A188462 Least number of 5th powers needed to represent n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Offset: 1

Author

Jean-François Alcover, Apr 01 2011

Keywords

• nonn

Comments

Vaughan & Wooley (1995) prove that a(n) <= 17 for large enough n; in fact it is conjectured that a(n) <= 6 for large enough n. The maximum value is a(223) = 37. - Charles R Greathouse IV, Jul 05 2017

Examples

			33 = 2^5 + 1^5 (least decomposition) hence a(33) = 2.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, Number theory for the millennium, III (Urbana, IL, 2000), 301-340, A K Peters, Natick, MA, 2002.
• Robert C. Vaughan and Trevor D. Wooley, Further improvements in Waring's problem, Acta Mathematica 174:2 (1995), pp. 147-240.

Crossrefs

Cf. A002828 (squares), A002376 (cubes), A002377 (4th powers), A374012 (6th powers).

Programs

Mathematica

Cnt5[n_] := Module[{k = 1}, While[Length[PowersRepresentations[n, k, 5]] == 0, k++]; k]; Array[Cnt5, 105] (* T. D. Noe, Apr 01 2011 *)

Python

from itertools import count
from sympy.solvers.diophantine.diophantine import power_representation
def A188462(n):
    if n == 1: return 1
    for k in count(1):
        try:
            next(power_representation(n,5,k))
        except:
            continue
        return k # Chai Wah Wu, Jun 25 2024