A010551 Multiply successively by 1,1,2,2,3,3,4,4,..., n >= 1, a(0) = 1.

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

Offset: 0

Author

Mark R. Diamond

Keywords

• nonn
• easy

Comments

From Emeric Deutsch, Dec 14 2008: (Start)

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) = A152666(n-1,1). (End)

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

From Daniel Forgues, May 20 2011: (Start)

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)

Partial products of A008619. - Reinhard Zumkeller, Apr 02 2012

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

Row sums of A246117. - Peter Bala, Aug 15 2014

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

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

Examples

			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
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..500
• Edinah K. Gnang and Isaac Wass, Growing Graceful and Harmonious Trees, arXiv:1808.05551 [math.CO], 2018-2020. See proposition 1.
• Frether Getachew Kebede and Fanja Rakotondrajao, Parity alternating permutations starting with an odd integer, arXiv:2101.09125 [math.CO], 2021.
• Steven Linton, James Propp, Tom Roby, and Julian West, Equivalence Classes of Permutations under Various Relations Generated by Constrained Transpositions, Journal of Integer Sequences, Vol. 15 (2012), #12.9.1.

Crossrefs

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

Column k=2 of A275062.

Programs

Haskell

a010551 n = a010551_list !! n
a010551_list = scanl (*) 1 a008619_list
-- Reinhard Zumkeller, Apr 02 2012

Magma

[Factorial(n div 2)*Factorial((n+1) div 2): n in [0..25]]; // Vincenzo Librandi Jan 17 2018

Maple

A010551 := proc(n)
    option remember;
    if n <= 1 then
        1
    else
        procname(n-1) *trunc( (n+1)/2 );
    fi;
end:

Mathematica

FoldList[ Times, 1, Flatten@ Array[ {#, #} &, 11]] (* Robert G. Wilson v, Jul 14 2010 *)

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

PARI

A010551(n)=(n\2)!*((n+1)\2)! \\ Michael Somos, Dec 29 2012, edited by M. F. Hasler, Nov 26 2017

Python

def O(f):
    c = 1
    while len(f) > 1:
        f.sort()
        m = abs(f[0] - f[1])
        c *= m
        f[0] = m
        f.pop(1)
    return c
a = lambda n: O(list(range(1, n+1)))
print([a(n) for n in range(0, 26)]) # Darío Clavijo, Aug 24 2024

Formula

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

Sum_{n >= 1} 1/a(n) = A130820. - Peter Bala, Jul 02 2016

a(n) ~ sqrt(Pi*n) * n! / 2^(n + 1/2). - Vaclav Kotesovec, Oct 02 2018

Sum_{n>=0} (-1)^n/a(n) = A229020. - Amiram Eldar, Apr 12 2021