A139031 Imaginary part of (4 + 3i)^n.

3, 24, 117, 336, -237, -10296, -76443, -354144, -922077, 1476984, 34867797, 242017776, 1064447283, 2465133864, -6890111163, -116749235904, -761741108157, -3175197967656, -6358056037323, 28515500892816, 387075408075603, 2383715742284424, 9392840736385317, 15549832333971936

Offset

1,1

Comments

Division of each term by 3 generates an integer sequence 1, 8, 39, 112, -79, -3432, -25481, -118048, -307359, 492328, ... - R. J. Mathar, Apr 08 2008

Links

Table of n, a(n) for n=1..24.

Index entries for linear recurrences with constant coefficients, signature (8,-25).

Formula

Imaginary part of (4 + 3i)^n.

Term (2,1) of [4,-3; 3,4]^n.

a(n)^2 + A139030(n)^2 = 5^(2*n).

a(n) = 8*a(n-1) - 25*a(n-2), n>2, given a(1) = 3, a(2) = 24.

(unsigned): Odd-indexed terms of A066771 interleaved with even-indexed terms of A066776.

O.g.f.: 3*x/(1-8*x+25*x^2). - R. J. Mathar, Apr 08 2008

a(n) = 5^n * sin(n * arctan(3/4)). - Amiram Eldar, Dec 12 2025

Example

a(3) = 117 since (4 + 3i)^3 = (-44 + 117i).
a(4) = 336 = 8*a(3) - 25*a(2) = 8*117 - 25*24.
a(3) = 117 = term (2,1) of [4,-3; 3,4]^3.

Mathematica

a[n_]:=Im[(4+3I)^n]; Array[a, 24] (* James C. McMahon, Jun 25 2025 *)

Prog

(PARI) a(n) = imag((4 + 3*I)^n); \\ Michel Marcus, Jun 25 2025

Crossrefs

Cf. A139030 (real part), A066771, A066776.

Sequence in context: A050545 A354676 A253944 * A065692 A127520 A175115

Adjacent sequences: A139028 A139029 A139030 * A139032 A139033 A139034

Keyword

sign,easy

Author

Gary W. Adamson, Apr 06 2008

Status

approved