A351064 Minimal number of positive perfect powers, with different exponents, whose sum is n (considering only minimal possible exponents for bases equal to 1).

1, 2, 3, 1, 2, 3, 4, 1, 1, 2, 3, 2, 3, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 2, 1, 2, 1, 2, 3, 4, 2, 1, 2, 3, 4, 1, 2, 3, 4, 2, 2, 3, 2, 2, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 5, 3, 2, 3, 2, 3, 4, 5, 2, 1, 2, 3, 4, 2, 3, 4, 5, 2, 2, 3, 3, 2, 3, 4, 3, 2, 1, 2, 3, 3, 2, 3, 4, 3, 2, 2, 2, 3, 3, 4, 3, 2, 2, 3, 4, 1, 2, 3, 4, 3, 3, 2

Offset: 1

Author

Alberto Zanoni, Feb 22 2022

Keywords

• nonn

Comments

Conjecture: the only numbers for which 5 addends are needed are 15, 23, 55, 62, 71.

The numbers mentioned in the conjecture are also the first five terms of A111151. - Omar E. Pol, Mar 01 2022

Examples

			a(1) = 1 because 1 can be represented with a single positive perfect power: 1 = 1^2.
a(2) = 2 because 2 can be represented with two (and not fewer) positive perfect powers with different exponents: 2 = 1^2 + 1^3.
a(6) = 3 because 6 can be represented with three (and not fewer) positive perfect powers with different exponents: 6 = 2^2 + 1^3 + 1^4.
a(7) = 4 because 7 can be represented with four (and not fewer) positive perfect powers with different exponents: 7 = 2^2 + 1^3 + 1^4 + 1^5.
a(15) = 5 because 15 can be represented with five (and not fewer) positive perfect powers with different exponents: 15 = 2^2 + 2^3 + 1^4 + 1^5 + 1^6.
		

Crossrefs

Cf. A111151, A351062, A351063, A351066.