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A010551
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Multiply successively by 1,1,2,2,3,3,4,4,..., n >= 1, a(0) = 1.
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31
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1, 1, 1, 2, 4, 12, 36, 144, 576, 2880, 14400, 86400, 518400, 3628800, 25401600, 203212800, 1625702400, 14631321600, 131681894400, 1316818944000, 13168189440000, 144850083840000, 1593350922240000, 19120211066880000, 229442532802560000, 2982752926433280000
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OFFSET
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0,4
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COMMENTS
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Number of permutations of {1,2,...,n-1} having a single run of odd entries. Example: a(5)=12 because we have 1324,1342,3124,3142,2134,4132,2314,4312, 2413, 4213, 2431 and 4231.
a(n+1) gives the permanent of the n X n matrix whose (i,j)-element is i+j-1 modulo 2. - John W. Layman, Jan 03 2011
a(0) = 1 since it is the empty product.
A010551(n-2), n >= 2, equal to (ceiling((n-2)/2))! * (floor((n-2)/2))!, gives the number of arrangements of n-2 entries from 2 to n-1, starting with an even entry and where the parity of adjacent entries alternates. This is the number of arrangements to investigate for row n of a prime pyramid (A051237). (End)
Also size of the equivalence class of S_n containing the identity permutation under transformations of positionally adjacent elements of the form abc <--> acb where a < b < c, cf. A210667 (equivalently under such transformations of the form abc <--> bac where a < b < c.) - Tom Roby, May 15 2012
a(n) is the number of parity-alternating permutations of size n. A permutation is parity-alternating if it sends even integers to even, and odd to odd. - Per W. Alexandersson, Jun 06 2022
n divides a(n) if and only if n is not prime. Since a(n) = floor(n/2)!*floor((n+1)/2)!, if n is prime then n is not a factor of a(n). All the prime factors of a(n) are in fact less than or equal to (n+1)/2. If n is composite, then it's possible to write it as p*q with p and q less than or equal to n/2. So p and q are factors of a(n). - Davide Oliveri, Apr 01 2023
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LINKS
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FORMULA
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a(n) = floor(n/2)!*floor((n+1)/2)! is the number of permutations p of {1, 2, 3, ..., n} such that for every i, i and p(i) have the same parity, i.e., p(i) - i is even. - Avi Peretz (njk(AT)netvision.net.il), Feb 22 2001
a(n) = n!/binomial(n, floor(n/2)). - Paul Barry, Sep 12 2004
G.f.: Sum_{n>=0} x^n/a(n) = besseli(0, 2*x) + x*besseli(1, 2*x). - Paul D. Hanna, Apr 07 2005
E.g.f.: 1/(1-x/2) + (1/2)/(1-x/2)*arccos(1-x^2/2)/sqrt(1-x^2/4). - Paul D. Hanna, Aug 28 2005
G.f.: G(0) where G(k) = 1 + (k+1)*x/(1 - x*(k+1)/(x*(k+1) + 1/G(k+1) )); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 28 2012
D-finite with recurrence: 4*a(n) - 2*a(n-1) - n*(n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
a(n) = a(n-1) * (a(n-2) + a(n-3)) / a(n-3) for all n >= 3. - Michael Somos, Dec 29 2012
G.f.: 1 + x + x^2*(1 + x*(G(0) - 1)/(x-1)) where G(k) = 1 - (k+2)/(1-x/(x - 1/(1 - (k+2)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - (k+1)/(1-x/(x - 1/(1 - (k+1)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x*G(0), where G(k) = 1 + x*(k+1)/(1 - x*(k+2)/(x*(k+2) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 08 2013
G.f.: Q(0), where Q(k) = 1 + x*(k+1)/(1 - x*(k+1)/(x*(k+1) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 08 2013
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EXAMPLE
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G.f. = 1 + x + x^2 + 2*x^3 + 4*x^4 + 12*x^5 + 36*x^6 + 144*x^7 + 576*x^8 + ...
For n = 7, a(n) = 1*1*2*2*3*3*4 (7 factors), which is 144. - Michael B. Porter, Jul 03 2016
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MAPLE
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option remember;
if n <= 1 then
1
else
procname(n-1) *trunc( (n+1)/2 );
fi;
end:
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MATHEMATICA
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PROG
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(PARI) {a(n)=local(X=x+x*O(x^n)); 1/polcoeff(besseli(0, 2*X)+X*besseli(1, 2*X), n, x)} \\ Paul D. Hanna, Apr 07 2005
(Haskell)
a010551 n = a010551_list !! n
a010551_list = scanl (*) 1 a008619_list
(Magma) [Factorial(n div 2)*Factorial((n+1) div 2): n in [0..25]]; // Vincenzo Librandi Jan 17 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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