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%I #23 Sep 08 2022 08:45:29
%S 1,1,1,4,8,128,2048,131072,2097152,536870912,137438953472,
%T 140737488355328,72057594037927936,295147905179352825856,
%U 1208925819614629174706176,19807040628566084398385987584,40564819207303340847894502572032,2658455991569831745807614120560689152
%N a(n) = 2^binomial(n+1,2)/A046161(n).
%C Apparently, also numerator of 2^(n*(n-1)/2)/n!. - _N. J. A. Sloane_, Dec 31 2010
%H G. C. Greubel, <a href="/A127943/b127943.txt">Table of n, a(n) for n = 0..82</a>
%F a(n) = 2^binomial(n+1,2)/denominator(binomial(2*n,n)/4^n).
%F a(n) = 2^A127944(n).
%p a:=n->2^(binomial(n+1,2))/denom(binomial(2*n,n)/4^n); seq(a(n),n=0..17); # _Muniru A Asiru_, Dec 10 2018
%t Table[2^Binomial[n+1,2]/Denominator[Binomial[2*n,n]/4^n], {n, 0, 25}] (* _G. C. Greubel_, May 01 2018 *)
%o (PARI) for(n=0,25, print1(2^(binomial(n+1,2))/denominator(binomial(2*n, n)/4^n), ", ")) \\ _G. C. Greubel_, May 01 2018
%o (PARI) a(n) = numerator(2^(n*(n-1)/2)/n!); \\ _Altug Alkan_, May 02 2018
%o (Magma) [2^(Binomial(n+1,2))/Denominator(Binomial(2*n, n)/4^n): n in [0..25]]; // _G. C. Greubel_, May 01 2018
%o (Sage) [2^binomial(n+1,2)/denominator(binomial(2*n,n)/4^n) for n in range(30)] # _G. C. Greubel_, Dec 09 2018
%o (GAP) List([0..30], n-> 2^(Binomial(n+1,2))/DenominatorRat(Binomial(2*n, n)/4^n)); # _G. C. Greubel_, Dec 09 2018
%K easy,nonn
%O 0,4
%A _Paul Barry_, Feb 08 2007
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