cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A000111 Euler or up/down numbers: e.g.f. sec(x) + tan(x). Also for n >= 2, half the number of alternating permutations on n letters (A001250).

Original entry on oeis.org

1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521, 353792, 2702765, 22368256, 199360981, 1903757312, 19391512145, 209865342976, 2404879675441, 29088885112832, 370371188237525, 4951498053124096, 69348874393137901, 1015423886506852352, 15514534163557086905, 246921480190207983616, 4087072509293123892361
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Number of linear extensions of the "zig-zag" poset. See ch. 3, prob. 23 of Stanley. - Mitch Harris, Dec 27 2005
Number of increasing 0-1-2 trees on n vertices. - David Callan, Dec 22 2006
Also the number of refinements of partitions. - Heinz-Richard Halder (halder.bichl(AT)t-online.de), Mar 07 2008
The ratio a(n)/n! is also the probability that n numbers x1,x2,...,xn randomly chosen uniformly and independently in [0,1] satisfy x1 > x2 < x3 > x4 < ... xn. - Pietro Majer, Jul 13 2009
For n >= 2, a(n-2) = number of permutations w of an ordered n-set {x_1 < ... x_n} with the following properties: w(1) = x_n, w(n) = x_{n-1}, w(2) > w(n-1), and neither any subword of w, nor its reversal, has the first three properties. The count is unchanged if the third condition is replaced with w(2) < w(n-1). - Jeremy L. Martin, Mar 26 2010
A partition of zigzag permutations of order n+1 by the smallest or the largest, whichever is behind. This partition has the same recurrent relation as increasing 1-2 trees of order n, by induction the bijection follows. - Wenjin Woan, May 06 2011
As can be seen from the asymptotics given in the FORMULA section, one has lim_{n->oo} 2*n*a(n-1)/a(n) = Pi; see A132049/A132050 for the simplified fractions. - M. F. Hasler, Apr 03 2013
a(n+1) is the sum of row n in triangle A008280. - Reinhard Zumkeller, Nov 05 2013
M. Josuat-Verges, J.-C. Novelli and J.-Y. Thibon (2011) give a far-reaching generalization of the bijection between Euler numbers and alternating permutations. - N. J. A. Sloane, Jul 09 2015
Number of treeshelves avoiding pattern T321. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link, see A278678 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016
Number of sequences (e(1), ..., e(n-1)), 0 <= e(i) < i, such that no three terms are equal. [Theorem 7 of Corteel, Martinez, Savage, and Weselcouch] - Eric M. Schmidt, Jul 17 2017
Number of self-dual edge-labeled trees with n vertices under "mind-body" duality. Also number of self-dual rooted edge-labeled trees with n vertices. See my paper linked below. - Nikos Apostolakis, Aug 01 2018
The ratio a(n)/n! is the volume of the convex polyhedron defined as the set of (x_1,...,x_n) in [0,1]^n such that x_i + x_{i+1} <= 1 for every 1 <= i <= n-1; see the solutions by Macdonald and Nelsen to the Amer. Math. Monthly problem referenced below. - Sanjay Ramassamy, Nov 02 2018
Number of total cyclic orders on {0,1,...,n} such that the triple (i-1,i,i+1) is positively oriented for every 1 <= i <= n-1; see my paper on cyclic orders linked below. - Sanjay Ramassamy, Nov 02 2018
The number of binary, rooted, unlabeled histories with n+1 leaves (following the definition of Rosenberg 2006). Also termed Tajima trees, Tajima genealogies, or binary, rooted, unlabeled ranked trees (Palacios et al. 2015). See Disanto & Wiehe (2013) for a proof. - Noah A Rosenberg, Mar 10 2019
From Gus Wiseman, Dec 31 2019: (Start)
Also the number of non-isomorphic balanced reduced multisystems with n + 1 distinct atoms and maximum depth. A balanced reduced multisystem is either a finite multiset, or a multiset partition with at least two parts, not all of which are singletons, of a balanced reduced multisystem. The labeled version is A006472. For example, non-isomorphic representatives of the a(0) = 1 through a(4) = 5 multisystems are (commas elided):
{1} {12} {{1}{23}} {{{1}}{{2}{34}}} {{{{1}}}{{{2}}{{3}{45}}}}
{{{12}}{{3}{4}}} {{{{1}}}{{{23}}{{4}{5}}}}
{{{{1}{2}}}{{{3}}{{45}}}}
{{{{1}{23}}}{{{4}}{{5}}}}
{{{{12}}}{{{3}}{{4}{5}}}}
Also the number of balanced reduced multisystems with n + 1 equal atoms and maximum depth. This is possibly the meaning of Heinz-Richard Halder's comment (see also A002846, A213427, A265947). The non-maximum-depth version is A318813. For example, the a(0) = 1 through a(4) = 5 multisystems are (commas elided):
{1} {11} {{1}{11}} {{{1}}{{1}{11}}} {{{{1}}}{{{1}}{{1}{11}}}}
{{{11}}{{1}{1}}} {{{{1}}}{{{11}}{{1}{1}}}}
{{{{1}{1}}}{{{1}}{{11}}}}
{{{{1}{11}}}{{{1}}{{1}}}}
{{{{11}}}{{{1}}{{1}{1}}}}
(End)
With s_n denoting the sum of n independent uniformly random numbers chosen from [-1/2,1/2], the probability that the closest integer to s_n is even is exactly 1/2 + a(n)/(2*n!). (See Hambardzumyan et al. 2023, Appendix B.) - Suhail Sherif, Mar 31 2024
The number of permutations of size n+1 that require exactly n passes through a stack (i.e. have reverse-tier n-1) with an algorithm that prioritizes outputting the maximum possible prefix of the identity in a given pass and reverses the remainder of the permutation for prior to the next pass. - Rebecca Smith, Jun 05 2024

