OFFSET
0,3
COMMENTS
For n > 0, the number of elements in the group of the Gaussian integers in a reduced system modulo (1+i)^n is 2^(n-1).
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2).
FORMULA
For n > 5, a(n) = 2^(floor(n/2) - 1).
For even n, a(n) = A227334(2^(n/2)).
G.f.: (1 + x + 2*x^3 - 4*x^5 - 4*x^6 - 4*x^7)/(1 - 2*x^2). - Andrew Howroyd, Nov 06 2025
From Amiram Eldar, Apr 03 2026: (Start)
Sum_{n>=0} 1/a(n) = 17/4.
Sum_{n>=0} (-1)^n/a(n) = 1/4. (End)
EXAMPLE
For Gaussian integer x such that (x, 1+i) = 1, x^4 - 1 = (x + 1)(x - 1)(x + i)(x - i) provides at least 7 factors of 1+i in total (and exactly 7 when x = 2+i), so a(7) = 4.
MATHEMATICA
Join[{1, 1, 2, 4, 4, 4}, Table[2^(Floor[n/2] - 1), {n, 6, 50}]] (* Vincenzo Librandi, Apr 04 2018 *)
PROG
(Magma) [1, 1, 2, 4, 4, 4] cat [2^(Floor(n div 2)-1): n in [6..50]]; // Vincenzo Librandi, Apr 04 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Apr 04 2018
STATUS
approved
