A351065 Number of different ways to obtain n as a sum of the minimal possible number of positive perfect powers with different exponents (considering only minimal possible exponents for bases equal to 1).

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 3, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 4, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 5, 2, 2, 2, 4, 1, 1, 1, 3, 4, 1, 2, 3, 1, 1, 2, 3, 2, 3, 3, 1, 1, 1, 1, 2, 6, 1, 4

Offset: 1

Author

Alberto Zanoni, Feb 22 2022

Keywords

• nonn

Comments

Every positive integer k appears in the sequence, as a(2^(2^k)) = k.

Examples

			a(4) = 1, because 4 = 2^2 is its only possible representation, and similarly for every power a^p, with a > 1 and p prime.
a(16) = 2, because 16 = 2^4 = 4^2. More generally, a^(p^2) -- with a > 1 and p prime -- can be written in exactly two ways.
a(17) = 3, because 17 = 1^2 + 2^4 = 3^2 + 2^3 = 4^2 + 1^3.
a(313) = 10, because 313 can be written in exactly 10 different ways (with three perfect powers): 4^2 + 6^3 + 3^4 = 5^2 + 2^5 + 2^8 = 5^2 + 4^4 + 2^5 = 7^2 + 2^3 + 2^8 = 7^2 + 2^3 + 4^4 = 9^2 + 6^3 + 2^4 = 11^2 + 2^6 + 2^7 = 11^2 + 4^3 + 2^7 = 13^2 + 2^4 + 2^7 = 17^2 + 2^3 + 2^4.
		

Crossrefs

Cf. A351062, A351063, A351064, A351066.