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A058034
Number of numbers whose cube root rounds to n.
1
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
0,2
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
EXAMPLE
a(2)=12 since the cube roots of 4, 5, 6, ..., 15 all lie between 1.5 and 2.5.
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 *)
PROG
(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
(Magma) [n mod 4 eq 0 select 3*n^2+1 else 3*n^2: n in [0..80]]; // Vincenzo Librandi, Dec 25 2015
CROSSREFS
Cf. A003215 for number whose floor (or ceiling) of the cube root is n, A004277 for number whose square root rounds to n.
Sequence in context: A033428 A213486 A018230 * A294416 A187273 A361847
KEYWORD
easy,nonn
AUTHOR
Henry Bottomley, Nov 22 2000
STATUS
approved