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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.

A001250 Number of alternating permutations of order n.

Original entry on oeis.org

1, 1, 2, 4, 10, 32, 122, 544, 2770, 15872, 101042, 707584, 5405530, 44736512, 398721962, 3807514624, 38783024290, 419730685952, 4809759350882, 58177770225664, 740742376475050, 9902996106248192, 138697748786275802, 2030847773013704704, 31029068327114173810
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

For n>1, a(n) is the number of permutations of order n with the length of longest run equal 2.
Boustrophedon transform of the Euler numbers (A000111). [Berry et al., 2013] - N. J. A. Sloane, Nov 18 2013
Number of inversion sequences of length n where all consecutive subsequences i,j,k satisfy i >= j < k or i < j >= k. a(4) = 10: 0010, 0011, 0020, 0021, 0022, 0101, 0102, 0103, 0112, 0113. - Alois P. Heinz, Oct 16 2019

Examples

			1 + x + 2*x^2 + 4*x^3 + 10*x^4 + 32*x^5 + 122*x^6 + 544*x^7 + 2770*x^8 + ...
From _Gus Wiseman_, Jun 21 2021: (Start)
The a(0) = 1 through a(4) = 10 permutations:
  ()  (1)  (1,2)  (1,3,2)  (1,3,2,4)
           (2,1)  (2,1,3)  (1,4,2,3)
                  (2,3,1)  (2,1,4,3)
                  (3,1,2)  (2,3,1,4)
                           (2,4,1,3)
                           (3,1,4,2)
                           (3,2,4,1)
                           (3,4,1,2)
                           (4,1,3,2)
                           (4,2,3,1)
(End)

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 261.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 262.
  • 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).

Links

Crossrefs

Cf. A000111. A diagonal of A010094.
Cf. A001251, A001252, A001253, A010026, A211318, A260786.
The version for permutations of prime indices is A345164.
The version for compositions is A025047, ranked by A345167.
The version for patterns is A345194.
A049774 counts permutations avoiding adjacent (1,2,3).
A344614 counts compositions avoiding adjacent (1,2,3) and (3,2,1).
A344615 counts compositions avoiding the weak adjacent pattern (1,2,3).
A344654 counts partitions without a wiggly permutation, ranked by A344653.
A345170 counts partitions with a wiggly permutation, ranked by A345172.
A345192 counts non-wiggly compositions, ranked by A345168.
Cf. A000041, A003242, A032020, A056986, A261962, A325534, A325535, A335452, A344652, A344740, A345165.
Row sums of A104345.

