A065619 -id:A065619 - cp's OEIS frontend

A009843 E.g.f. x/cos(x) (odd powers only).

1, 3, 25, 427, 12465, 555731, 35135945, 2990414715, 329655706465, 45692713833379, 7777794952988025, 1595024111042171723, 387863354088927172625, 110350957750914345093747, 36315529600705266098580265, 13687860690719716241164167451, 5858139922124796551409938058945

Offset: 0

R. H. Hardin

Expanding x/cosh(x) gives alternated signed values at odd positions.

Related to the formulas sum(k>0,sin(kx)/k^(2n+1))=(-1)^(n+1)/2*x^(2n+1)/(2n+1)!*sum(i=0,2n,(2Pi/x)^i*B(i)*C(2n+1,i)) and if x=Pi/2 sum(k>0,(-1)^(k+1)/k^(2n+1))=(-1)^n*E(2n)*Pi^(2n+1)/2^(2n+2)/(2n)!. - Benoit Cloitre, May 01 2002

			x/cos(x) = x + 1/2*x^3 + 5/24*x^5 + 61/720*x^7 + 277/8064*x^9 + ...

R. P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
Sylvie Corteel, Alexander Lazar, and Anna Vanden Wyngaerd, Decorated square paths at q = -1, arXiv:2408.10640 [math.CO], 2024. See p. 3.
Bishal Deb and Alan D. Sokal, Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers, arXiv:2212.07232 [math.CO], 2022. See p. 12.
Peter Luschny, The lost Bernoulli numbers.
Wikipedia, Bernoulli Polynomials

Bisection of A009391, A009392, A065619, A083008.

Cf. A099028. Cf. A002111, A069852, A069994.

seq((2*i+1)!*coeff(series(x/cos(x),x,32),x,2*i+1),i=0..13);
A009843 := n -> (-1)^n*(2*n+1)*euler(2*n): # Peter Luschny

c = CoefficientList[Series[1/MittagLefflerE[2,z^2],{z,0,40}],z]; Table[(-1)^n* Factorial[2*n+1]*c[[2*n+1]], {n,0,16}] (* Peter Luschny, Jul 03 2016 *)

a(n)=(-1)^(n+1)*sum(i=0,2*n+1,binomial(2*n+1,i)*bernfrac(i)*4^i)

a(n)=subst(bernpol(2*n+1),'x,1/4)*4^(2*n+1)*(-1)^(n+1) \\ Charles R Greathouse IV, Dec 10 2014

# The objective of this implementation is efficiency.
# n -> [a(0), a(1), ..., a(n)] for n > 0.
def A009843_list(n):
    S = [0 for i in range(n+1)]
    S[0] = 1
    for k in range(1, n+1):
        S[k] = k*S[k-1]
    for k in range(1, n+1):
        for j in range(k, n+1):
            S[j] = (j-k)*S[j-1]+(j-k+1)*S[j]
        S[k] = (2*k+1)*S[k]
    return S
print(A009843_list(10)) # Peter Luschny, Aug 09 2011

