A058034 Number of numbers whose cube root rounds to n.

1, 3, 12, 27, 49, 75, 108, 147, 193, 243, 300, 363, 433, 507, 588, 675, 769, 867, 972, 1083, 1201, 1323, 1452, 1587, 1729, 1875, 2028, 2187, 2353, 2523, 2700, 2883, 3073, 3267, 3468, 3675, 3889, 4107, 4332, 4563, 4801, 5043, 5292, 5547, 5809, 6075, 6348

Offset: 0

Author

Henry Bottomley, Nov 22 2000

Keywords

• easy
• nonn

Examples

			a(2)=12 since the cube roots of 4, 5, 6, ..., 15 all lie between 1.5 and 2.5.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

Crossrefs

Cf. A003215 for number whose floor (or ceiling) of the cube root is n, A004277 for number whose square root rounds to n.

Programs

Magma

[n mod 4 eq 0 select 3*n^2+1 else 3*n^2: n in [0..80]]; // Vincenzo Librandi, Dec 25 2015

Maple

seq(1 + floor((n+1/2)^3) - ceil((n-1/2)^3), n = 0 .. 100);

Mathematica

Table[SeriesCoefficient[-(3 x^5 + 6 x^4 + 6 x^3 + 7 x^2 + x + 1)/((x - 1)^3 (x + 1) (x^2 + 1)), {x, 0, n}], {n, 0, 46}] (* Michael De Vlieger, Dec 24 2015 *)
LinearRecurrence[{2, -1, 0, 1, -2, 1}, {1, 3, 12, 27, 49, 75}, 50] (* Vincenzo Librandi, Dec 25 2015 *)

PARI

Vec(-(3*x^5+6*x^4+6*x^3+7*x^2+x+1)/((x-1)^3*(x+1)*(x^2+1)) + O(x^100)) \\ Colin Barker, Jul 04 2014

Formula

a(n) = 3n^2+1 if n == 0 (mod 4), 3n^2 otherwise.

a(n) = A033428(n)+A011765(n) = A034131(n-1)-A034131(n-2).

a(n) = (1+(-1)^n+(-i)^n+i^n+12*n^2)/4 where i=sqrt(-1). - Colin Barker, Jul 04 2014

G.f.: -(3*x^5+6*x^4+6*x^3+7*x^2+x+1) / ((x-1)^3*(x+1)*(x^2+1)). - Colin Barker, Jul 04 2014