A125229 a(n) = j such that i^j is maximized subject to i+j = n (i >= 0, j >= 0).

0, 0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 23, 24, 25, 25, 26, 27, 28, 28, 29, 30, 31, 32, 32, 33, 34, 35, 35, 36, 37, 38, 39, 39, 40, 41, 42, 42, 43, 44, 45, 46, 46, 47, 48, 49, 50, 50, 51, 52

Offset: 0

Author

Sébastien Dumortier, Jan 15 2007

Keywords

• easy
• nonn

Comments

For n = 2, there is not a unique maximum because 2^0 = 1^1; we choose j = 0. - T. D. Noe, Apr 08 2014

Examples

			If the sum is 5, the powers are: 0^5, 1^4, 2^3, 3^2, 4^1 and 5^0. The highest is 3^2 so a(5) = 2.
		

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A056155.

Programs

Mathematica

Join[{0,0},Table[SortBy[{#[[1]],#[[2]],#[[1]]^#[[2]]}&/@Flatten[ Permutations /@ IntegerPartitions[n,{2}],1],Last][[-1,2]],{n,3,80}]] (* Harvey P. Dale, Sep 01 2013 *)
Join[{0, 0, 0}, Flatten[Table[s = Table[(n - k)^k, {k, n}]; Position[s, Max[s]], {n, 3, 80}]]] (* T. D. Noe, Apr 08 2014 *)

Extensions

a(0) added by T. D. Noe, Apr 08 2014