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Multiply successively by 1,1,2,2,3,3,4,4,..., n >= 1, a(0) = 1.
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%I #171 Aug 25 2024 11:23:57

%S 1,1,1,2,4,12,36,144,576,2880,14400,86400,518400,3628800,25401600,

%T 203212800,1625702400,14631321600,131681894400,1316818944000,

%U 13168189440000,144850083840000,1593350922240000,19120211066880000,229442532802560000,2982752926433280000

%N Multiply successively by 1,1,2,2,3,3,4,4,..., n >= 1, a(0) = 1.

%C From _Emeric Deutsch_, Dec 14 2008: (Start)

%C 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.

%C a(n) = A152666(n-1,1). (End)

%C 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

%C From _Daniel Forgues_, May 20 2011: (Start)

%C a(0) = 1 since it is the empty product.

%C 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)

%C Partial products of A008619. - _Reinhard Zumkeller_, Apr 02 2012

%C 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

%C Row sums of A246117. - _Peter Bala_, Aug 15 2014

%C 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

%C 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

%C Number of permutations of {1, 2, ..., n-1} where each entry is not greater than twice the previous entry. - _Dewangga Putra Sheradhien_, Jul 13 2024

%H Reinhard Zumkeller, <a href="/A010551/b010551.txt">Table of n, a(n) for n = 0..500</a>

%H Edinah K. Gnang and Isaac Wass, <a href="http://arxiv.org/abs/1808.05551">Growing Graceful and Harmonious Trees</a>, arXiv:1808.05551 [math.CO], 2018-2020. See proposition 1.

%H Frether Getachew Kebede and Fanja Rakotondrajao, <a href="https://arxiv.org/abs/2101.09125">Parity alternating permutations starting with an odd integer</a>, arXiv:2101.09125 [math.CO], 2021.

%H Steven Linton, James Propp, Tom Roby, and Julian West, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Roby/roby4.html">Equivalence Classes of Permutations under Various Relations Generated by Constrained Transpositions</a>, Journal of Integer Sequences, Vol. 15 (2012), #12.9.1.

%F 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

%F a(n) = n!/binomial(n, floor(n/2)). - _Paul Barry_, Sep 12 2004

%F G.f.: Sum_{n>=0} x^n/a(n) = besseli(0, 2*x) + x*besseli(1, 2*x). - _Paul D. Hanna_, Apr 07 2005

%F 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

%F 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

%F D-finite with recurrence: 4*a(n) - 2*a(n-1) - n*(n-1)*a(n-2) = 0. - _R. J. Mathar_, Dec 03 2012

%F a(n) = a(n-1) * (a(n-2) + a(n-3)) / a(n-3) for all n >= 3. - _Michael Somos_, Dec 29 2012

%F 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

%F 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

%F 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

%F 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

%F Sum_{n >= 1} 1/a(n) = A130820. - _Peter Bala_, Jul 02 2016

%F a(n) ~ sqrt(Pi*n) * n! / 2^(n + 1/2). - _Vaclav Kotesovec_, Oct 02 2018

%F Sum_{n>=0} (-1)^n/a(n) = A229020. - _Amiram Eldar_, Apr 12 2021

%e G.f. = 1 + x + x^2 + 2*x^3 + 4*x^4 + 12*x^5 + 36*x^6 + 144*x^7 + 576*x^8 + ...

%e For n = 7, a(n) = 1*1*2*2*3*3*4 (7 factors), which is 144. - _Michael B. Porter_, Jul 03 2016

%p A010551 := proc(n)

%p option remember;

%p if n <= 1 then

%p 1

%p else

%p procname(n-1) *trunc( (n+1)/2 );

%p fi;

%p end:

%t FoldList[ Times, 1, Flatten@ Array[ {#, #} &, 11]] (* _Robert G. Wilson v_, Jul 14 2010 *)

%o (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

%o (PARI) A010551(n)=(n\2)!*((n+1)\2)! \\ _Michael Somos_, Dec 29 2012, edited by _M. F. Hasler_, Nov 26 2017

%o (Haskell)

%o a010551 n = a010551_list !! n

%o a010551_list = scanl (*) 1 a008619_list

%o -- _Reinhard Zumkeller_, Apr 02 2012

%o (Magma) [Factorial(n div 2)*Factorial((n+1) div 2): n in [0..25]]; // _Vincenzo Librandi_ Jan 17 2018

%o (Python)

%o def O(f):

%o c = 1

%o while len(f) > 1:

%o f.sort()

%o m = abs(f[0] - f[1])

%o c *= m

%o f[0] = m

%o f.pop(1)

%o return c

%o a = lambda n: O(list(range(1, n+1)))

%o print([a(n) for n in range(0, 26)]) # _Darío Clavijo_, Aug 24 2024

%Y Cf. A008619, A064044, A246117, A130820, A229020.

%Y Column k=2 of A275062.

%K nonn,easy

%O 0,4

%A _Mark R. Diamond_