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A001250
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Number of alternating permutations of order n.
(Formerly M1235 N0472)
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24
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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
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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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.
C. Davis, Problem 4755, Amer. Math. Monthly, 64 (1957) 596; solution by W. J. Blundon, 65 (1958), 533-534.
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).
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LINKS
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Max Alekseyev and Alois P. Heinz, Table of n, a(n) for n = 0..500 (terms n=1..100 from Max Alekseyev)
Max A. Alekseyev, On the number of permutations with bounded run lengths, arXiv:1205.4581 [math.CO], 2012-2013.
D. Berry, J. Broom, D. Dixon, A. Flaherty, Umbral Calculus and the Boustrophedon Transform, 2013.
Chandler Davis, Problem 4755: A Permutation Problem, Amer. Math. Monthly, 64 (1957) 596; solution by W. J. Blundon, 65 (1958), 533-534. [Denoted by P_n in solution.] [Annotated scanned copy]
S. Kitaev, Multi-avoidance of generalized patterns, Discrete Math., 260 (2003), 89-100. (See p. 100.)
S. T. Thompson, Problem E754: Skew Ordered Sequences, Amer. Math. Monthly, 54 (1947), 416-417. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Alternating Permutation
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FORMULA
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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(A109449(n-1,k)*A000111(k): k = 0..n-1). - Reinhard Zumkeller, Sep 17 2014
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EXAMPLE
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1 + x + 2*x^2 + 4*x^3 + 10*x^4 + 32*x^5 + 122*x^6 + 544*x^7 + 2770*x^8 + ...
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MAPLE
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# 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
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MATHEMATICA
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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 *)
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PROG
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(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
(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
(PARI)
x='x+O('x^66);
egf=2*(tan(x)+1/cos(x))-2-x;
Vec(serlaplace(egf))
/* Joerg Arndt, May 28 2012 */
(Haskell)
a001250 n = if n == 1 then 1 else 2 * a000111 n
-- Reinhard Zumkeller, Sep 17 2014
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CROSSREFS
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Cf. A000111. A diagonal of A010094.
Cf. A001251, A001252, A001253, A010026, A211318, A260786.
Sequence in context: A176006 A263664 A263665 * A013032 A098830 A121277
Adjacent sequences: A001247 A001248 A001249 * A001251 A001252 A001253
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane
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EXTENSIONS
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Edited by Max Alekseyev, May 04 2012
a(0)=1 prepended by Alois P. Heinz, Nov 29 2015
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STATUS
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approved
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