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a(n) is the number of equations in the set E_{n,b} := {x+2^b*y=n^b, 2^b*x+3^b*y=n^b, ..., k^b*x+(k+1)^b*y=n^b, ..., n^b*x+(n+1)^b*y=n^b} which admit at least one nonnegative integer solution when b is sufficiently large.
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%I #13 Dec 11 2022 10:19:37

%S 1,2,3,4,3,5,4,6,5,6,4,8,5,7,7,8,5,9,5,9,8,8,6,12,7,8,8,10,6,12,7,11,

%T 9,9,9,14,7,9,9,13,7,13,8,12,12,10,8,16,9,12,10,12,8,14,10,14,11,11,9,

%U 19,9,11,13,14,11,15,9,13,11,15,9,19,10,12,14,14,12,16,10,18,13

%N a(n) is the number of equations in the set E_{n,b} := {x+2^b*y=n^b, 2^b*x+3^b*y=n^b, ..., k^b*x+(k+1)^b*y=n^b, ..., n^b*x+(n+1)^b*y=n^b} which admit at least one nonnegative integer solution when b is sufficiently large.

%C Defining a(n,b) as the number of equations of the set E_{n,b} which admit at least one nonnegative integer solution, it's possible to prove the existence of b_0 such that for all b > b_0, a(n,b) = a(n) whose value does not depend on b anymore.

%C a(n) is the number of positive integers k such that k(k+1) <= n or k divides n or k+1 divides n.

%F a(n) = ceiling(sqrt(n) - 3/2) + A000005(n).

%F a(n) ~ A356770(n)/2 as n->infinity.

%F a(n) <= A356770(n) for all n >= 1.

%e a(11) = 4 since for all b >= 29 the number of equations of the set E_{11,b} which admit at least one nonnegative integer solution is exactly equal to 4.

%e a(4) = 4 since for all b >= 1 the number of equations of the set E_{11,b} which admit at least one nonnegative integer solution is exactly equal to 4.

%t Table[Ceiling[Sqrt[n] - 3/2] + Length[Divisors[n]], {n, 1, 100}]

%Y Cf. A000005, A356770.

%K nonn

%O 1,2

%A _Luca Onnis_, Oct 10 2022