a(n) = (2n+1)*A000364(n) = sum(i=0, 2n, B(i)*C(2n+1, i)*4^i)=(2n+1)*E(2n) where B(i) are the Bernoulli numbers, C(2n, i) the binomial numbers and E(2n) the Euler numbers. - Benoit Cloitre, May 01 2002
Recurrence: a(n) = -(-1)^n*Sum[i=0..n-1, (-1)^i*a(i)*C(2n+1, 2i+1) ]. - Ralf Stephan, Feb 24 2005
a(n) = 4^n |E_{2n}(1/2)+E_{2n}(1)| (2n+1) for n > 0; E_{n}(x) Euler polynomial. - Peter Luschny, Nov 25 2010
a(n) = (2*n+1)! * [x^(2*n+1)] x/cos(x).
From Sergei N. Gladkovskii, Nov 15 2011, Oct 19 2012, Nov 10 2012, Jan 14 2013, Apr 10 2013, Oct 13 2013, Dec 01 2013: (Start) Continued fractions:
E.g.f.: x / cos(x) = x+x^3/Q(0); Q(k) = 8k+2-x^2/(1+(2k+1)*(2k+2)/Q(k+1)).
E.g.f.: x + x^3/U(0) where U(k) = (2*k+1)*(2*k+2) - x^2 + x^2*(2*k+1)*(2*k+2)/U(k+1).
G.f.: 1/G(0) where G(k) = 1 - x*(8*k^2+8*k+3)-16*x^2*(k+1)^4/G(k+1).
E.g.f.: 2*x/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))).
Let A(x) = S_{n>=0}a(n)*x^n/(2*n+1)! then A(x) = 1 + Q(0)*x/(2-x) where Q(k) = 1 - x*(2*k+1)*(2*k+2)/(x*(2*k+1)*(2*k+2) + ((2*k+1)*(2*k+2) - x)*((2*k+3)*(2*k+4) - x)/Q(k+1)).
G.f.: T(0)/(1-3*x) where T(k) = 1 - 16*x^2*(k+1)^4/(16*x^2*(k+1)^4 - (1 - x*(8*k^2 +8*k+3)) *(1 - x*(8*k^2+24*k+19))/T(k+1)).
G.f.: 1/T(0) where T(k) = 1 + x - x*(2*k+2)^2/(1 - x*(2*k+2)^2/T(k+1)). (End)
a(n) = (-1)^n*2^(4*n+1)*(2*n+1)*(zeta(-2*n,1/4)-zeta(-2*n,3/4)). - Peter Luschny, Jul 22 2013
From Peter Bala, Mar 02 2015: (Start)
a(n) = (-1)^(n+1)*4^(2*n + 1)*B(2*n + 1,1/4), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A069852 and A069994.
Conjecturally, a(n) = the unsigned numerator of B(2*n+1,1/4).
G.f. for signed version of sequence: Sum_{n >= 0} { 1/(n + 1) * Sum_{k = 0..n} (-1)^k*binomial(n,k)/( (1 - (4*k + 1)*x)*(1 - (4*k + 3)*x) ) } = 1 - 3*x^2 + 25*x^4 - 427*x^6 + .... (End)
a(n) ~ (2*n+1)! * 2^(2*n+2)/Pi^(2*n+1). - Vaclav Kotesovec, Jul 04 2016
G.f.: 1/(1 + x - 4*x/(1 - 4*x/(1 + x - 16*x/(1 - 16*x/(1 + x - 36*x/(1 - 36*x/(1 + x - ...))))))). Cf. A005439. - Peter Bala, May 07 2017

Extended and signs tested by Olivier Gérard, Mar 15 1997

A009752 Expansion of e.g.f. tan(x)*x (even powers only).

Original entry on oeis.org

0, 2, 8, 96, 2176, 79360, 4245504, 313155584, 30460116992, 3777576173568, 581777702256640, 108932957168730112, 24370173276164456448, 6419958484945407574016, 1967044844910430876860416, 693575525634287935244206080, 278846808228005417477465964544, 126799861926498005417315327279104

Offset: 0

R. H. Hardin

nonn

			2*x/(1+e^(2*x)) = 0 + x - 2/2!*x^2 + 8/4!*x^4 - 96/6!*x^6 + 2176/8!*x^8 ...

G. C. Greubel, Table of n, a(n) for n = 0..240
Bishal Deb and Alan D. Sokal, Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers, arXiv:2212.07232 [math.CO], 2022. See p. 12.

A diagonal of A232933.

Cf. A009725, A024255, A065619, A099028, A110501, A117513.

a := n -> 4^n*n*`if`(n=0,0,abs(euler(2*n-1, 0))): # Peter Luschny, Jun 09 2016

nn = 30; t = Range[0, nn]! CoefficientList[Series[x*Tan[x], {x, 0, nn}], x]; Take[t, {1, nn + 1, 2}] (* T. D. Noe, Sep 20 2012 *)
Table[(-1)^n 4 n PolyLog[1 - 2 n, -I], {n, 0, 19}] (* Peter Luschny, Aug 17 2021 *)

my(x='x+O('x^50)); v=Vec(serlaplace(x*tan(x))); concat([0], vector(#v\2,n,v[2*n-1])) \\ G. C. Greubel, Feb 12 2018