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 5*x^4 + 16*x^5 + 61*x^6 + 272*x^7 + 1385*x^8 + ...
Sequence starts 1,1,2,5,16,... since possibilities are {}, {A}, {AB}, {ACB, BCA}, {ACBD, ADBC, BCAD, BDAC, CDAB}, {ACBED, ADBEC, ADCEB, AEBDC, AECDB, BCAED, BDAEC, BDCEA, BEADC, BECDA, CDAEB, CDBEA, CEADB, CEBDA, DEACB, DEBCA}, etc. - _Henry Bottomley_, Jan 17 2001

References

  • M. D. Atkinson: Partial orders and comparison problems, Sixteenth Southeastern Conference on Combinatorics, Graph Theory and Computing, (Boca Raton, Feb 1985), Congressus Numerantium 47, 77-88.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 34, 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 258-260, section #11.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 110.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 262.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 66.
  • O. Heimo and A. Karttunen, Series help-mates in 8, 9 and 10 moves (Problems 2901, 2974-2976), Suomen Tehtavaniekat (Proceedings of the Finnish Chess Problem Society) vol. 60, no. 2/2006, pp. 75, 77.
  • L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 238.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • E. Netto, Lehrbuch der Combinatorik. 2nd ed., Teubner, Leipzig, 1927, p. 110.
  • C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 184.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1997 and Vol. 2, 1999; see Problem 5.7.

Links

Crossrefs

Cf. A000364 (secant numbers), A000182 (tangent numbers).
Related triangles: A008280, A008281, A008282, A008970, A010094, A059720, A177762.
Cf. A181937 for n-alternating permutations.
Cf. A109449 for an extension to an exponential Riordan array.
Column k=1 of A010094, A229892, A258829, A262124, A275784.
Column k=2 of A250261.
Cf. A002105, A186365.
For 0-1-2 trees with n nodes and k leaves, see A301344.
Matula-Goebel numbers of 0-1-2 trees are A292050.
An overview over generalized Euler numbers gives A349264.
Cf. A000258, A000311, A002846, A005121, A318813, A320270, A330474, A330665, A330679, A340315.

