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%I #10 Jan 02 2022 21:20:29
%S 1,8,800,18783360,28634752192836096,
%T 187118328452563147377366903401859072,
%U 22533823529098462258163079522899558155141642796614195116180863201125539840
%N a(n) = (binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2)))/2^n.
%H G. C. Greubel, <a href="/A158817/b158817.txt">Table of n, a(n) for n = 2..11</a>
%H Gyula O. H. Katona and Leonid Makar-Limanov, <a href="http://real.mtak.hu/id/eprint/21056">A problem for abelian groups</a>, Asian-Eur. J. Math. 1 (2008), no. 2, 237--241. (Reviewer: Thomas Britz) 20K01 (05B40 94B65).
%F a(n) = ( binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2)) )/2^n.
%t Table[(Binomial[2^n, 2^(n-1)] -Binomial[2^(n-1), 2^(n-2)])/2^n, {n, 2, 12}]
%o (Sage) [( binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2)) )/2^n for n in (2..12)] # _G. C. Greubel_, Dec 22 2021
%Y Cf. A069954.
%K nonn
%O 2,2
%A _Tanya Khovanova_ and Leonid Makar-Limanov, Mar 27 2009