a(n) = n 4^n |E_{2n-1}(1/2)+E_{2n-1}(1)| for n > 0; E_{n}(x) Euler polynomials. - Peter Luschny, Nov 25 2010
a(n) = (2*n)! * [x^(2*n)] tan(x)*x.
a(n) = 2*(2*n)!*Pi^(-2*n)*(4^n-1)*Li{2*n}(1) for n > 0. - Peter Luschny, Jun 29 2012
E.g.f.: sqrt(x)*tan(sqrt(x))= sum(n>=0,  a(n)*x^n/(2*n)! ) = x/T(0) where T(k)= 1 - 4*k^2 + x*(1 - 4*k^2)/T(k+1) ; (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 19 2012
E.g.f.: -1 - x^(1/2)- Q(0),where Q(k) = 4*k -1 - x/( 1 - x/ (4*k+1 + x/( 1 + x/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 24 2013
From Peter Luschny, Jun 09 2016: (Start)
a(n) = (4^n-16^n)*Sum_{k=0..2*n} (-1)^(n-k)*Stirling2(2*n, k)*k!/(k+1).
2*a(n)/4^n = A110501(n) for n>=1.
a(n) / 2^n = A117513(n) for n>=1. (End)
a(n) ~ (4*(4^(2*n)-2^(2*n)))*Pi*(n/(Pi*e))^(2*n+1/2)*exp(1/2+1/(24*n)-1/(2880*n^3) +1/(40320*n^5)-...). - Peter Luschny, Jan 16 2017
a(n) = (-1)^n*4*n*PolyLog(1 - 2*n, -i). - Peter Luschny, Aug 17 2021
a(n) = 2*A024255(n). - Alois P. Heinz, Aug 17 2021

Extended and signs tested by Olivier Gérard, Mar 15 1997

A249402 The number of 3-alternating permutations of [n].

Original entry on oeis.org

1, 1, 1, 2, 3, 11, 40, 99, 589, 3194, 11259, 92159, 666160, 3052323, 31799041, 287316122, 1620265923, 20497038755, 222237912664, 1488257158851, 22149498351205, 280180369563194, 2172534146099019, 37183508549366519, 537546603651987424, 4736552519729393091

Offset: 0

R. J. Mathar, Oct 27 2014

nonn

A sequence a(1),a(2),... is called k-alternating if a(i) > a(i+1) iff i=1 (mod k). a(n) gives the number of 3-alternating permutations of [n].

			The a(4)=3 3-alternating permutations of [4] are: [2 1 3 4 ] [3 1 2 4 ] and [4 1 2 3 ].
The a(5)=11 3-alternating permutations of [5] are: [2 1 3 5 4 ] [2 1 4 5 3 ] [3 1 2 5 4 ] [3 1 4 5 2 ] [3 2 4 5 1 ] [4 1 2 5 3 ] [4 1 3 5 2 ] [4 2 3 5 1 ] [5 1 2 4 3 ] [5 1 3 4 2 ] and [5 2 3 4 1 ].

Alois P. Heinz, Table of n, a(n) for n = 0..500
R. P. Stanley, A survey of alternating permutations, arXiv:0912.4240, page 17.

Cf. A065619 (2-alternating).
Cf. A178963, A249583 (alternative definitions of 3-alternating permutations).
Column k=3 of A250261.

b:= proc(u, o, t) option remember; `if`(u+o=0, 1,
     `if`(t=1, add(b(u-j, o+j-1, irem(t+1, 3)), j=1..u),
               add(b(u+j-1, o-j, irem(t+1, 3)), j=1..o)))
    end:
a:= n-> b(0, n, 0):
seq(a(n), n=0..35);  # Alois P. Heinz, Oct 27 2014

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, If[t == 1, Sum[b[u-j, o+j-1, Mod[t+1, 3]], {j, 1, u}], Sum[b[u+j-1, o-j, Mod[t+1, 3]], {j, 1, o}]]]; a[n_] := b[0, n, 0]; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Jun 22 2015, after Alois P. Heinz *)

a(16)-a(25) from Alois P. Heinz, Oct 27 2014

A104345 Triangle read by rows: T(n,k) is the number of alternating permutations on [n+1] with 1 in position k+1, 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 2, 5, 8, 6, 8, 5, 16, 25, 20, 20, 25, 16, 61, 96, 75, 80, 75, 96, 61, 272, 427, 336, 350, 350, 336, 427, 272, 1385, 2176, 1708, 1792, 1750, 1792, 1708, 2176, 1385, 7936, 12465, 9792, 10248, 10080, 10080, 10248, 9792, 12465, 7936