Programs

  • Haskell
    a000111 0 = 1
a000111 n = sum $ a008280_row (n - 1)
-- Reinhard Zumkeller, Nov 01 2013
  • Maple
    A000111 := n-> n!*coeff(series(sec(x)+tan(x),x,n+1), x, n);
s := series(sec(x)+tan(x), x, 100): A000111 := n-> n!*coeff(s, x, n);
A000111:=n->piecewise(n mod 2=1,(-1)^((n-1)/2)*2^(n+1)*(2^(n+1)-1)*bernoulli(n+1)/(n+1),(-1)^(n/2)*euler(n)):seq(A000111(n),n=0..30); A000111:=proc(n) local k: k:=floor((n+1)/2): if n mod 2=1 then RETURN((-1)^(k-1)*2^(2*k)*(2^(2*k)-1)*bernoulli(2*k)/(2*k)) else RETURN((-1)^k*euler(2*k)) fi: end:seq(A000111(n),n=0..30); (C. Ronaldo)
T := n -> 2^n*abs(euler(n,1/2)+euler(n,1)): # Peter Luschny, Jan 25 2009
S := proc(n,k) option remember; if k=0 then RETURN(`if`(n=0,1,0)) fi; S(n,k-1)+S(n-1,n-k) end:
A000364 := n -> S(2*n,2*n);
A000182 := n -> S(2*n+1,2*n+1);
A000111 := n -> S(n,n); # Peter Luschny, Jul 29 2009
a := n -> 2^(n+2)*n!*(sum(1/(4*k+1)^(n+1), k = -infinity..infinity))/Pi^(n+1):
1, seq(a(n), n = 1..22); # Emeric Deutsch, Aug 17 2009
# alternative Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 29 2015
  • Mathematica
    n=22; CoefficientList[Series[(1+Sin[x])/Cos[x], {x, 0, n}], x] * Table[k!, {k, 0, n}] (* Jean-François Alcover, May 18 2011, after Michael Somos *)
a[n_] := If[EvenQ[n], Abs[EulerE[n]], Abs[(2^(n+1)*(2^(n+1)-1)*BernoulliB[n+1])/(n+1)]]; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Oct 09 2012, after C. Ronaldo *)
ee = Table[ 2^n*EulerE[n, 1] + EulerE[n] - 1, {n, 0, 26}]; Table[ Differences[ee, n] // First // Abs, {n, 0, 26}] (* Jean-François Alcover, Mar 21 2013, after Paul Curtz *)
a[ n_] := If[ n < 0, 0, (2 I)^n If[ EvenQ[n], EulerE[n, 1/2], EulerE[n, 0] I]]; (* Michael Somos, Aug 15 2015 *)
a[ n_] := If[ n < 1, Boole[n == 0], With[{m = n - 1}, m! SeriesCoefficient[ 1 / (1 - Sin[x]), {x, 0, m}]]]; (* Michael Somos, Aug 15 2015 *)
s[0] = 1; s[] = 0; t[n, 0] := s[n]; t[n_, k_] := t[n, k] = t[n, k-1] + t[n-1, n-k]; a[n_] := t[n, n]; Array[a, 30, 0](* Jean-François Alcover, Feb 12 2016 *)
a[n_] := If[n == 0, 1, 2*Abs[PolyLog[-n, I]]]; (* Jean-François Alcover, Dec 02 2023, after M. F. Hasler *)
a[0] := 1; a[1] := 1; a[n_] := a[n] = Sum[Binomial[n - 2, k] a[k] a[n - 1 - k], {k, 0, n - 2}]; Map[a, Range[0, 26]] (* Oliver Seipel, May 24 2024 after Peter Bala *)
