A001586 -id:A001586 - cp's OEIS frontend

A242195 Least prime divisor of the n-th tangent number T_n which does not divide any T_k with k < n, or 1 if such a primitive prime divisor of T_n does not exist.

1, 2, 1, 17, 31, 691, 43, 257, 73, 41, 89, 103, 2731, 113, 151, 37, 43691, 109, 174763, 61681, 337, 59, 178481, 97, 251, 157, 39409, 113161, 67, 1321, 266689, 641, 839, 101, 281, 433, 223, 229, 121369, 631

Offset: 1

Zhi-Wei Sun, May 07 2014

nonn

Conjecture: (i) a(n) is prime for any n > 3.
(ii) For the n-th Springer number S_n given by A001586, if n is greater than one and not equal to 5, then S_n has a prime divisor which does not divide any S_k with k < n.
See also A242193 and A242194 for similar conjectures involving Bernoulli numbers and Euler numbers.

			a(4) = 17 since T_4 = 2^4*17 with 17 dividing none of T_1 = 1, T_2 = 2 and T_3 = 2^4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..54

Cf. A000040, A000182, A001586, A242169, A242170, A242171, A242173, A242193, A242194.

t[n_]:=(-1)^(n-1)*2^(2n)(2^(2n)-1)BernoulliB[2n]/(2n)
f[n_]:=FactorInteger[t[n]]
p[n_]:=Table[Part[Part[f[n],k],1],{k,1,Length[f[n]]}]
Do[If[t[n]<2,Goto[cc]];Do[Do[If[Mod[t[i],Part[p[n],k]]==0,Goto[aa]],{i,1,n-1}];Print[n," ",Part[p[n],k]];Goto[bb];Label[aa];Continue,{k,1,Length[p[n]]}];Label[cc];Print[n," ",1];Label[bb];Continue,{n,1,40}]

# uses[LPDtransform from A242193]
def Tnum(n): return (-1)^(n-1)*2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n)
A242195list = lambda sup: [LPDtransform(n, Tnum) for n in (1..sup)]
print(A242195list(40)) # Peter Luschny, Jul 26 2019

A154343 S(n,k) an additive decomposition of the Springer number (generalized Euler number), (triangle read by rows).

Original entry on oeis.org

1, 3, -2, 9, -16, 4, 27, -98, 60, 0, 81, -544, 616, 0, -96, 243, -2882, 5400, 0, -3360, 960, 729, -14896, 43564, 0, -72480, 46080, -5760, 2187, -75938, 334740, 0, -1246560, 1323840, -362880, 0, 6561, -384064, 2495056, 0, -18801216, 29675520

Offset: 0

Peter Luschny, Jan 07 2009

The Swiss-Knife polynomials A153641 can be understood as a sum of polynomials. Evaluated at x=1/2 and multiplied by 2^n these polynomials result in a decomposition of the Springer numbers A001586.

			Triangle begins:
  1,
  3,    -2,
  9,    -16,     4,
  27,   -98,     60,      0,
  81,   -544,    616,     0, -96,
  243,  -2882,   5400,    0, -3360,     960,
  729,  -14896,  43564,   0, -72480,    46080,    -5760,
  2187, -75938,  334740,  0, -1246560,  1323840,  -362880,   0,
  6561, -384064, 2495056, 0, -18801216, 29675520, -13386240, 0, 645120,
  ...

G. C. Greubel, Table of n, a(n) for n = 0..1274
Peter Luschny, The Swiss-Knife polynomials.

Cf. A001586, A188458.

Cf. A153641, A154341, A154342, A154344, A154345.

S := proc(n,k) local v,c; c := m -> if irem(m+1,4) = 0 then 0 else 1/((-1)^iquo(m+1,4)*2^iquo(m,2)) fi; add((-1)^(v)*binomial(k,v)*2^n*c(k)*(v+3/2)^n,v=0..k) end: seq(print(seq(S(n,k),k=0..n)),n=0..8);

c[m_] := If[Mod[m+1, 4] == 0, 0, 1/((-1)^Quotient[m+1, 4]*2^Quotient[m, 2])]; s[n_, k_] := Sum[(-1)^v*Binomial[k, v]*2^n*c[k]*(v+3/2)^n, {v, 0, k}]; Table[s[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2013, after Maple *)

Let c(k) = ((-1)^floor(k/4) / 2^floor(k/2)) * [4 not div k+1] (Iverson notation).
S(n,k) = Sum_{v=0..k} (-1)^v*binomial(k,v)*2^n*c(k)*(v+3/2)^n.
A188458(n) = Sum_{k=0..n} S(n,k).