Offset: 0

David Callan, Mar 02 2005

			Table begins
\ k..0....1....2....3....4....
n
0 |..1
1 |..1....1
2 |..1....2....1
3 |..2....3....3....2
4 |..5....8....6....8....5
5 |.16...25...20...20...25...16
6 |.61...96...75...80...75...96...61
7 |272..427..336..350..350..336..427..272
For example, T(3,1) counts 2143, 3142, 4132 - the alternating permutations on [4] with 1 in position 2.

Alois P. Heinz, Rows n = 0..150, flattened

Cf. A104346. Row sums are A001250; column k=0 and main diagonal are the up-down numbers (A000111); column k=1 is A065619.

T(2n,n) gives A362581.

b:= proc(u, o) option remember; `if`(u+o=0, 1,
      add(b(o-1+j, u-j), j=1..u))
    end:
T:= (n, k)-> binomial(n, k)*b(k, 0)*b(n-k, 0):
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 25 2023

b[u_, o_] := b[u, o] = If[u+o == 0, 1, Sum[b[o-1+j, u-j], {j, 1, u}]];
T[n_, k_] := Binomial[n, k]*b[k, 0]*b[n-k, 0];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz *)

The mixed o.g.f./e.g.f. is Sum_{k=0..n} T(n, k)*x^n/n!*y^k = (sec(x) + tan(x))*(sec(x*y) + tan(x*y)).

T(n,k) = binomial(n,k)*A000111(k)*A000111(n-k). - Alois P. Heinz, Apr 25 2023

A099028 Euler-Seidel matrix T(k,n) with start sequence e.g.f. 2x/(1+e^(2x)), read by antidiagonals.

Original entry on oeis.org

0, 1, 1, 0, -1, -2, -3, -3, -2, 0, 0, 3, 6, 8, 8, 25, 25, 22, 16, 8, 0, 0, -25, -50, -72, -88, -96, -96, -427, -427, -402, -352, -280, -192, -96, 0, 0, 427, 854, 1256, 1608, 1888, 2080, 2176, 2176, 12465, 12465, 12038, 11184, 9928, 8320, 6432, 4352, 2176, 0

Offset: 0

Ralf Stephan, Sep 27 2004

In an Euler-Seidel matrix, the rows are consecutive pairwise sums and the columns consecutive differences.

			Seidel matrix:
[    0     1    -2     0     8     0   -96     0  2176     0]
[    1    -1    -2     8     8   -96   -96  2176  2176     .]
[    0    -3     6    16   -88  -192  2080  4352     .     .]
[   -3     3    22   -72  -280  1888  6432     .     .     .]
[    0    25   -50  -352  1608  8320     .     .     .     .]
[   25   -25  -402  1256  9928     .     .     .     .     .]
[    0  -427   854 11184     .     .     .     .     .     .]
[ -427   427 12038     .     .     .     .     .     .     .]
[    0 12465     .     .     .     .     .     .     .     .]
[12465     .     .     .     .     .     .     .     .     .]

D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.

First column (odd part) is A009843, main diagonal is in A099029. Antidiagonal sums are in A065619. Cf. A009752.

T[k_, n_] := T[k, n] = If[k == 0, SeriesCoefficient[2x/(1 + E^(2x)), {x, 0, n}] n!, T[k-1, n] + T[k-1, n+1]];
Table[T[k-n, n], {k, 0, 9}, {n, 0, k}] (* Jean-François Alcover, Jun 11 2019 *)

def SeidelMatrixA099028(dim):
    E = matrix(ZZ, dim)
    t = taylor(2*x/(1+exp(2*x)), x, 0, dim + 1)
    for k in (0..dim-1):
        E[0, k] = factorial(k) * t.coefficient(x, k)
    R = [0]
    for n in (1..dim-1):
        for k in (0..dim-n-1):
            E[n, k] = E[n-1, k] + E[n-1, k+1]
        R.extend([E[n-k,k] for k in (0..n)])
    return R
print(SeidelMatrixA099028(10)) # Peter Luschny, Jul 02 2016

Recurrence: T(k, n) = T(k-1, n) + T(k-1, n+1).