a[0] := 1; a[1] := 1; a[n_] := a[n] = 1/(n (n-1)) Sum[a[n-1-k] a[k] k, {k, 1, n-1}]; Map[#! a[#]&, Range[0, 26]] (* Oliver Seipel, May 27 2024 *)
  • Maxima
    a(n):=sum((if evenp(n+k) then (-1)^((n+k)/2)*sum(j!*stirling2(n,j)*2^(1-j)*(-1)^(n+j-k)*binomial(j-1,k-1),j,k,n) else 0),k,1,n); /* Vladimir Kruchinin, Aug 19 2010 */
  • Maxima
    a(n):=if n<2 then 1 else 2*sum(4^m*(sum((i-(n-1)/2)^(n-1)*binomial(n-2*m-1,i-m)*(-1)^(n-i-1),i,m,(n-1)/2)),m,0,(n-2)/2); /* Vladimir Kruchinin, Aug 09 2011 */
  • PARI
    {a(n) = if( n<1, n==0, n--; n! * polcoeff( 1 / (1 - sin(x + x * O(x^n))), n))}; \\ Michael Somos, Feb 03 2004
  • PARI
    {a(n) = local(v=[1], t); if( n<0, 0, for(k=2, n+2, t=0; v = vector(k, i, if( i>1, t+= v[k+1-i]))); v[2])}; \\ Michael Somos, Feb 03 2004
  • PARI
    {a(n) = local(an); if( n<1, n>=0, an = vector(n+1, m, 1); for( m=2, n, an[m+1] = sum( k=0, m-1, binomial(m-1, k) * an[k+1] * an[m-k]) / 2); an[n+1])}; \\ Michael Somos, Feb 03 2004
  • PARI
    z='z+O('z^66); egf = (1+sin(z))/cos(z); Vec(serlaplace(egf)) \\ Joerg Arndt, Apr 30 2011
  • PARI
    A000111(n)={my(k);sum(m=0,n\2,(-1)^m*sum(j=0,k=n+1-2*m,binomial(k,j)*(-1)^j*(k-2*j)^(n+1))/k>>k)}  \\ M. F. Hasler, May 19 2012
  • PARI
    A000111(n)=if(n,2*abs(polylog(-n,I)),1)  \\ M. F. Hasler, May 20 2012
  • Python
    # requires python 3.2 or higher
from itertools import accumulate
A000111_list, blist = [1,1], [1]
for n in range(10**2):
    blist = list(reversed(list(accumulate(reversed(blist))))) + [0] if n % 2 else [0]+list(accumulate(blist))
    A000111_list.append(sum(blist)) # Chai Wah Wu, Jan 29 2015
  • Python
    from mpmath import *
mp.dps = 150
l = chop(taylor(lambda x: sec(x) + tan(x), 0, 26))
[int(fac(i) * li) for i, li in enumerate(l)]  # Indranil Ghosh, Jul 06 2017
  • Python
    from sympy import bernoulli, euler
def A000111(n): return abs(((1<Chai Wah Wu, Nov 13 2024
  • Sage
    # Algorithm of L. Seidel (1877)
def A000111_list(n) :
    R = []; A = {-1:0, 0:1}; k = 0; e = 1
    for i in (0..n) :
        Am = 0; A[k + e] = 0; e = -e
        for j in (0..i) : Am += A[k]; A[k] = Am; k += e
        R.append(Am)
    return R
A000111_list(22) # Peter Luschny, Mar 31 2012 (revised Apr 24 2016)

