A213243 Number of nonzero elements in GF(2^n) that are cubes.

1, 1, 7, 5, 31, 21, 127, 85, 511, 341, 2047, 1365, 8191, 5461, 32767, 21845, 131071, 87381, 524287, 349525, 2097151, 1398101, 8388607, 5592405, 33554431, 22369621, 134217727, 89478485, 536870911, 357913941, 2147483647, 1431655765, 8589934591, 5726623061, 34359738367, 22906492245

Offset

1,3

Comments

This is a strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for n, m >= 1. In particular, this is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Jul 06 2026

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,5,0,-4).

Formula

a(n) = M / gcd( M, 3 ), where M = 2^n - 1.

Conjectures from Colin Barker, Aug 23 2014, verified by Robert Israel, Apr 22 2016: (Start)

a(n) = (-1)*((-2+(-1)^n)*(-1+2^n))/3.

a(n) = 5*a(n-2) - 4*a(n-4).

G.f.: x*(2*x^2+x+1) / ((x-1)*(x+1)*(2*x-1)*(2*x+1)). (End)

E.g.f.: (-1 + exp(x) - 2*exp(3*x) + 2*exp(4*x))*exp(-2*x)/3. - Ilya Gutkovskiy, Apr 22 2016

a(n) = A051176(2^n - 1). - Peter Bala, Jul 06 2026

Maple

A213243:=n->(2^n-1)/gcd(2^n-1, 3): seq(A213243(n), n=1..50); # Wesley Ivan Hurt, Aug 23 2014

Mathematica

Table[(2^n - 1)/GCD[2^n - 1, 3], {n, 50}] (* Vincenzo Librandi, Mar 16 2013 *)
(* Alternative: *)
LinearRecurrence[{0, 5, 0, -4}, {1, 1, 7, 5}, 40] (* Harvey P. Dale, Jan 05 2017 *)

Prog

(Magma) [(2^n - 1) / GCD (2^n - 1, 3): n in [1..40]]; // Vincenzo Librandi, Mar 16 2013
(PARI) a(n)=(2^n-1)/gcd(2^n-1, 3) \\ Edward Jiang, Sep 04 2014

Crossrefs

Cf. A051176, A213244 (5th powers), A213245 (7th powers), A213246 (9th powers), A213247 (11th powers), A213248 (13th powers).

Sequence in context: A146619 A059990 A213246 * A185269 A344917 A328758

Adjacent sequences: A213240 A213241 A213242 * A213244 A213245 A213246

Keyword

nonn,easy,changed

Author

Joerg Arndt, Jun 07 2012

Status

approved