cp's OEIS Frontend

A323225 a(n) = ((2^n*n + i*(1 - i)^n - i*(1 + i)^n))/4, where i is the imaginary unit.

%I A323225 #20 Mar 27 2020 03:04:39
%S A323225 0,1,3,7,16,38,92,220,512,1160,2576,5648,12288,26592,57280,122816,
%T A323225 262144,557184,1179904,2490624,5242880,11009536,23067648,48233472,
%U A323225 100663296,209717248,436211712,905973760,1879048192,3892305920,8053047296,16642981888,34359738368
%N A323225 a(n) = ((2^n*n + i*(1 - i)^n - i*(1 + i)^n))/4, where i is the imaginary unit.
%C A323225 Related to Clifford algebras (see A323100 and A323346).
%H A323225 Robert Israel, <a href="/A323225/b323225.txt">Table of n, a(n) for n = 0..3308</a>
%H A323225 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6,-14,16,-8).
%F A323225 a(n) = Sum_{k = 0..n} A323346(n - k, k - 1).
%F A323225 a(n) = (A001787(n) + A009545(n))/2.
%F A323225 a(n) = [x^n] (x*(3*x^2 - 3*x + 1))/((2*x - 1)^2*(2*x^2 - 2*x + 1)).
%F A323225 a(n) = n! [x^n] (exp(2*x)*x + exp(x)*sin(x))/2.
%F A323225 a(n) = (4*n*a(n-3) + (2 - 6*n)*a(n-2) + (4*n - 2)*a(n-1))/(n - 1) for n >= 3.
%p A323225 a := n -> ((2^n*n + I*(1 - I)^n - I*(1 + I)^n))/4:
%p A323225 seq(a(n), n=0..32);
%t A323225 LinearRecurrence[{6, -14, 16, -8}, {0, 1, 3, 7}, 40] (* _Jean-François Alcover_, Mar 20 2019 *)
%t A323225 Table[((2^n n + I (1 - I)^n - I (1 + I)^n))/4, {n, 0, 29}] (* _Alonso del Arte_, Mar 27 2020 *)
%Y A323225 Antidiagonal sums of A323346.
%Y A323225 Cf. A001787, A009545, A321959, A323100.
%K A323225 nonn,easy
%O A323225 0,3
%A A323225 _Peter Luschny_, Mar 18 2019