Formula

E.g.f.: (1+sin(x))/cos(x) = tan(x) + sec(x).
E.g.f. for a(n+1) is 1/(cos(x/2) - sin(x/2))^2 = 1/(1-sin(x)) = d/dx(sec(x) + tan(x)).
E.g.f. A(x) = -log(1-sin(x)), for a(n+1). - Vladimir Kruchinin, Aug 09 2010
O.g.f.: A(x) = 1+x/(1-x-x^2/(1-2*x-3*x^2/(1-3*x-6*x^2/(1-4*x-10*x^2/(1-... -n*x-(n*(n+1)/2)*x^2/(1- ...)))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
E.g.f. A(x) = y satisfies 2y' = 1 + y^2. - Michael Somos, Feb 03 2004
a(n) = P_n(0) + Q_n(0) (see A155100 and A104035), defining Q_{-1} = 0. Cf. A156142.
2*a(n+1) = Sum_{k=0..n} binomial(n, k)*a(k)*a(n-k).
Asymptotics: a(n) ~ 2^(n+2)*n!/Pi^(n+1). For a proof, see for example Flajolet and Sedgewick.
a(n) = (n-1)*a(n-1) - Sum_{i=2..n-2} (i-1)*E(n-2, n-1-i), where E are the Entringer numbers A008281. - Jon Perry, Jun 09 2003
a(2*k) = (-1)^k euler(2k) and a(2k-1) = (-1)^(k-1)2^(2k)(2^(2k)-1) Bernoulli(2k)/(2k). - C. Ronaldo (aga_new_ac(AT)hotmail.com), Jan 17 2005
|a(n+1) - 2*a(n)| = A000708(n). - Philippe Deléham, Jan 13 2007
a(n) = 2^n|E(n,1/2) + E(n,1)| where E(n,x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
a(n) = 2^(n+2)*n!*S(n+1)/(Pi)^(n+1), where S(n) = Sum_{k = -inf..inf} 1/(4k+1)^n (see the Elkies reference). - Emeric Deutsch, Aug 17 2009
a(n) = i^(n+1) Sum_{k=1..n+1} Sum_{j=0..k} binomial(k,j)(-1)^j (k-2j)^(n+1) (2i)^(-k) k^{-1}. - Ross Tang (ph.tchaa(AT)gmail.com), Jul 28 2010
a(n) = sum((if evenp(n+k) then (-1)^((n+k)/2)*sum(j!*Stirling2(n,j)*2^(1-j)*(-1)^(n+j-k)*binomial(j-1,k-1),j,k,n) else 0),k,1,n), n>0. - Vladimir Kruchinin, Aug 19 2010
If n==1(mod 4) is prime, then a(n)==1(mod n); if n==3(mod 4) is prime, then a(n)==-1(mod n). - Vladimir Shevelev, Aug 31 2010
For m>=0, a(2^m)==1(mod 2^m); If p is prime, then a(2*p)==1(mod 2*p). - Vladimir Shevelev, Sep 03 2010
From Peter Bala, Jan 26 2011: (Start)
a(n) = A(n,i)/(1+i)^(n-1), where i = sqrt(-1) and {A(n,x)}n>=1 = [1,1+x,1+4*x+x^2,1+11*x+11*x^2+x^3,...] denotes the sequence of Eulerian polynomials.
Equivalently, a(n) = i^(n+1)*Sum_{k=1..n} (-1)^k*k!*Stirling2(n,k) * ((1+i)/2)^(k-1) = i^(n+1)*Sum_{k = 1..n} (-1)^k*((1+i)/2)^(k-1)* Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^n.
This explicit formula for a(n) can be used to obtain congruence results. For example, for odd prime p, a(p) = (-1)^((p-1)/2) (mod p), as noted by Vladimir Shevelev above.
For the corresponding type B results see A001586. For the corresponding results for plane increasing 0-1-2 trees see A080635.
For generalized Eulerian, Stirling and Bernoulli numbers associated with the zigzag numbers see A145876, A147315 and A185424, respectively. For a recursive triangle to calculate a(n) see A185414.
(End)
a(n) = I^(n+1)*2*Li_{-n}(-I) for n > 0. Li_{s}(z) is the polylogarithm. - Peter Luschny, Jul 29 2011
a(n) = 2*Sum_{m=0..(n-2)/2} 4^m*(Sum_{i=m..(n-1)/2} (i-(n-1)/2)^(n-1)*binomial(n-2*m-1,i-m)*(-1)^(n-i-1)), n > 1, a(0)=1, a(1)=1. - Vladimir Kruchinin, Aug 09 2011
