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A069954 a(n) = binomial(2^(n+1), 2^n)/2 = binomial(2^(n+1) - 1, 2^n) = binomial(2^(n+1) - 1, 2^n-1).

%I A069954 #32 Oct 11 2024 14:47:48
%S A069954 1,3,35,6435,300540195,916312070471295267,
%T A069954 11975573020964041433067793888190275875,
%U A069954 2884329411724603169044874178931143443870105850987581016304218283632259375395
%N A069954 a(n) = binomial(2^(n+1), 2^n)/2 = binomial(2^(n+1) - 1, 2^n) = binomial(2^(n+1) - 1, 2^n-1).
%C A069954 Terms are always odd. a(1) = A061548(2), a(2) = A061548(3), a(3) = A061548(5), a(4) = A061548(9), a(5) = A061548(17), ... Hence it seems that a(n) = A061548(A000051(n)).
%C A069954 C(2*k, k)/2 = C(2*k-1, k) = C(2*k-1, k-1) is odd if and only if k = 2^n. - _Michael Somos_, Mar 12 2014
%H A069954 Vincenzo Librandi, <a href="/A069954/b069954.txt">Table of n, a(n) for n = 0..10</a>
%F A069954 From _Harry Richman_, May 18 2023: (Start)
%F A069954 a(n) = A001790(2^n).
%F A069954 a(n) = 1/2 * A000984(2^n).
%F A069954 a(n) = 1/2 * (2^n + 1) * A000108(2^n).
%F A069954 log log a(n) ~ (log 2) * (n + 1) + log log 2 + O(n / 2^n). (End)
%F A069954 a(n) = A037293(n+1) / 2. - _Tilman Piesk_, Oct 11 2024
%e A069954 C(2,1)/2 = C(1,0) = C(1,1) = 1. C(4,2)/2 = C(3,1) = C(3,2) = 3. C(8,4)/2 = C(7,3) = C(7,4) = 35. - _Michael Somos_, Mar 12 2014
%t A069954 Table[Binomial[2^(n+1) -1, 2^n -1], {n, 0, 10}] (* _Vincenzo Librandi_, Mar 14 2014 *)
%o A069954 (Magma) [Binomial(2^(n+1)-1, 2^n-1): n in [0..10]]; // _Vincenzo Librandi_, Mar 14 2014
%o A069954 (SageMath) [binomial(2^(n+1) -1, 2^n) for n in (0..9)] # _G. C. Greubel_, Aug 16 2022
%Y A069954 Cf. A000051, A000108, A000984, A001790, A061548, A037293.
%K A069954 easy,nonn
%O A069954 0,2
%A A069954 _Benoit Cloitre_, Apr 27 2002
%E A069954 a(0) = 1 added by _Michael Somos_, Mar 12 2014