A185356 Triangle read by rows: number of type B alternating permutations according to their last value.

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 0, 1, 2, 4, 4, 3, 0, 3, 2, 0, 0, 4, 8, 11, 0, 11, 14, 16, 16, 80, 80, 76, 68, 57, 0, 57, 46, 32, 16, 0, 0, 80, 160, 236, 304, 361, 0, 361, 418, 464, 496, 512, 512, 3904, 3904, 3824, 3664, 3428, 3124, 2763, 0, 2763, 2402, 1984, 1520, 1024, 512, 0

Offset: 0

N. J. A. Sloane, Dec 22 2011

"The table counting type B alternating permutations by their last value is obtained by the following algorithm: first separate the picture by the column p = 0 and then compute two triangles. Put 1 at the top of each triangle and compute the rest as follows: fill the second row of the left (resp. right) triangle as the sum of the elements of the first row (resp. strictly) to their left. Then fill the third row of the right (resp. left) triangle as the sum of the elements of the previous row (resp. strictly) to their right. Compute all rows successively by reading from left to right and right to left alternately." [Joshua-Verges et al.]

			Triangle begins:
                       0
                     1 0 1
                 0   1 0 1   2
             4   4   3 0 3   2   0
         0   4   8  11 0 11  14  16  16
    80  80  76  68  57 0 57  46  32  16  0
  0 80 160 236 304 361 0 361 418 464 496 512 512

M. Josuat-Vergès, J.-C. Novelli and J.-Y. Thibon, The algebraic combinatorics of snakes, arXiv preprint arXiv:1110.5272 [math.CO], 2011.

See A202690 for another version.
See A010094 and A008281 for type A permutations.
Cf. A000831 (row sums, for n>0).
Cf. A001586 for the middle coefficients.

T(n,k) = {if ((k==0), return(0)); if (n==1, if (abs(k)==1, return(1))); if (n%2, if (k<0, sum(j=k+1, n-1, T(n-1,j)), sum(j=k, n-1, T(n-1,j))), if (k<0, sum(j=-n+1, k, T(n-1,j)), sum(j=-n+1, k-1, T(n-1,j))));}
tabf(nn) = {for (n=0, nn, for (k=-n, n, print1(T(n, k), ", ");); print;);} \\ Michel Marcus, Jun 03 2020

More terms from Michel Marcus, Jun 03 2020

A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 11, 26, 12, 8, 57, 120, 136, 32, 16, 361, 970, 760, 560, 80, 32, 2763, 7052, 8860, 3680, 2000, 192, 64, 24611, 72530, 72884, 58520, 15120, 6496, 448, 128, 250737, 716528, 976464, 538048, 314720, 55552, 19712, 1024, 256

Offset: 1

Peter Bala, Jan 28 2011

Define a polynomial sequence S(n,x) recursively by
(1)... S(n+1,x) = x*S(n,x-1)+(x+1)*S(n,x+1) with S(0,x) = 1.
This table lists the coefficients of these polynomials (for n>=1) in ascending powers of x.
The first few polynomials are
S(0,x) = 1
S(1,x) = 2*x+1
S(2,x) = 4*x^2+4*x+3
S(3,x) = 8*x^3+12*x^2+26*x+11.
The sequence [1,1,3,11,57,...] of constant terms of the polynomials is the sequence of Springer numbers A001586. The zeros of the polynomials S(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
Compare the recurrence (1) with the recurrence relation satisfied by the coefficients T(n,k) of the polynomials of A104035, namely
(2)... T(n+1,k) = k*T(n,k-1)+(k+1)*T(n,k+1).

			Table begin
n\k|.....0.....1.....2.....3.....4.....5......6
===============================================
0..|.....1
1..|.....1.....2
2..|.....3.....4.....4
3..|....11....26....12.....8
4..|....57...120...136....32...16
5..|...361...970...760...560...80.....32
6..|..2763..7052..8860..3680..2000...192....64
...