A178616 Triangle by columns, odd columns of Pascal's triangle A007318, otherwise (1, 0, 0, 0, ...).

Original entry on oeis.org

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

Offset: 0

Gary W. Adamson, May 30 2010

Row sums = a variant of A052950, starting (1, 1, 3, 4, 9, 16, 33, ...); whereas A052950 starts (2, 1, 3, 4, 9, ...).

Column 1 of the inverse of A178616 is a signed variant of A065619 prefaced with a 0; where A065619 = (1, 2, 3, 8, 25, 96, 427, ...).

			First few rows of the triangle:
  1,
  0,  1;
  0,  2, 1;
  0,  3, 0,   1
  0,  4, 0,   4, 1;
  0,  5, 0,  10, 0,   1;
  0,  6, 0,  20, 0,   6, 1;
  0,  7, 0,  35, 0,  21, 0,   1;
  0,  8, 0,  56, 0,  56, 0,   8, 1;
  0,  9, 0,  84, 0, 126, 0,  36, 0,  1;
  0, 10, 0, 120, 0, 252, 0, 120, 0, 10, 1;
  0, 11, 0, 165, 0, 462, 0, 330, 0, 55, 0, 1;
  ...

Cf. A162109, A065619, A052950, A095704.

Triangle, odd columns of Pascal's triangle; (1, 0, 0, 0, ...) as even columns k.
Alternatively, (since A178616 + A162169 - Identity matrix) = Pascal's triangle,
we can begin with Pascal's triangle, subtract A162169, then add the Identity
matrix to obtain A178616.

A250259 The number of 4-alternating permutations of [n].

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 19, 78, 217, 496, 3961, 25442, 105963, 349504, 3908059, 34227438, 190065457, 819786496, 11785687921, 130746521282, 907546301523, 4835447317504, 84965187064099, 1141012634368398, 9504085749177097, 60283564499562496, 1251854782837499881

Offset: 0

R. J. Mathar, Nov 15 2014

nonn

A sequence a(1),a(2),... is called k-alternating if a(i) > a(i+1) iff i=1 (mod k).

Alois P. Heinz, Table of n, a(n) for n = 0..500
R. P. Stanley, A survey of alternating permutations, arXiv:0912.4240 [math.CO], 2009, page 17.

Cf. A249402 (3-alternating), A065619 (2-alternating), A250260 (5-alternating).

Column k=4 of A250261.

onestep := proc(n::integer,ups::integer,downs::integer,uplen::integer)
    local thisstep,left,doup,tak,ret ;
    option remember;
    left := ups+downs ;
    if left = 0 then
        return 1;
    end if;
    thisstep := n-left+1 ;
    if modp(thisstep-2,uplen+1) = 0 then
        doup := false;
    else
        doup := true;
    end if;
    if doup then
        ret := 0 ;
        for tak from 1 to ups do
            ret := ret+procname(n,ups-tak,downs+tak-1,uplen) ;
        end do:
        return ret ;
    else
        ret := 0 ;
        for tak from 1 to downs do
            ret := ret+procname(n,ups+tak-1,downs-tak,uplen) ;
        end do:
        return ret ;
    end if;
end proc:
downupP := proc(n::integer,uplen::integer)
    local ret,tak;
    if n = 0 then
        return 1;
    end if;
    ret := 0 ;
    for tak from 1 to n do
        ret := ret+onestep(n,n-tak,tak-1,uplen) ;
    end do:
    return ret ;
end proc:
A250259 :=proc(n)
    downupP(n,3) ;
end proc:
seq(A250259(n),n=0..20) ; # R. J. Mathar, Nov 15 2014
# second Maple program:
b:= proc(u, o, t) option remember; `if`(u+o=0, 1,
     `if`(t=1, add(b(u-j, o+j-1, irem(t+1, 4)), j=1..u),
               add(b(u+j-1, o-j, irem(t+1, 4)), j=1..o)))
    end:
a:= n-> b(0, n, 0):
seq(a(n), n=0..35);  # Alois P. Heinz, Nov 15 2014

