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

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).

Links

Robert Israel, Table of n, a(n) for n = 0..3308

Index entries for linear recurrences with constant coefficients, signature (6,-14,16,-8).

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.

Cf. A001787, A009545, A321959, A323100.

Sequence in context: A052967 A297498 A239040 * A293065 A211278 A364625

Adjacent sequences: A323222 A323223 A323224 * A323226 A323227 A323228

Keyword

nonn,easy

Author

Peter Luschny, Mar 18 2019

Status

approved