Peter Bala, The zeros of the row polynomials of A185417

Cf A001586 (1st column and row sums), A104035, A126156, A147309, A185415, A185418, A185419

#A185417
S := proc(n,x) option remember;
description 'polynomials S(n,x)'
if n = 0 return 1 else return x*S(n-1,x-1)+(x+1)*S(n-1,x+1)
end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(S(n,x),x); end do;

S[0, ] = 1; S[n, x_] := S[n, x] = x*S[n-1, x-1] + (x+1)*S[n-1, x+1]; Table[ CoefficientList[S[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 15 2015 *)

E.g.f: F(x,t) = 1/(cos(t)-sin(t))*(tan(2*t)+sec(2*t))^x
= (cos(t)+sin(t))^x/(cos(t)-sin(t))^(x+1)
= 1 + (2*x+1)*t + (4*x^2+4*x+3)*t^2/2! + ....
Note that (tan(t)+sec(t))^x is the e.g.f for table A147309.
ROW POLYNOMIALS
The easily checked identity d/dt F(x,t) = x*F(x-1,t)+(x+1)*F(x+1,t) shows that the row generating polynomials of this table are the polynomials S(n,x) described in the Comments section above.
The polynomials S(n,-x) satisfy a Riemann hypothesis: that is, the zeros of S(n,-x) lie on the vertical line Re(x) = 1/2 in the complex plane - see the link.
RELATION WITH OTHER SEQUENCES
1st column [1,1,3,11,57,...] is A001586.
Row sums sequence [1,3,11,57,...] is also A001586.
For n>=1, the values 1/2^n*P(2*n,-1/2) = [1,7,139,5473,...] appear to be A126156.

A185419 Table of coefficients of a polynomial sequence of binomial type related to the enumeration of minimax trees A080795.

Original entry on oeis.org

1, 3, 1, 10, 9, 1, 42, 67, 18, 1, 248, 510, 235, 30, 1, 1992, 4378, 2835, 605, 45, 1, 19600, 44268, 34888, 10605, 1295, 63, 1, 222288, 524748, 461748, 178913, 31080, 2450, 84, 1, 2851712, 7103088, 6728428, 3069612, 690753, 77112, 4242, 108, 1

Offset: 1

Peter Bala, Feb 07 2011

DEFINITION
Define a sequence of polynomials M(n,x) by means of the recurrence relation
(1)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)},
with starting value M(0,x) = 1. We call these the minimax polynomials.
The first few polynomials are
M(1,x) = x
M(2,x) = x*(x   +  3)
M(3,x) = x*(x^2 +  9*x   +  10)
M(4,x) = x*(x^3 + 18*x^2 +  67*x   +  42)
M(5,x) = x*(x^4 + 30*x^3 + 235*x^2 + 510*x + 248).
This triangle lists the coefficients of these polynomials (apart from M(0,x)) in ascending powers of x.
RELATION TO MINIMAX TREES
The value M(n,1) equals the number of minimax trees on n nodes - A080795(n). This result can be used to recursively calculate the entries of A080795 - see A185420.
In addition, the minimax polynomials M(n,x) occur in the formula for the number T(n,k) of forests of k minimax trees on n nodes. ... T(n,k) = 1/k!*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*M(n,j).
ANALOGIES WITH THE MONOMIALS
{M(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting M(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
(2)... 1^[p]+...+(n-1)^[p] = -2*n^[p]+ 1/(p+1)*Sum_{k = 0..floor(p/2)} 8^k*binomial(p+1,2k)*B_(2k)*n^[p+1-2k], where {B_k}k>=0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For other polynomial sequences defined by recurrences similar to (1), and related to the zigzag numbers A000111 and the Springer numbers A001586, see A147309 and A185417, respectively. See also A185415.
The Bell transform of A143523(n). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
n\k|.....1......2......3......4......5......6......7
====================================================
..1|.....1
..2|.....3......1
..3|....10......9......1
..4|....42.....67.....18......1
..5|...248....510....235.....30......1
..6|..1992...4378...2835....605.....45......1
..7|.19600..44268..34888..10605...1295.....63......1
..
Example of the generalized Bernoulli summation formula:
The second row of the triangle gives x^[2] = 3*x+x^2.
Then 1^[2]+2^[2]+...+(n-1)^[2] = (n^3+3*n^2-4*n)/3 = 1/3*(MB(3,n)-MB(3,0)).
From _R. J. Mathar_, Mar 15 2013: (Start)
The matrix inverse starts
       1;
      -3,       1;
      17,      -9,        1;
    -147,      95,      -18,      1;
    1697,   -1245,      305,    -30,      1;
  -24483,   19687,    -5670,    745,    -45,    1;
  423857, -365757,   118237, -18690,   1540,  -63,   1;
-8560947, 7819287, -2761122, 498197, -50190, 2842, -84, 1; (End)

