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a(n) = 2^A327492(n).
6

%I #12 Oct 03 2019 06:16:53

%S 1,4,8,32,128,512,1024,4096,32768,131072,262144,1048576,4194304,

%T 16777216,33554432,134217728,2147483648,8589934592,17179869184,

%U 68719476736,274877906944,1099511627776,2199023255552,8796093022208,70368744177664,281474976710656,562949953421312

%N a(n) = 2^A327492(n).

%F a(n) = denominator(b(n)) where b(n) = n!/(2^n*floor(n/2)!)^2 is the normalized swinging factorial (A056040).

%p A327493 := n -> 2^(A327492_list(n+1)[n+1]):

%p seq(A327493(n), n = 0..26);

%o (PARI) seq(n)={my(a=vector(n+1)); a[1]=1; for(n=1, n, a[n+1] = a[n] * 2^if(n%4, n%2 + 1, valuation(n,2))); a} \\ _Andrew Howroyd_, Sep 28 2019

%o (PARI) a(n)={ denominator(sum(j=0, n, j!/(2^j*(j\2)!)^2)) } \\ _Andrew Howroyd_, Sep 28 2019

%o (Julia)

%o bitcount(n) = sum(digits(n, base = 2))

%o A327493(n) = 2^(2n - bitcount(n) + mod(n, 2))

%o [A327493(n) for n in 0:26] |> println # _Peter Luschny_, Oct 03 2019

%Y Cf. A327492, A327491, A327494, A327495, A056040.

%K nonn

%O 0,2

%A _Peter Luschny_, Sep 27 2019