Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #14 Feb 17 2024 15:03:09
%S 0,-2,-16,-88,-384,-1312,-2816,3712,86016,613888,3190784,13248512,
%T 42172416,72409088,-264175616,-3561586688,-23209181184,-114441715712,
%U -451350102016,-1321966501888,-1548729974784,14049490239488,143370521411584,865974366502912,4060384503791616,15163588700274688
%N a(n) is the numerator of the imaginary part of 1/(1+i/2)^n, where i is the imaginary unit.
%C The corresponding denominators are 5^n.
%H Hugo Pfoertner, <a href="/A370192/b370192.txt">Table of n, a(n) for n = 0..300</a>
%H Hugo Pfoertner <a href="https://oeis.org/plot2a?name1=A370191&name2=A370192&tform1=asinh&tform2=asinh&shift=0&radiop1=xy&drawpoints=true">Plot of asinh(A370192(n)) vs asinh(A370191(n))</a>, using Plot 2.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-20).
%F From _Stefano Spezia_, Feb 17 2024: (Start)
%F G.f.: -2*x/(1 - 8*x + 20*x^2).
%F E.g.f.: -exp(4*x)*sin(2*x). (End)
%e n (5/(1 + i/2))^n
%e A370191(n) a(n)
%e 0 1 +0 *i
%e 1 4 -2 *i
%e 2 12 -16 *i
%e 3 16 -88 *i
%e 4 -112 -384 *i
%e 5 -1216 -1312 *i
%e 6 -7488 -2816 *i
%e 7 -35584 +3712 *i
%e 8 -134912 +86016 *i
%t LinearRecurrence[{8, -20}, {0, -2}, 26]
%o (PARI) a370192(n) = numerator(imag(1/(1+I/2)^n))
%Y Cf. A000351 (denominators), A370191.
%K sign,frac,easy
%O 0,2
%A _Hugo Pfoertner_, Feb 17 2024