Cf. A080253 (coeffs. of delta operator), A080795 (row sums), A143523 (column 1), A147309, A185415, A185417, A185420.

#A185419
M := proc(n,x) option remember;
if n = 0 then
return 1
else return
x*(2*M(n-1,x+1)-M(n-1,x-1))
end if;
end proc:
with(PolynomialTools):
for n from 1 to 10 do
CoefficientList(M(n,x),x);
end do;

M[0, ] = 1; M[n, x_] := M[n, x] = x (2 M[n-1, x+1] - M[n-1, x-1]);
Table[CoefficientList[M[n, x], x] // Rest, {n, 1, 10}] (* Jean-François Alcover, Jun 26 2019 *)

# uses[bell_matrix from A264428]
# Adds a column 1,0,0,0, ... at the left side of the triangle.
bell_matrix(lambda n: A143523(n), 10) # Peter Luschny, Jan 18 2016

GENERATING FUNCTION
Let a = 3-2*sqrt(2). Let f(t) = (1/2)*sqrt(2)*((1+a*exp(2*sqrt(2)*t))/ (1-a*exp(2*sqrt(2)*t))) = 1 + t + 4*t^2/2! + 20*t^3/3! + ... be the e.g.f. for A080795. Then the e.g.f. for the current table, including a constant 1, is
(1)... F(x,t) = f(t)^x = Sum_{n>=0} M(n,x)*t^n/n! = 1 + x*t + (3*x+x^2)*t^2/2! + (10*x+9*x^2+x^3)*t^3/3! + ....
ROW POLYNOMIALS
One easily checks that d/dt(F(x,t)) = x*(2*F(x+1,t)-F(x-1,t)) and hence the row generating polynomials M(n,x) satisfy the recurrence relation
(2)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)}.
The form of the e.g.f shows that the row polynomials are a polynomial sequence of binomial type. The associated delta operator D* is given by
(3)... D* = sqrt(2)/4*log((3+2*sqrt(2))*(sqrt(2)*exp(D)-1)/(sqrt(2)*exp(D)+1)),
where D is the derivative operator d/dx. This expands to
(4)... D* = D - 3*D^2/2! + 17*D^3/3! - 147*D^4/4! + ....
The sequence of coefficients [1,3,17,147,...] is A080253.
The delta operator D* acts as a lowering operator for the minimax polynomials
(5)...(D*) M(n,x) = n*M(n-1,x).
In what follows it will be convenient to denote M(n,x) by x^[n].
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(6)... x^[p] = (-1)^((p^2-1)/8)*x (mod p).
More generally, for k = 1,2,...
(7)... x^[p+k-1] = (-1)^((p^2-1)/8)*x^[k] (mod p).
GENERALIZED BERNOULLI POLYNOMIALS ASSOCIATED WITH THE MINIMAX POLYNOMIALS
The generalized Bernoulli polynomial MB(k,x) associated with the minimax polynomial x^[k] (= M(k,x)) may be defined as the result of applying the differential operator D*/(exp(D)-1) to the polynomial x^[k]:
(8)... MB(k,x) := {D*/(exp(D)-1)} x^[k].
The first few generalized Bernoulli polynomials are
MB(0,x) = 1,
MB(1,x) = x   -  2,
MB(2,x) = x^2 -    x   + 4/3,
MB(3,x) = x^3 +  3*x^2 - 4*x,
MB(4,x) = x^4 + 10*x^3 + 3*x^2 - 14*x - 32/15.
Since exp(D)-1 is the forward difference operator it follows from (5) and (8) that
(9)... MB(k,x+1) - MB(k,x) = k*x^[k-1].
Summing (9) from x = 1 to x = n-1 and telescoping we find a closed form expression for the finite sums
(10)... 1^[p]+2^[p]+...+(n-1)^[p] = 1/(p+1)*{MB(p+1,n)-MB(p+1,1)}.
The generalized Bernoulli polynomials can be expanded in terms of the minimax polynomials x^[k]. Use (3) to express exp(D)-1 in terms of D*.
Substitute the resulting expression in (8) and expand as a power series in D* to arrive at the expansion:
(11)... MB(k,x) = -2*k*x^[k-1] + Sum_{j=0..floor(k/2)} 2^(3*j) * binomial(k,2j)*B_(2j)*x^[k-2j], where {B_j}j>=0 = [1,-1/2,1/6,0,-1/30,...] denotes the Bernoulli number sequence.
RELATION WITH OTHER SEQUENCES
Column 1 [1, 3, 10, 42, 248, ...] = A143523 with an offset of 1.
Row sums [1, 1, 4, 20, 128, 1024, ...] = A080795.

