cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A034121 Fractional part of cube root of a(n) starts with digit 5.

Original entry on oeis.org

4, 16, 17, 43, 44, 45, 46, 92, 93, 94, 95, 96, 97, 167, 168, 169, 170, 171, 172, 173, 174, 175, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 422, 423, 424, 425, 426, 427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 615, 616, 617
Offset: 1

Views

Author

Patrick De Geest, Sep 15 1998

Keywords

Crossrefs

Cf. A034131.

Programs

  • Mathematica
    Select[Range[800],NumberDigit[CubeRoot[#],-1]==5&] (* Harvey P. Dale, Sep 06 2024 *)

A058034 Number of numbers whose cube root rounds to n.

Original entry on oeis.org

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

Views

Author

Henry Bottomley, Nov 22 2000

Keywords

Examples

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

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
Showing 1-2 of 2 results.