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%I #75 Apr 13 2024 16:17:02
%S 1,1,2,6,3,15,5,35,35,35,7,77,77,1001,143,143,143,2431,2431,46189,
%T 46189,46189,4199,96577,96577,96577,7429,7429,7429,215441,215441,
%U 6678671,6678671,6678671,392863,392863,392863,14535931,765049,765049,765049
%N a(n) = Product_{p in P_n} where P_n = {p prime, n/2 < p <= n }.
%C Old name: Product of primes p for which p divides n! but p^2 does not (i.e. ord_p(n!)=1). - Dion Gijswijt (gijswijt(AT)science.uva.nl), Jan 07 2007
%C Squarefree part of n! divided by gcd(Q,F), where Q is the largest square divisor and F is the squarefree part of n!. - _Labos Elemer_, Jul 12 2000
%C a(1) = 1, a(n) = n*a(n-1) if n is a prime else a(n) = least integer multiple of a(n-1)/n. - _Amarnath Murthy_, Apr 29 2004
%C Let P(i) denote the primorial number A034386(i). Then a(n) = P(n)/P(floor(n/2)). - _Peter Luschny_, Mar 05 2011
%C Letting H(n) = 1 + 1/2 + ... + 1/n denote the n-th harmonic number, it is known that a(n) is equal to the denominator (in lowest terms) of H(n)^2*n! for n >= 6 (see below example). - _John M. Campbell_, Mar 27 2016
%C For all n satisfying 6 <= n < 897, a(n) = A130087(n). - _John M. Campbell_, Mar 27 2016
%C It is also known that a(n) is equal to lcm^2(1, 2, ..., n)/gcd(lcm^2(1, 2, ..., n), n!). - _John M. Campbell_, Apr 04 2016
%H Vincenzo Librandi, <a href="/A055773/b055773.txt">Table of n, a(n) for n = 0..200</a> (corrected by Michel Marcus, Jan 19 2019)
%H J. M. Campbell et al., <a href="http://math.stackexchange.com/q/1694721/">A problem involving the product prod_{k=1..n} k^mu(k), where mu denotes the Möbius function</a>, Mathematics Stack Exchange (2016).
%H F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Sloane/sloane55.html">A Slow-Growing Sequence Defined by an Unusual Recurrence</a>, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
%H <a href="/index/Ge#Gijswijt">Index entries for sequences related to Gijswijt's sequence</a>
%F a(n) = numerator(A056040(n)^2/n!).
%F a(n) = numerator(A056040(n)/floor(n/2)!^2).
%F a(n) = numerator(n!/floor(n/2)!^4). - _Peter Luschny_, Jul 30 2011
%F a(n) = product of primes p such that n/2 < p <= n. - _Klaus Brockhaus_, May 02 2004
%F a(n) = A055204(n)/A055230(n)=A055231(n!)=n!/([A007913(n!)*A055229[n]).
%F a(n) = Product_{i=pi(n/2)+1..pi(n)} p(i), where pi denotes the prime counting function and p(i) denotes the i-th prime number. - _John M. Campbell_, Mar 27 2016
%e n = 13, P_n = {7, 11, 13}, a(13) = 7*11*13 = 1001.
%e Letting n = 14, the denominator (in lowest terms) of H(n)^2*n! = 131803989435744/143 is a(14)=143. - _John M. Campbell_, Mar 27 2016
%p a := n -> mul(k,k=select(isprime,[$iquo(n,2)+1..n])); # _Peter Luschny_, Jun 20 2009
%p A055773 := n -> numer(n!/iquo(n,2)!^4); # _Peter Luschny_, Jul 30 2011
%t Table[Numerator[n!/Floor[n/2]!^4], {n, 0, 40}] (* _Michael De Vlieger_, Mar 27 2016 *)
%o (PARI) q=1;for(n=2,41,print1(q,",");q=if(isprime(n),q*n,q/gcd(q,n))) \\ _Klaus Brockhaus_, May 02 2004
%o (PARI) a(n) = k=1;forprime(p=nextprime(n\2+1),precprime(n),k=k*p);k \\ _Klaus Brockhaus_, May 02 2004
%o (PARI) a(n) = prod(i=primepi(n/2)+1,primepi(n),prime(i)) \\ _John M. Campbell_, Mar 27 2016
%o (Python)
%o from math import prod
%o from sympy import primerange
%o def A055773(n): return prod(primerange((n>>1)+1,n+1)) # _Chai Wah Wu_, Apr 13 2024
%Y Cf. A000188, A008833, A007913, A055229, A055231 (for n), A055071, A055204, A055230, A094299, A094302, A193477, A130087.
%K nonn,changed
%O 0,3
%A _Labos Elemer_, Jul 12 2000
%E Entry revised by _N. J. A. Sloane_, Jan 07 2007
%E Simpler definition based on a comment of _Klaus Brockhaus_, set offset to 0 and prepended 1 to data. - _Peter Luschny_, Mar 09 2013
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