A185420 Square array, read by antidiagonals, used to recursively calculate the number of minimax trees A080795.

Original entry on oeis.org

1, 4, 1, 20, 5, 1, 128, 32, 6, 1, 1024, 256, 46, 7, 1, 9856, 2464, 432, 62, 8, 1, 110720, 27680, 4784, 662, 80, 9, 1, 1421312, 355328, 60864, 8224, 952, 100, 10, 1, 20525056, 5131264, 873664, 116128, 13048, 1308, 122, 11, 1

Offset: 1

Peter Bala, Jan 30 2011

The table entries T(n,k), for n,k>=1, are defined by means of the recurrence relation
(1)... T(n+1,k) = (2*k+2)*T(n,k+1)-(k-1)*T(n,k-1),
with boundary condition T(1,k) = 1.
The first column of the table gives A080795.
For similarly defined tables used to calculate the zigzag numbers A000111 and the Springer numbers A001586 see A185414 and A185418, respectively.
See also A185416.

			Square array begins
n\k|......1.......2.......3........4.......5.........6
======================================================
..1|......1.......1.......1........1........1........1
..2|......4.......5.......6........7........8........9
..3|.....20......32......46.......62.......80......100
..4|....128.....256.....432......662......952.....1308
..5|...1024....2464....4784.....8224....13048....19544
..6|...9856...27680...60864...116128...201632...327096
..7|.110720..355328..873664..1833728..3460640..6046720
..
Examples of recurrence relation:
T(4,3) = 432 = 8*T(3,4) - 2*T(3,2) = 8*62 - 2*32;
T(6,2) = 27680 = 6*T(5,3) - 1*T(5,1) = 6*4784 - 1*1024.

Cf. A080795, A185414, A185416, A185418, A185419.

#A185420
M := proc(n,x) option remember;
description 'minimax polynomials M(n,x)'
if n = 0
return 1
else return
x*(2*M(n-1,x+1)-M(n-1,x-1))
end proc:
for n from 1 to 10 do
seq(M(n,k)/k, k = 1..10);
end do;

M[n_, x_] := M[n, x] = If[n == 0, 1, x (2 M[n - 1, x + 1] - M[n - 1, x - 1])];
T[n_, k_] := M[n, k]/k;
Table[T[d - k + 1, k], {d, 1, 9}, {k, 1, d}] // Flatten (* Jean-François Alcover, Sep 24 2022 *)

{T(n,k)=if(n<1||k<1,0,if(n==1,1,(2*k+2)*T(n-1,k+1)-(k-1)*T(n-1,k-1)))}

(1)... T(n,k) = M(n,k)/k with M(n,x) the polynomials described in A185419.
(2)... First column: T(n,1) = A080795(n).
(3)... Second column: T(n,2) = (1/4)*A080795(n+1).

A227544 Expansion of e.g.f. 1/(1 - sin(6*x))^(1/6).

