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%I #45 Mar 04 2018 08:27:01
%S 1,1,2,2,3,4,4,8,5,16,6,32,7,64,8,128,9,256,10,512,11,1024,12,2048,13,
%T 4096,14,8192,15,16384,16,32768,17,65536,18,131072,19,262144,20,
%U 524288,21,1048576,22,2097152,23,4194304,24,8388608,25,16777216,26,33554432
%N Alternate n+1, 2^n.
%C The offset 0 is a choice. Another sequence could begin with A001477 instead of A000027. The Akiyama-Tanigawa transform applied to 1/(n+1) and 1/2^n are the second Bernoulli numbers A164555(n)/A027642(n) and the second (fractional) Euler numbers A198631(n)/A006519(n+1). (The first Euler numbers are not in the OEIS). Hence a(n).
%C a(2n+1) - a(2n) = 2^n -n -1 = 0, 0, 1, 4, 11,... = A000295(n) (Eulerian numbers).
%C a(2n+1) + a(2n) = 2^n +n +1 = A005126(n).
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,4,0,-5,0,2).
%F a(n) = ((n+1) mod 2) * (n/2 + 1) + (n mod 2) * 2^((n-1)/2). - _Wesley Ivan Hurt_, Jun 29 2014
%F G.f.: (1 + x - x^2) * (1 - x^2 - x^3) / ((1 - x^2)^2 * (1 - 2*x^2)). - _Michael Somos_, Jun 30 2014
%e G.f. = 1 + x + 2*x^2 + 2*x^3 + 3*x^4 + 4*x^5 + 4*x^6 + 8*x^7 + 5*x^8 + ...
%p A242971:=n->((n+1) mod 2)*(n/2 + 1) + (n mod 2) * 2^((n-1)/2); seq(A242971(n), n=0..50); # _Wesley Ivan Hurt_, Jun 29 2014
%t Table[Mod[n + 1, 2] (n/2 + 1) + Mod[n, 2] 2^((n - 1)/2), {n, 0, 50}] (* _Wesley Ivan Hurt_, Jun 29 2014 *)
%Y Cf. A000027, A000079, A209308, A227577.
%K nonn
%O 0,3
%A _Paul Curtz_, Jun 22 2014
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