b[u_, o_, t_] := b[u, o, t] = If[u + o == 0, 1, If[t == 1, Sum[b[u - j, o + j - 1, Mod[t + 1, 4]], {j, 1, u}], Sum[b[u + j - 1, o - j, Mod[t + 1, 4]], {j, 1, o}]]]; a[n_] := b[0, n, 0]; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Jul 10 2017, after Alois P. Heinz *)

A250260 The number of 5-alternating permutations of [n].

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 5, 29, 133, 412, 1041, 2300, 22991, 170832, 822198, 3114489, 10006375, 141705439, 1457872978, 9522474417, 48094772656, 202808749375, 3716808948931, 48860589990687, 403131250565618, 2545098156762649, 13287626090593750

Offset: 0

R. J. Mathar, Nov 15 2014

nonn

A sequence a(1), a(2),... is called k-alternating if a(i) > a(i+1) iff i=1 (mod k).

Alois P. Heinz, Table of n, a(n) for n = 0..500
R. P. Stanley, A survey of alternating permutations, arXiv:0912.4240, page 17.

Cf. A065619 (2-alternating), A249402 (3-alternating), A250259 (4-alternating).

Column k=5 of A250261.

# dowupP defined in A250259.
A250260 :=proc(n)
    downupP(n,4) ;
end proc:
seq(A250260(n),n=0..20) ;
# second Maple program:
b:= proc(u, o, t) option remember; `if`(u+o=0, 1,
     `if`(t=1, add(b(u-j, o+j-1, irem(t+1, 5)), j=1..u),
               add(b(u+j-1, o-j, irem(t+1, 5)), j=1..o)))
    end:
a:= n-> b(0, n, 0):
seq(a(n), n=0..35);  # Alois P. Heinz, Nov 15 2014

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, 1, If[t == 1, Sum[b[u-j, o+j-1, Mod[t+1, 5]], {j, 1, u}], Sum[b[u+j-1, o-j, Mod[t+1, 5]], {j, 1, o}]]]; a[n_] := b[0, n, 0]; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Jun 24 2015, after Alois P. Heinz *)

A243963 a(n) = n4^n(-Z(1-n, 1/4)/2 + Z(1-n, 3/4)/2 - Z(1-n, 1)*(1 - 2^(-n))) for n > 0 and a(0) = 0, where Z(n, c) is the Hurwitz zeta function.

Original entry on oeis.org

0, 0, 2, 3, -8, -25, 96, 427, -2176, -12465, 79360, 555731, -4245504, -35135945, 313155584, 2990414715, -30460116992, -329655706465, 3777576173568, 45692713833379, -581777702256640, -7777794952988025, 108932957168730112, 1595024111042171723, -24370173276164456448

Offset: 0

Paul Curtz, Jun 16 2014

sign

Previous name was: 0 followed by -(n+1)*A163747(n).
Difference table of a(n):
0,    0,    2,     3,    -8,   -25,...
0,    2,    1,   -11,   -17,   121,...
2,   -1,  -12,    -6,   138,   210,...
-3, -11,    6,   144,    72, -3144,...
-8,  17,  138,   -72, -3216, -1608,...
25, 121, -210, -3144,  1608,...
a(n) is an autosequence of second kind. Its inverse binomial transform is the signed sequence. Its main diagonal is the first upper diagonal multiplied by 2.

Cf. A132049, A065619.

a := n -> `if`(n=0, 0, n*4^n*(-Zeta(0, 1-n, 1/4)/2 + Zeta(0, 1-n, 3/4)/2 + Zeta(1-n)*(2^(-n)-1))): seq(a(n), n=0..24); # Peter Luschny, Jul 21 2020

a[0] = 0; a[n_] := -n*SeriesCoefficient[(2*E^x*(1 - E^x))/(1 + E^(2*x)), {x, 0, n-1}]*(n-1)!; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Jun 17 2014 *)

a(n) = 0, 0, followed by (period 4: repeat 1, 1, -1, -1)*A065619(n+2).

a(2n) = (-1)^(n+1)A009752(n). a(2n+1) = (-1)^n*A009843(n+1).

