login
A323225
a(n) = ((2^n*n + i*(1 - i)^n - i*(1 + i)^n))/4, where i is the imaginary unit.
1
0, 1, 3, 7, 16, 38, 92, 220, 512, 1160, 2576, 5648, 12288, 26592, 57280, 122816, 262144, 557184, 1179904, 2490624, 5242880, 11009536, 23067648, 48233472, 100663296, 209717248, 436211712, 905973760, 1879048192, 3892305920, 8053047296, 16642981888, 34359738368
OFFSET
0,3
COMMENTS
Related to Clifford algebras (see A323100 and A323346).
FORMULA
a(n) = Sum_{k = 0..n} A323346(n - k, k - 1).
a(n) = (A001787(n) + A009545(n))/2.
a(n) = [x^n] (x*(3*x^2 - 3*x + 1))/((2*x - 1)^2*(2*x^2 - 2*x + 1)).
a(n) = n! [x^n] (exp(2*x)*x + exp(x)*sin(x))/2.
a(n) = (4*n*a(n-3) + (2 - 6*n)*a(n-2) + (4*n - 2)*a(n-1))/(n - 1) for n >= 3.
MAPLE
a := n -> ((2^n*n + I*(1 - I)^n - I*(1 + I)^n))/4:
seq(a(n), n=0..32);
MATHEMATICA
LinearRecurrence[{6, -14, 16, -8}, {0, 1, 3, 7}, 40] (* Jean-François Alcover, Mar 20 2019 *)
Table[((2^n n + I (1 - I)^n - I (1 + I)^n))/4, {n, 0, 29}] (* Alonso del Arte, Mar 27 2020 *)
CROSSREFS
Antidiagonal sums of A323346.
Sequence in context: A052967 A297498 A239040 * A293065 A211278 A364625
KEYWORD
nonn,easy
AUTHOR
Peter Luschny, Mar 18 2019
STATUS
approved