login
A018886
Waring's problem: least positive integer requiring maximum number of terms when expressed as a sum of positive n-th powers.
2
1, 7, 23, 79, 223, 703, 2175, 6399, 19455, 58367, 176127, 528383, 1589247, 4767743, 14319615, 42991615, 129105919, 387186687, 1161822207, 3486515199, 10458497023, 31377588223, 94136958975, 282427654143, 847282962431, 2541815332863
OFFSET
1,2
COMMENTS
a(n) = (Q-1)*(2^n) + (2^n-1)*(1^n) is a sum of Q + 2^n - 2 terms, Q = trunc(3^n / 2^n).
REFERENCES
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 393.
LINKS
P. Pollack, Analytic and Combinatorial Number Theory Course Notes, exercise 7.1.1. p. 277.
Eric Weisstein's World of Mathematics, Waring's Problem.
FORMULA
a(n) = 2^n*floor((3/2)^n) - 1 = 2^n*A002379(n) - 1.
EXAMPLE
a(3) = 23 = 16 + 7 = 2*(2^3) + 7*(1^3) is a sum of 9 cubes;
a(4) = 79 = 64 + 15 = 4*(2^4) + 15*(1^4) is a sum of 19 biquadrates.
MAPLE
A018886 := proc(n)
2^n*floor((3/2)^n)-1
end proc: # R. J. Mathar, May 07 2015
MATHEMATICA
a[n_]:=-1+2^n*Floor[(3/2)^n]
a[Range[1, 20]] (* Julien Kluge, Jul 21 2016 *)
PROG
(Python)
def a(n): return (3**n//2**n-1)*2**n + (2**n-1)
print([a(n) for n in range(1, 27)]) # Michael S. Branicky, Dec 17 2021
(Python)
def A018886(n): return (3**n&-(1<<n))-1 # Chai Wah Wu, Jun 25 2024
CROSSREFS
KEYWORD
nonn,easy,nice
STATUS
approved