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%I #26 Apr 30 2022 01:14:56
%S 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,
%T 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,
%U 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
%N Minimal number of positive perfect powers, with different exponents, whose sum is n (considering only minimal possible exponents for bases equal to 1).
%C Conjecture: the only numbers for which 5 addends are needed are 15, 23, 55, 62, 71.
%C The numbers mentioned in the conjecture are also the first five terms of A111151. - _Omar E. Pol_, Mar 01 2022
%e a(1) = 1 because 1 can be represented with a single positive perfect power: 1 = 1^2.
%e a(2) = 2 because 2 can be represented with two (and not fewer) positive perfect powers with different exponents: 2 = 1^2 + 1^3.
%e 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.
%e 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.
%e 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.
%Y Cf. A111151, A351062, A351063, A351066.
%K nonn
%O 1,2
%A _Alberto Zanoni_, Feb 22 2022
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