Original entry on oeis.org

1, 1, 7, 55, 721, 11761, 240247, 5801095, 162512161, 5171130721, 184337942887, 7275081518935, 314918762166001, 14834964193292881, 755507853144691927, 41362173671901329575, 2422478811455080626241, 151132171549872325122241, 10006051653759338150151367, 700695219796759105368529015

Offset: 0

Paul D. Hanna, Dec 20 2013

nonn

Generally, for e.g.f. 1/(1-sin(p*x))^(1/p) we have a(n) ~ n! * 2^(n+3/p) * p^n / (Gamma(2/p) * n^(1-2/p) * Pi^(n+2/p)). - Vaclav Kotesovec, Jan 03 2014

			E.g.f.: A(x) = 1 + x + 7*x^2/2! + 55*x^3/3! + 721*x^4/4! + 11761*x^5/5! + ...
where A(x)^3 = 1 + 3*x + 27*x^2/2! + 297*x^3/3! + 4617*x^4/4! + 87723*x^5/5! + ...
and 1/A(x)^3 = 1 - 3*x - 9*x^2/2! + 27*x^3/3! + 81*x^4/4! - 243*x^5/5! + ...
which illustrates 1/A(x)^3 = cos(3*x) - sin(3*x).
O.g.f.: 1/(1-x - 6*1*1*x^2/(1-7*x - 6*2*4*x^2/(1-13*x - 6*3*7*x^2/(1-19*x - 6*4*10*x^2/(1-25*x - 6*5*13*x^2/(1-...)))))), a continued fraction.

Cf. A001586 (p=2), A007788 (p=3), A144015 (p=4), A230134 (p=5), A235128 (p=7), A230114 (p=8).

Cf. A008542, A136630.

CoefficientList[Series[1/(1-Sin[6*x])^(1/6), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jan 03 2014 *)

{a(n)=local(X=x+x*O(x^n)); n!*polcoeff((cos(3*X)-sin(3*X))^(-1/3), n)}
for(n=0,20,print1(a(n),", "))

{a(n)=local(A=1+x+x*O(x^n)); for(i=0, n, A=exp(intformal(A^3/subst(A^3, x, -x)))); n!*polcoeff(A, n)}
for(n=0,20,print1(a(n),", "))

a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
a008542(n) = prod(k=0, n-1, 6*k+1);
a(n) = sum(k=0, n, a008542(k)*(6*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

E.g.f. A(x) satisfies:
(1) A(x) = (cos(3*x) - sin(3*x))^(-1/3).
(2) A(x)^3/A(-x)^3 = 1/cos(6*x) + tan(6*x).
(3) A(x) = exp( Integral A(x)^3/A(-x)^3 dx ).
O.g.f.: 1/G(0) where G(k) = 1 - (6*k+1)*x - 6*(k+1)*(3*k+1)*x^2/G(k+1) [continued fraction formula from A144015 due to Sergei N. Gladkovskii].
a(n) ~ n! * 2^(2*n+1/2) * 3^n / (Gamma(1/3) * n^(2/3) * Pi^(n+1/3)). - Vaclav Kotesovec, Jan 03 2014
a(n) = Sum_{k=0..n} A008542(k) * (6*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A230114 Expansion of e.g.f. 1/(1 - sin(8*x))^(1/8).

Original entry on oeis.org

1, 1, 9, 89, 1521, 32401, 869049, 27608489, 1019581281, 42824944801, 2017329504489, 105299243488889, 6032850630082641, 376363074361201201, 25396689469918450329, 1843101478742259481289, 143145930384321475601601, 11846611289341729822881601, 1040750126963789832859930569

Offset: 0

Paul D. Hanna, Dec 20 2013

nonn

Generally, for e.g.f. 1/(1-sin(p*x))^(1/p) is a(n) ~ n! * 2^(n+3/p) * p^n / (Gamma(2/p) * n^(1-2/p) * Pi^(n+2/p)). - Vaclav Kotesovec, Jan 03 2014

			E.g.f.: A(x) = 1 + x + 9*x^2/2! + 89*x^3/3! + 1521*x^4/4! + 32401*x^5/5! + ...
where A(x)^4 = 1 + 4*x + 48*x^2/2! + 704*x^3/3! + 14592*x^4/4! + 369664*x^5/5! + ...
and 1/A(x)^4 = 1 - 4*x - 16*x^2/2! + 64*x^3/3! + 256*x^4/4! - 1024*x^5/5! + ...
which illustrates 1/A(x)^4 = cos(4*x) - sin(4*x).
O.g.f.: 1/(1-x - 8*1*1*x^2/(1-9*x - 8*2*5*x^2/(1-17*x - 8*3*9*x^2/(1-25*x - 8*4*13*x^2/(1-33*x - 8*5*17*x^2/(1-...)))))), a continued fraction.

