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%I #19 Nov 23 2023 02:53:54
%S 1,3,6,15,30,70,140,315,630,1386,2772,6006,12012,25740,51480,109395,
%T 218790,461890,923780,1939938,3879876,8112468,16224936,33801950,
%U 67603900,140408100,280816200,581690700,1163381400,2404321560,4808643120,9917826435,19835652870
%N a(n) = 2^n*binomial((n + 1 + (n mod 2))/2, 1/2).
%F a(2*n) = A002457(n).
%F a(2*n+1) = A033876(n).
%F a(2*n+2)/2 = a(2*n+1).
%F Conjecture: (n+1)*a(n) -2*a(n-1) +4*(-n-1)*a(n-2)=0. - _R. J. Mathar_, May 11 2014
%F a(n) = A100071(n+2)/2. - _Michel Marcus_, Sep 14 2015
%F Sum_{n>=0} 1/a(n) = 2*Pi/sqrt(3) - 2. - _Amiram Eldar_, Mar 04 2023
%F a(n) = (n+2)*binomial(n+1,ceiling(n/2))/2. - _Wesley Ivan Hurt_, Nov 23 2023
%p a := n -> 2^n*binomial((n+1+(n mod 2))/2, 1/2); seq(a(n), n=0..29);
%t a[n_] := 2^n*Binomial[(n + 1 + Mod[n, 2])/2, 1/2]; Array[a, 33, 0] (* _Amiram Eldar_, Mar 04 2023 *)
%Y Cf. A100071, A002457, A033876.
%K nonn
%O 0,2
%A _Peter Luschny_, May 06 2014
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