cp's OEIS Frontend

A057898 Largest number such that n = m^a(n) - a(n) with m a positive integer; i.e., where (n + a(n))^(1/a(n)) is a positive integer.

%I A057898 #12 Sep 04 2020 21:22:57
%S A057898 1,2,1,1,3,1,2,1,1,1,1,4,1,2,1,1,1,1,1,1,1,1,2,3,1,1,5,1,1,1,1,1,1,2,
%T A057898 1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,6,1,1,3,2,1,1,1,1,1,1,
%U A057898 1,1,1,1,1,1,1,1,4,1,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1
%N A057898 Largest number such that n = m^a(n) - a(n) with m a positive integer; i.e., where (n + a(n))^(1/a(n)) is a positive integer.
%C A057898 It may be that positive integers can be written as n = m^k - k (with m and k > 1) in at most one way [checked up to 10000] as well as with k = 1 and m = n+1.
%H A057898 Robert Israel, <a href="/A057898/b057898.txt">Table of n, a(n) for n = 1..10000</a>
%e A057898 a(5) = 3 since 5 = 2^3 - 3.
%p A057898 N:= 200: # for a(1)..a(N)
%p A057898 V:= Vector(N,1):
%p A057898 for k from 2 while 2^k-k <= N do
%p A057898   for m from 2 do
%p A057898     v:= m^k-k;
%p A057898     if v > N then break fi;
%p A057898     V[v]:= k;
%p A057898   od;
%p A057898 od:
%p A057898 convert(V,list); # _Robert Israel_, Sep 04 2020
%Y A057898 Cf. A057897, A057899.
%K A057898 nonn
%O A057898 1,2
%A A057898 _Henry Bottomley_, Sep 26 2000