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A081125 a(n) = n! / floor(n/2)!. 11

%I #79 Jul 22 2022 16:44:43

%S 1,1,2,6,12,60,120,840,1680,15120,30240,332640,665280,8648640,

%T 17297280,259459200,518918400,8821612800,17643225600,335221286400,

%U 670442572800,14079294028800,28158588057600,647647525324800,1295295050649600

%N a(n) = n! / floor(n/2)!.

%C Product of the largest parts in the partitions of n+1 into exactly two parts, n > 0. - _Wesley Ivan Hurt_, Jan 26 2013 (Clarified on Apr 20 2016)

%H Vincenzo Librandi, <a href="/A081125/b081125.txt">Table of n, a(n) for n = 0..400</a>

%H Peter Luschny, <a href="/A180000/a180000.pdf">Die schwingende Fakultät und Orbitalsysteme</a>, August 2011.

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>.

%F E.g.f.: (1+x)*exp(x^2). - _Vladeta Jovovic_, Sep 24 2003

%F From _Peter Luschny_, Aug 07 2009: (Start)

%F a(n) = sqrt(n!*n$) where n$ denotes the swinging factorial (A056040).

%F a(n) = 2^n Gamma((n+1+(n mod 2))/2)/sqrt(Pi). (End)

%F E.g.f.: E(0) where E(k) = 1 + x/(1 - x/(x + (k+1)/E(k+1))) ; (continued fraction, 3rd kind, 3-step). - _Sergei N. Gladkovskii_, Sep 20 2012

%F G.f.: G(0) where G(k) = 1 + x*(2*k+1)/(1 - 2*x/(2*x + 1/G(k+1))); (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Nov 18 2012

%F Conjecture: a(n) +2*a(n-1) -2*n*a(n-2) +4*(-n+2)*a(n-3) = 0. - _R. J. Mathar_, Nov 26 2012

%F From _Wesley Ivan Hurt_, Jun 06 2013: (Start)

%F a(n) = n!/(n-floor((n+1)/2))!.

%F a(n) = Product_{i = ceiling(n/2)..(n-1)} i. [Note: empty product = 1]

%F a(n) = P( n, floor((n+1)/2) ), where P(n,k) are the number of k-permutations of n objects. (End)

%F a(n) = n$*floor(n/2)! where n$ denotes the swinging factorial (A056040). - _Peter Luschny_, Oct 28 2013

%F From _Amiram Eldar_, Mar 10 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 1 + (3/2)*exp(1/4)*sqrt(Pi)*erf(1/2).

%F Sum_{n>=0} (-1)^n/a(n) = 1 - (1/2)*exp(1/4)*sqrt(Pi)*erf(1/2). (End)

%e a(3) = 6 since 3+1 = 4 has two partitions into two parts, (3,1) and (2,2), and the product of the largest parts is 6. - _Wesley Ivan Hurt_, Jan 26 2013 (Clarified on Apr 20 2016)

%p Method 1) a:=n->n!/floor(n/2)!; seq(a(k),k=0..40); # _Wesley Ivan Hurt_, Jun 03 2013

%p Method 2) with(combinat, numbperm); seq(numbperm(k, floor((k+1)/2)), k = 0..40); # _Wesley Ivan Hurt_, Jun 06 2013

%t Table[n!/Floor[n/2]!, {n, 0, 30}] (* _Wesley Ivan Hurt_, Apr 20 2016 *)

%o (Magma) [Factorial(n)/(Factorial(Floor(n/2))): n in [0..30]]; // _Vincenzo Librandi_, Sep 13 2011

%o (PARI) a(n)=n!/(n\2)! \\ _Charles R Greathouse IV_, Sep 13 2011

%o (Sage)

%o def a(n): return rising_factorial(ceil(n/2),floor(n/2))

%o [a(n) for n in range(26)] # _Peter Luschny_, Oct 09 2013

%o (Python)

%o from sympy import rf

%o def A081125(n): return rf((m:=n+1>>1)+(n+1&1),m) # _Chai Wah Wu_, Jul 22 2022

%Y Cf. A004526, A056040, A081123.

%K nonn,easy

%O 0,3

%A _Paul Barry_, Mar 07 2003

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Last modified March 28 09:04 EDT 2024. Contains 371240 sequences. (Running on oeis4.)