x,y) = (exp(i*x) + exp(i*x*y)) / (1 + exp(i*(x + x*y))), where i^2 = -1.
O.g.f.: 1/(1 - 1*y*x/(1 - (1+y)^2*x/(1 - (1+2*y)*(2+1*y)*x/(1 - (2+2*y)^2*x/(1 - (2+3*y)*(3+2*y)*x/(1 - (3+3*y)^2*x/(1 - (3+4*y)*(4+3*y)*x/(1 - (4+4*y)^2*x/(1 - (4+5*y)*(5+4*y)*x/(1 - (5+5*y)^2*x/(1 - ...))))))))))), a continued fraction.

A296836 Expansion of e.g.f. exp(x*tanh(x/2)) (even powers only).

Original entry on oeis.org

1, 1, 2, 3, -3, 20, 105, -5271, 133826, -2714517, 25525845, 2131781300, -235250824479, 17527695547713, -1124258412169438, 58383380825728035, -975024061456732035, -398903577787777972396, 97649546210035758250281, -17069419358223320552890167

Offset: 0

Ilya Gutkovskiy, Dec 21 2017

sign

			exp(x*tanh(x/2)) = 1 + x^2/2! + 2*x^4/4! + 3*x^6/6! - 3*x^8/8! + ...

Cf. A001469, A003723, A006229, A009252, A009273, A110501, A296835.

nmax = 19; Table[(CoefficientList[Series[Exp[x Tanh[x/2]], {x, 0, 2 nmax}], x] Range[0, 2 nmax]!)[[n]], {n, 1, 2 nmax + 1, 2}]

a(n) = (2*n)! * [x^(2*n)] exp(x*tanh(x/2)).

A166355 Diagonal sums of exponential Riordan array [1+x*tan(x/2),x], A166353.

Original entry on oeis.org

1, 2, 5, 15, 64, 443, 4887, 78996, 1745995, 50333929, 1829758158, 81753825477, 4399497764477, 280491321580150, 20898005984605281, 1798558057748753171, 177034863818072607020, 19758697171102806823327

Offset: 0

Paul Barry, Oct 12 2009

Aerated sequence gives diagonal sums of A166353.

Cf. A110501, A166353.

(* The function RiordanArray is defined in A256893. *)
nmax = 17; R = RiordanArray[1 + # Tan[#/2]&, #&, 2 nmax + 1, True];
a[n_] := Sum[R[[i, 2 n - i + 2]], {i, 2 n + 1, n + 1, -1}];
Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Jul 20 2019 *)

a(n)=sum{k=0..n, C(n+k,2k)*G(k)} where G(n)=0^n+2(-1)^n*(1-4^n)*sum{k=0..2n, sum{j=0..k, (-1)^j*C(k,j)*j^(2n)/(k+1)}}.

A211194 G.f.: Sum_{n>=0} n! * (x/2)^n * Product_{k=1..n} (3k-1) / (1 + k(3k-1)/2x).

Original entry on oeis.org

1, 1, 4, 31, 394, 7441, 195544, 6822451, 305075254, 17010802021, 1157048302084, 94291964597671, 9069435785880514, 1016607721798423801, 131360503523334458224, 19382685928544981625691, 3239003918648541605116174, 608539911518928818091672781

Offset: 0

Paul D. Hanna, Feb 03 2013

nonn

O.g.f. is related to pentagonal numbers A000326. If b(n) = A000326(n)*x/(1+A000326(n)x), we have A(x) = 1 +b(1) +b(1)b(2) +b(1)b(2)b(3) +b(1)b(2)b(3)b(4) + ... . Philippe Deléham, Feb 04 2013

			G.f.: A(x) = 1 + x + 5*x^2 + 49*x^3 + 797*x^4 + 19417*x^5 + 661829*x^6 +...
where
A(x) = 1 + 1*x/(1+x) + 1*5*x^2/((1+x)*(1+5*x)) + 1*5*12*x^3/((1+x)*(1+5*x)*(1+12*x)) + 1*5*12*22*x^4/((1+x)*(1+5*x)*(1+12*x)*(1+22*x)) + 1*5*12*22*35*x^5/((1+x)*(1+5*x)*(1+12*x)*(1+22*x)*(1+35*x)) + 1*5*12*22*35*51*x^6/((1+x)*(1+5*x)*(1+12*x)*(1+22*x)*(1+35*x)*(1+51*x)) +...

Cf. A110501, A024283, A221972, A211183, A084939.

{a(n)=polcoeff(sum(m=0, n, m!*(x/2)^m*prod(k=1, m, (3*k-1)/(1+(3*k-1)/2*k*x+x*O(x^n)))), n)}
for(n=0,21,print1(a(n),", "))

G.f.: Sum_{n>=0} A084939(n) * x^n / Product_{k=1..n} (1 + k*(3*k-1)/2*x).

a(n) = Sum_{k, 0<=k<=n} A211183(n,k)*3^(n-k). - Philippe Deléham, Feb 03 2013

cp's OEIS Frontend

A166353 Exponential Riordan array [1+x*tan(x/2),x].

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A166354 Row sums of exponential Riordan array [1+x*tan(x/2),x], A166353.

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A221971 G.f.: A(x,y) = Sum_{n>=0} n! * x^n*y^n * Product_{k=1..n} (1 + k*x) / (1 + k*x*y + k^2*x^2*y).

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A223925 a(2n+1) = 2*n-1; a(2n)= 4^n.

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A236935 The infinite Seidel matrix H read by antidiagonals upwards; H = (H(n,k): n,k >= 0).

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PARI

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A240485 a(n) = -Zeta(1-n)*n*(2^(n+1) - 4) - Zeta(-n)*(n+1)*(2^(n+2) - 2), for n = 0 the limit is understood.

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Maple

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Sage

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A272481 E.g.f. A(x,y) = cos((x - x*y)/2) / cos((x + x*y)/2) represented as a triangle, read by rows, where row n lists of coefficients T(n,k) of x^(2*n)*y^k/n! in A(x,y), for k=0..2*n.

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A296836 Expansion of e.g.f. exp(x*tanh(x/2)) (even powers only).

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A221971 G.f.: A(x,y) = Sum_{n>=0} n! * x^ny^n Product_{k=1..n} (1 + kx) / (1 + kx*y + k^2x^2y).

A240485 a(n) = -Zeta(1-n)n(2^(n+1) - 4) - Zeta(-n)(n+1)(2^(n+2) - 2), for n = 0 the limit is understood.

A272481 E.g.f. A(x,y) = cos((x - xy)/2) / cos((x + xy)/2) represented as a triangle, read by rows, where row n lists of coefficients T(n,k) of x^(2n)y^k/n! in A(x,y), for k=0..2*n.

A211194 G.f.: Sum_{n>=0} n! * (x/2)^n * Product_{k=1..n} (3k-1) / (1 + k(3k-1)/2x).