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%I A058034 #26 Mar 18 2024 12:18:54
%S A058034 1,3,12,27,49,75,108,147,193,243,300,363,433,507,588,675,769,867,972,
%T A058034 1083,1201,1323,1452,1587,1729,1875,2028,2187,2353,2523,2700,2883,
%U A058034 3073,3267,3468,3675,3889,4107,4332,4563,4801,5043,5292,5547,5809,6075,6348
%N A058034 Number of numbers whose cube root rounds to n.
%H A058034 Robert Israel, <a href="/A058034/b058034.txt">Table of n, a(n) for n = 0..10000</a>
%H A058034 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,1,-2,1).
%F A058034 a(n) = 3n^2+1 if n == 0 (mod 4), 3n^2 otherwise.
%F A058034 a(n) = A033428(n)+A011765(n) = A034131(n-1)-A034131(n-2).
%F A058034 a(n) = (1+(-1)^n+(-i)^n+i^n+12*n^2)/4 where i=sqrt(-1). - _Colin Barker_, Jul 04 2014
%F A058034 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
%e A058034 a(2)=12 since the cube roots of 4, 5, 6, ..., 15 all lie between 1.5 and 2.5.
%p A058034 seq(1 + floor((n+1/2)^3) - ceil((n-1/2)^3), n = 0 .. 100);
%t A058034 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 *)
%t A058034 LinearRecurrence[{2, -1, 0, 1, -2, 1}, {1, 3, 12, 27, 49, 75}, 50] (* _Vincenzo Librandi_, Dec 25 2015 *)
%o A058034 (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
%o A058034 (Magma) [n mod 4 eq 0 select 3*n^2+1 else 3*n^2: n in [0..80]]; // _Vincenzo Librandi_, Dec 25 2015
%Y A058034 Cf. A003215 for number whose floor (or ceiling) of the cube root is n, A004277 for number whose square root rounds to n.
%K A058034 easy,nonn
%O A058034 0,2
%A A058034 _Henry Bottomley_, Nov 22 2000

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