Cf. A001586 (p=2), A007788 (p=3), A144015 (p=4), A230134 (p=5), A227544 (p=6), A235128 (p=7).

Cf. A045755, A136630.

CoefficientList[Series[1/(1-Sin[8*x])^(1/8), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jan 03 2014 *)

{a(n)=local(X=x+x*O(x^n)); n!*polcoeff((cos(4*X)-sin(4*X))^(-1/4), n)}
for(n=0, 20, print1(a(n), ", "))

{a(n)=local(A=1+x+x*O(x^n)); for(i=0, n, A=exp(intformal(A^4/subst(A^4, x, -x)))); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))

a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
a045755(n) = prod(k=0, n-1, 8*k+1);
a(n) = sum(k=0, n, a045755(k)*(8*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

E.g.f. A(x) satisfies:
(1) A(x) = (cos(4*x) - sin(4*x))^(-1/4).
(2) A(x)^4/A(-x)^4 = 1/cos(8*x) + tan(8*x).
(3) A(x) = exp( Integral A(x)^4/A(-x)^4 dx ).
O.g.f.: 1/G(0) where G(k) = 1 - (8*k+1)*x - 8*(k+1)*(4*k+1)*x^2/G(k+1) [continued fraction formula from A144015 due to Sergei N. Gladkovskii].
a(n) ~ n! * 2^(4*n+3/8) / (Gamma(1/4) * n^(3/4) * Pi^(n+1/4)). - Vaclav Kotesovec, Jan 03 2014
a(n) = Sum_{k=0..n} A045755(k) * (8*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A230134 Expansion of e.g.f. 1/(1 - sin(5*x))^(1/5).

Original entry on oeis.org

1, 1, 6, 41, 456, 6301, 108576, 2207981, 52012416, 1390239481, 41593598976, 1376769180401, 49955931795456, 1971671764875541, 84095262825824256, 3854514200269774901, 188942180401957502976, 9863099585213327293681, 546266997049408050364416, 31993839349571172423492281

Offset: 0

Paul D. Hanna, Dec 20 2013

nonn

			E.g.f.: A(x) = 1 + x + 6*x^2/2! + 41*x^3/3! + 456*x^4/4! + 6301*x^5/5! +...
O.g.f.: 1/(1-x - 5*1*2/2*x^2/(1-6*x - 5*2*7/2*x^2/(1-11*x - 5*3*12/2*x^2/(1-16*x - 5*4*17/2*x^2/(1-21*x - 5*5*22/2*x^2/(1-...)))))), a continued fraction.

Cf. A001586, A007788, A144015, A227544, A230114.

Cf. A008548, A136630.

CoefficientList[Series[1/(1-Sin[5*x])^(1/5), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jan 03 2014 *)

{a(n)=local(X=x+x*O(x^n)); n!*polcoeff((1-sin(5*X))^(-1/5), n)}
for(n=0, 20, print1(a(n), ", "))

{a(n)=local(A=1+x+x*O(x^n)); for(i=0, n, A=exp(intformal(A^(5/2)/subst(A^(5/2), x, -x)))); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))

a136630(n, k) = 1/(2^k*k!)*sum(j=0, k, (-1)^(k-j)*(2*j-k)^n*binomial(k, j));
a008548(n) = prod(k=0, n-1, 5*k+1);
a(n) = sum(k=0, n, a008548(k)*(5*I)^(n-k)*a136630(n, k)); \\ Seiichi Manyama, Jun 24 2025

E.g.f. A(x) satisfies: A(x) = (cos(5*x/2) - sin(5*x/2))^(-2/5).
O.g.f.: 1/G(0) where G(k) = 1 - (5*k+1)*x - 5*(k+1)*(5*k+2)/2*x^2/G(k+1) [continued fraction formula from A144015 due to Sergei N. Gladkovskii].
a(n) ~ n! * sqrt(5+sqrt(5)) * Gamma(3/5) * 2^(n-9/10) * 5^n / (n^(3/5) * Pi^(n+7/5)). - Vaclav Kotesovec, Jan 03 2014
a(n) = Sum_{k=0..n} A008548(k) * (5*i)^(n-k) * A136630(n,k), where i is the imaginary unit. - Seiichi Manyama, Jun 24 2025

A007836 Springer numbers associated with symplectic group.