New name by Peter Luschny, Jul 21 2020

A294033 Triangle read by rows, expansion of exp(xz)z*(tanh(z) + sech(z)), T(n, k) for n >= 1 and 0 <= k <= n-1.

Original entry on oeis.org

1, 2, 2, -3, 6, 3, -8, -12, 12, 4, 25, -40, -30, 20, 5, 96, 150, -120, -60, 30, 6, -427, 672, 525, -280, -105, 42, 7, -2176, -3416, 2688, 1400, -560, -168, 56, 8, 12465, -19584, -15372, 8064, 3150, -1008, -252, 72, 9, 79360, 124650, -97920, -51240, 20160, 6300, -1680, -360, 90, 10, -555731, 872960, 685575, -359040, -140910, 44352, 11550, -2640, -495, 110, 11

Offset: 1

Peter Luschny, Oct 24 2017

			Triangle starts:
  [1][   1]
  [2][   2,   2]
  [3][  -3,   6,    3]
  [4][  -8, -12,   12,    4]
  [5][  25, -40,  -30,   20,    5]
  [6][  96, 150, -120,  -60,   30,  6]
  [7][-427, 672,  525, -280, -105, 42, 7]

T(n, 0) = signed A065619. Row sums of abs(T(n,k)) = A231179.
Diagonals A000027, A002378, A027480, A162668.
A003506 (m=1), this seq. (m=2), A294034 (m=3).
Cf. A000111, A247453.

gf := exp(x*z)*z*(tanh(z)+sech(z)):
s := n -> n!*coeff(series(gf,z,n+2),z,n):
C := n -> PolynomialTools:-CoefficientList(s(n),x):
ListTools:-FlattenOnce([seq(C(n), n=1..7)]);
# Alternatively:
T := (n, k) -> `if`(n = k+1, n,
(k+1)*binomial(n,k+1)*2^(n-k-1)*(euler(n-k-1, 1/2)+euler(n-k-1, 1))):
for n from 1 to 7 do seq(T(n,k), k=0..n-1) od;

L[0] := 1; L[n_] := (-1)^Binomial[n, 2] 2 Abs[PolyLog[-n, -I]];
p[n_] := n Sum[Binomial[n - 1, k - 1] L[k - 1] x^(n - k), {k, 0, n}];
Table[CoefficientList[p[n], x], {n, 1, 11}] // Flatten

T(n, k) = (k+1)*binomial(n,k+1)*2^(n-k-1)*(Euler(n-k-1, 1/2) + Euler(n-k-1, 1)) for 0 <= k <= n-2.

T(n, k) is the coefficient of x^k of the polynomial p(n) = n*Sum_{k=1..n} binomial(n-1, k-1)*L(k-1)*x^(n-k) and L(n) = (-1)^binomial(n,2)*A000111(n). In particular n divides T(n, k).

cp's OEIS Frontend

A009843 E.g.f. x/cos(x) (odd powers only).

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A009752 Expansion of e.g.f. tan(x)*x (even powers only).

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A249402 The number of 3-alternating permutations of [n].

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A104345 Triangle read by rows: T(n,k) is the number of alternating permutations on [n+1] with 1 in position k+1, 0<=k<=n.

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A099028 Euler-Seidel matrix T(k,n) with start sequence e.g.f. 2x/(1+e^(2x)), read by antidiagonals.

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A178616 Triangle by columns, odd columns of Pascal's triangle A007318, otherwise (1, 0, 0, 0, ...).

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A250259 The number of 4-alternating permutations of [n].

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A243963 a(n) = n4^n(-Z(1-n, 1/4)/2 + Z(1-n, 3/4)/2 - Z(1-n, 1)*(1 - 2^(-n))) for n > 0 and a(0) = 0, where Z(n, c) is the Hurwitz zeta function.

A294033 Triangle read by rows, expansion of exp(xz)z*(tanh(z) + sech(z)), T(n, k) for n >= 1 and 0 <= k <= n-1.