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A350693 Number of b > 0 which permit n^3 to be written as a sum of powers of b in n parts. Each exponent c is an integer >= 0, n^3 = b^c_1 + b^c_2 + ... + b^c_n.

%I A350693 #32 Feb 24 2022 23:22:27
%S A350693 3,5,8,7,10,13,17,19,12,20,16,18,18,25,25,21,14,28,31,34,19,22,29,34,
%T A350693 28,33,29,38,19,33,30,31,34,51,44,30,20,41,38,44,18,37,42,52,27,30,37,
%U A350693 59,39,50,28,35,37,82,64,44,19,36,27,36,27,52,85,65,35,40,29
%N A350693 Number of b > 0 which permit n^3 to be written as a sum of powers of b in n parts. Each exponent c is an integer >= 0, n^3 = b^c_1 + b^c_2 + ... + b^c_n.
%C A350693 If n^3 is written in different number bases, a(n) is an upper limit for the count of number bases which allow n^3 to be written as a base-b number with a digit sum of n (generalized Dudeney numbers).
%C A350693 a(n) has an upper limit in the number of divisors of n^3-n. Let d be one of these divisors, then it appears that a lower limit can be found by excluding all divisors d where d+1 does not share all its prime divisors with binomial(n^3, n) (A107444).
%F A350693 a(n) <= A000005(n^3-n). Conjectured to become a(n) = A000005(n^3-n), if the definition would permit negative values for b and only the absolute value of the sum needs to be equal to n^3.
%e A350693 a(2) = 3 because 2^3 = 2^2 + 2^2 = 4^1 + 4^1 = 7^1 + 7^0.
%o A350693 (PARI) a(n) = sum(d=2, n^3, s=sumdigits(n^3, d); s<=n&&(n-s)%(d-1)==0); \\ _Jinyuan Wang_, Jan 15 2022
%Y A350693 Cf. A000005, A046459, A092517, A107444.
%K A350693 nonn
%O A350693 2,1
%A A350693 _Thomas Scheuerle_, Jan 12 2022
%E A350693 More terms from _Jinyuan Wang_, Jan 15 2022