Programs

  • Haskell
    a001250 n = if n == 1 then 1 else 2 * a000111 n
-- Reinhard Zumkeller, Sep 17 2014
  • Maple
    # With Eulerian polynomials:
A := (n, x) -> `if`(n<2, 1/2/(1+I)^(1-n), add(add((-1)^j*binomial(n+1, j)*(m+1-j)^n, j=0..m)*x^m, m=0..n-1)):
A001250 := n -> 2*(I-1)^(1-n)*exp(I*(n-1)*Pi/2)*A(n,I);
seq(A001250(i), i=0..22); # Peter Luschny, May 27 2012
# second 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-> `if`(n<2, 1, 2)*b(n, 0):
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 29 2015
  • Mathematica
    a[n_] := 4*Abs[PolyLog[-n, I]]; a[0] = a[1] = 1; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 09 2016, after M. F. Hasler *)
Table[Length[Select[Permutations[Range[n]],And@@(!(OrderedQ[#]||OrderedQ[Reverse[#]])&/@Partition[#,3,1])&]],{n,8}] (* Gus Wiseman, Jun 21 2021 *)
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}]; Join[{a[0], a[1]}, Map[2 #! a[#]&, Range[2,24]]] (* Oliver Seipel, May 27 2024 *)
  • PARI
    {a(n) = local(v=[1], t); if( n<0, 0, for( k=2, n+3, t=0; v = vector( k, i, if( i>1, t += v[k+1 - i]))); v[3])} /* Michael Somos, Feb 03 2004 */
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( (tan(x + x * O(x^n)) + 1 / cos(x + x * O(x^n)))^2, n))} /* Michael Somos, Feb 05 2011 */
  • PARI
    A001250(n)=sum(m=0,n\2,my(k);(-1)^m*sum(j=0,k=n+1-2*m,binomial(k,j)*(-1)^j*(k-2*j)^(n+1))/k>>k)*2-(n==1)  \\ M. F. Hasler, May 19 2012
  • PARI
    A001250(n)=4*abs(polylog(-n,I))-(n==1)  \\ M. F. Hasler, May 20 2012
  • PARI
    x='x+O('x^66);
egf=2*(tan(x)+1/cos(x))-2-x;
Vec(serlaplace(egf))
/* Joerg Arndt, May 28 2012 */
  • Python
    from itertools import accumulate, islice
def A001250_gen(): # generator of terms
    yield from (1,1)
    blist = (0,2)
    while True:
        yield (blist := tuple(accumulate(reversed(blist),initial=0)))[-1]
A001250_list = list(islice(A001250_gen(),40)) # Chai Wah Wu, Jun 09-11 2022
  • Python
    from sympy import bernoulli, euler
def A001250(n): return 1 if n<2 else abs(((1<Chai Wah Wu, Nov 13 2024
  • Sage
    # Algorithm of L. Seidel (1877)
def A001250_list(n) :
    R = [1]; A = {-1:0, 0:2}; 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
        if i > 1 : R.append(A[-i//2] if i%2 == 0 else A[i//2])
    return R
A001250_list(22) # Peter Luschny, Mar 31 2012

Formula

a(n) = coefficient of x^(n-1)/(n-1)! in power series expansion of (tan(x) + sec(x))^2 = (tan(x)+1/cos(x))^2.
a(n) = coefficient of x^n/n! in power series expansion of 2*(tan(x) + sec(x)) - 2 - x. - Michael Somos, Feb 05 2011
For n>1, a(n) = 2 * A000111(n). - Michael Somos, Mar 19 2011
a(n) = 4*|Li_{-n}(i)| - [n=1] = Sum_{m=0..n/2} (-1)^m*2^(1-k)*Sum_{j=0..k} binomial(k,j)*(-1)^j*(k-2*j)^(n+1)/k - [n=1], where k = k(m) = n+1-2*m and [n=1] equals 1 if n=1 and zero else; Li denotes the polylogarithm (and i^2 = -1). - M. F. Hasler, May 20 2012
From Sergei N. Gladkovskii, Jun 18 2012: (Start)
Let E(x) = 2/(1-sin(x))-1 (essentially the e.g.f.), then
E(x) = -1 + 2*(-1/x + 1/(1-x)/x - x^3/((1-x)*((1-x)*G(0) + x^2))) where G(k) = (2*k+2)*(2*k+3)-x^2+(2*k+2)*(2*k+3)*x^2/G(k+1); (continued fraction, Euler's 1st kind, 1-step).
E(x) = -1 + 2*(-1/x + 1/(1-x)/x - x^3/((1-x)*((1-x)*G(0) + x^2))) where G(k) = 8*k + 6 - x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)); (continued fraction, Euler's 2nd kind, 2-step).
E(x) = (tan(x) + sec(x))^2 = -1 + 2/(1-x*G(0)) where 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))); (continued fraction, 3rd kind, 3-step).
(End)
G.f.: conjecture: 2*T(0)/(1-x) -1, 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))/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2013
a(n) ~ 2^(n+3) * n! / Pi^(n+1). - Vaclav Kotesovec, Sep 06 2014
a(n) = Sum_{k=0..n-1} A109449(n-1,k)*A000111(k). - Reinhard Zumkeller, Sep 17 2014

Extensions

Edited by Max Alekseyev, May 04 2012
a(0)=1 prepended by Alois P. Heinz, Nov 29 2015

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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