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A001250
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Number of alternating permutations of order n.
(Formerly M1235 N0472)
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122
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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.
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
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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.
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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C. Davis, Problem 4755, Amer. Math. Monthly, 64 (1957) 596; solution by W. J. Blundon, 65 (1958), 533-534.
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]
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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
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
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
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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 + ...
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)
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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);
# 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):
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MATHEMATICA
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Table[Length[Select[Permutations[Range[n]], And@@(!(OrderedQ[#]||OrderedQ[Reverse[#]])&/@Partition[#, 3, 1])&]], {n, 8}] (* Gus Wiseman, Jun 21 2021 *)
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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
(Sage) # Algorithm of L. Seidel (1877)
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
(PARI)
x='x+O('x^66);
egf=2*(tan(x)+1/cos(x))-2-x;
Vec(serlaplace(egf))
(Haskell)
a001250 n = if n == 1 then 1 else 2 * a000111 n
(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]
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CROSSREFS
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The version for permutations of prime indices is A345164.
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.
Cf. A000041, A003242, A032020, A056986, A261962, A325534, A325535, A335452, A344652, A344740, A345165.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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