Original entry on oeis.org

1, 1, 1, 5, 23, 151, 1141, 10205, 103823, 1190191, 15151981, 212222405, 3242472023, 53670028231, 956685677221, 18271360434605, 372221031054623, 8056751598834271, 184647141575344861, 4466900836910758805

Offset: 0

N. J. A. Sloane

Comments from F. Chapoton, Oct 30 2009: To compute this sequence, I used something similar to the Boustrophedon definition of the Euler numbers, but with two triangles instead of one. This is described (page 94) in Arnold's article in "Leçons de mathématiques d'aujourd'hui, volume 1" Editions Cassini. This is very similar to A001586, except that the initial conditions ( (0,1) at top of the two triangles ) are exchanged.

V. I. Arnold, The calculus of snakes and the combinatorics of Bernoulli, Euler and Springer numbers of Coxeter groups, Uspekhi Mat. Nauk., 47 (#1, 1992), 3-45 = Russian Math. Surveys, Vol. 47 (1992), 1-51.
V. I. Arnold, Nombres d'Euler, de Bernoulli et de Springer pour les groupes de Coxeter et les espaces de morsification : le calcul des serpents, in "Leçons de mathématiques d'aujourd'hui, volume 1", Editions Cassini.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, Volume 6.1 (1999): Research paper R21, 13 p.
M. Josuat-Verges, J.-C. Novelli and J.-Y. Thibon, The algebraic combinatorics of snakes, arXiv preprint arXiv:1110.5272 [math.CO], 2011.
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011-2012.

Cf. A001586, A155100, A104035, A156142.

p[n_, u_] := D[Tan[x], {x, n}] /. Tan[x] -> u /. Sec[x] -> Sqrt[1+u^2] // Expand; p[-1, u_] = 1; t[n_, k_] := t[n, k] = k*t[n-1, k-1]+(k+1)*t[n-1, k+1]; t[0, 0] = 1; t[0, ] = 0; t[-1, ] = 0; q[n_, u_] := Sum[t[n, k]*u^k, {k, 0, n}]; a[n_] := p[n, 1]-q[n, 1]; a[0]=1; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Feb 05 2014 *)
nmax = 20; CoefficientList[Series[1 + (Sin[x] + Cos[x] - 1) / (Cos[x] - Sin[x]), {x, 0, nmax}], x] * Range[0,nmax]! (* Vaclav Kotesovec, Dec 08 2020 *)

a(n) = P_n(1) - Q_n(1) (see A155100 and A104035), defining Q_{-1} = 0. Cf. A156142.
From Vaclav Kotesovec, Dec 08 2020: (Start)
E.g.f.: (2*cos(x) - 1) / (cos(x) - sin(x)).
a(n) ~ (2 - sqrt(2)) * 2^(2*n + 3/2) * n^(n + 1/2) / (Pi^(n + 1/2) * exp(n)). (End)

More terms from F. Chapoton, Oct 30 2009

cp's OEIS Frontend

A242195 Least prime divisor of the n-th tangent number T_n which does not divide any T_k with k < n, or 1 if such a primitive prime divisor of T_n does not exist.

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A154343 S(n,k) an additive decomposition of the Springer number (generalized Euler number), (triangle read by rows).

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A185356 Triangle read by rows: number of type B alternating permutations according to their last value.

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A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

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A185419 Table of coefficients of a polynomial sequence of binomial type related to the enumeration of minimax trees A080795.

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A185420 Square array, read by antidiagonals, used to recursively calculate the number of minimax trees A080795.

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A227544 Expansion of e.g.f. 1/(1 - sin(6*x))^(1/6).

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