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Least positive integer k such that n divides k^5-j^5 for some j in [1,k-1].
3

%I #19 May 14 2021 16:46:52

%S 2,3,4,4,6,7,8,4,6,11,3,8,14,15,16,4,18,9,20,12,22,3,24,8,6,27,6,16,

%T 30,31,2,4,4,35,36,12,38,39,40,12,7,43,44,5,18,47,48,8,14,11,52,28,54,

%U 9,7,16,58,59,60,32,7,4,24,6,66,8,68,36,70,71,4,12,74,75,16,40

%N Least positive integer k such that n divides k^5-j^5 for some j in [1,k-1].

%C For a guide to related sequences, see A204892.

%C a(n) <= n+1. If n is divisible by p^2 then a(n) <= p+n/p. - _Robert Israel_, May 14 2021

%H Robert Israel, <a href="/A205791/b205791.txt">Table of n, a(n) for n = 1..3000</a>

%e 1 divides 2^5-1^5 -> k=2, j=1

%e 2 divides 3^5-1^5 -> k=3, j=1

%e 3 divides 4^5-1^5 -> k=4, j=1

%e 4 divides 4^5-2^5 -> k=4, j=2

%e 5 divides 6^5-1^5 -> k=6, j=1

%e 6 divides 7^5-1^5 -> k=7, j=1

%p N:= 100: # for a(1)..a(N)

%p V:= Vector(N):

%p count:= 0:

%p for k from 1 while count < N do

%p for j from 1 to k-1 while count < N do

%p Q:= select(t -> t <= N and V[t] = 0, numtheory:-divisors(k^5-j^5));

%p if Q <> {} then

%p newcount:= nops(Q);

%p count:= count + newcount;

%p V[convert(Q,list)]:= k;

%p fi

%p od od:

%p convert(V,list); # _Robert Israel_, May 14 2021

%t s = Table[n^4, {n, 1, 120}] ;

%t lk = Table[

%t NestWhile[# + 1 &, 1,

%t Min[Table[Mod[s[[#]] - s[[j]], z], {j, 1, # - 1}]] =!= 0 &], {z, 1,

%t Length[s]}]

%t Table[NestWhile[# + 1 &, 1,

%t Mod[s[[lk[[j]]]] - s[[#]], j] =!= 0 &], {j, 1, Length[lk]}]

%t (* _Peter J. C. Moses_, Jan 27 2012 *)

%t Array[(k=1;While[FreeQ[Mod[Table[k^5-j^5,{j,k-1}],#],0],k++];k)&,100] (* _Giorgos Kalogeropoulos_, May 14 2021 *)

%Y Cf. A204892, A344308.

%K nonn,look

%O 1,1

%A _Clark Kimberling_, Feb 01 2012