a(n) = D^(n-1)(1/(1-x)) evaluated at x = 0, where D is the operator sqrt(1-x^2)*d/dx. Cf. A006154. a(n) equals the alternating sum of the nonzero elements of row n-1 of A196776. This leads to a combinatorial interpretation for a(n); for example, a(4*n+2) gives the number of ordered set partitions of 4*n+1 into k odd-sized blocks, k = 1 (mod 4), minus the number of ordered set partitions of 4*n+1 into k odd-sized blocks, k = 3 (mod 4). Cf A002017. - Peter Bala, Dec 06 2011
From Sergei N. Gladkovskii, Nov 14 2011 - Dec 23 2013: (Start)
Continued fractions:
E.g.f.: tan(x) + sec(x) = 1 + x/U(0); U(k) = 4k+1-x/(2-x/(4k+3+x/(2+x/U(k+1)))).
E.g.f.: for a(n+1) is E(x) = 1/(1-sin(x)) = 1 + x/(1 - x + x^2/G(0)); G(k) = (2*k+2)*(2*k+3)-x^2+(2*k+2)*(2*k+3)*x^2/G(k+1).
E.g.f.: for a(n+1) is E(x) = 1/(1-sin(x)) = 1/(1 - x/(1 + x^2/G(0))) ; G(k) = 8*k+6-x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)).
E.g.f.: for a(n+1) is E(x) = 1/(1 - sin(x)) = 1/(1 - x*G(0)); G(k) = 1 - x^2/(2*(2*k+1)*(4*k+3) - 2*x^2*(2*k+1)*(4*k+3)/(x^2 - 4*(k+1)*(4*k+5)/G(k+1))).
E.g.f.: for a(n+1) is E(x) = 1/(1 - sin(x)) = 1/(1 - x*G(0)) where G(k)= 1 - x^2/( (2*k+1)*(2*k+3) - (2*k+1)*(2*k+3)^2/(2*k+3 - (2*k+2)/G(k+1))).
E.g.f.: tan(x) + sec(x) = 1 + 2*x/(U(0)-x) where U(k) = 4k+2 - x^2/U(k+1).
E.g.f.: tan(x) + sec(x) = 1 + 2*x/(2*U(0)-x) where U(k) = 4*k+1 - x^2/(16*k+12 - x^2/U(k+1)).
E.g.f.: tan(x) + sec(x) = 4/(2-x*G(0))-1 where G(k) = 1 - x^2/(x^2 - 4*(2*k+1)*(2*k+3)/G(k+1)).
G.f.: 1 + x/Q(0), m=+4, u=x/2, where Q(k) = 1 - 2*u*(2*k+1) - m*u^2*(k+1)*(2*k+1)/(1 - 2*u*(2*k+2) - m*u^2*(k+1)*(2*k+3)/Q(k+1)).
G.f.: conjecture: 1 + T(0)*x/(1-x), where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-x*(k+1))*(1-x*(k+2))/T(k+1)).
E.g.f.: 1+ 4*x/(T(0) - 2*x), where T(k) = 4*(2*k+1) - 4*x^2/T(k+1):
E.g.f.: T(0)-1, where T(k) = 2 + x/(4*k+1 - x/(2 - x/( 4*k+3 + x/T(k+1)))). (End)
E.g.f.: tan(x/2 + Pi/4). - Vaclav Kotesovec, Nov 08 2013
Asymptotic expansion: 4*(2*n/(Pi*e))^(n+1/2)*exp(1/2+1/(12*n) -1/(360*n^3) + 1/(1260*n^5) - ...). (See the Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 10 2015: (Start)
The e.g.f. A(x) = tan(x) + sec(x) satisfies A''(x) = A(x)*A'(x), hence the recurrence a(0) = 1, a(1) = 1, else a(n) = Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i).
Note, the same recurrence, but with the initial conditions a(0) = 0 and a(1) = 1, produces the sequence [0,1,0,1,0,4,0,34,0,496,...], an aerated version of A002105. (End)
a(n) = A186365(n)/n for n >= 1. - Anton Zakharov, Aug 23 2016
From Peter Luschny, Oct 27 2017: (Start)
a(n) = abs(2*4^n*(H(((-1)^n - 3)/8, -n) - H(((-1)^n - 7)/8, -n))) where H(z, r) are the generalized harmonic numbers.
a(n) = (-1)^binomial(n + 1, 2)*2^(2*n + 1)*(zeta(-n, 1 + (1/8)*(-7 + (-1)^n)) - zeta(-n, 1 + (1/8)*(-3 + (-1)^n))). (End)
a(n) = i*(i^n*Li_{-n}(-i) - (-i)^n*Li_{-n}(i)), where i is the imaginary unit and Li_{s}(z) is the polylogarithm. - Peter Luschny, Aug 28 2020
Sum_{n>=0} 1/a(n) = A340315. - Amiram Eldar, May 29 2021
a(n) = n!*Re([x^n](1 + I^(n^2 - n)*(2 - 2*I)/(exp(x) + I))). - Peter Luschny, Aug 09 2021

Extensions

Edited by M. F. Hasler, Apr 04 2013
Title corrected by Geoffrey Critzer, May 18 2013

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