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a(n) = product(i >= 0, P(n, i)^(2^i)) where P(n, i) = product(p prime, n/2^(i+1) < p <= n/2^i).
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%I #15 May 28 2014 02:08:57

%S 1,1,2,6,12,60,180,1260,5040,5040,25200,277200,2494800,32432400,

%T 227026800,227026800,3632428800,61751289600,61751289600,1173274502400,

%U 29331862560000,29331862560000,322650488160000,7420961227680000,601097859442080000,601097859442080000

%N a(n) = product(i >= 0, P(n, i)^(2^i)) where P(n, i) = product(p prime, n/2^(i+1) < p <= n/2^i).

%C a(n) are the partial products of A219964(n).

%C a(n) divides n!, n!/a(n) = 1, 1, 1, 1, 2, 2, 4, 4, 8, 72, 144, 144, 192...

%C The swinging factorial (A056040) divides a(n), a(n)/n$ = 1, 1, 1, 1, 2,...

%C The primorial of n (A034386) divides a(n), a(n)/n# = 1, 1, 1, 1, 2, 2, 6,..

%C If p^k is the largest power of a prime dividing a(n) then k is 2^n for some n >= 0.

%C a(n) / A055773(n) is the largest square dividing a(n), a(n) / A055773(n) = A008833(a(n)).

%p a := proc(n) local k; `if`(n < 2, 1,

%p mul(k, k = select(isprime, [$iquo(n, 2)+1..n]))*a(iquo(n,2))^2) end:

%p seq(a(i), i=0..25);

%o (Sage)

%o def a(n) :

%o if n < 2 : return 1

%o return mul(k for k in prime_range(n//2+1,n+1))*a(n//2)^2

%o [a(n) for n in (0..25)]

%Y Cf. A055773.

%K nonn

%O 0,3

%A _Peter Luschny_, Mar 30 2013