A242871 -id:A242871 - cp's OEIS frontend

A242870 Numbers n such that (n^n-2^2)/(n-2) is an integer.

1, 3, 4, 6, 8, 14, 20, 22, 38, 44, 56, 62, 86, 102, 110, 128, 158, 164, 182, 222, 254, 296, 302, 326, 344, 380, 422, 470, 488, 502, 542, 590, 622, 662, 686, 758, 782, 822, 884, 902, 974, 1028, 1094, 1102, 1136, 1262, 1316, 1334, 1406, 1460, 1502, 1622, 1766, 1808

Offset: 1

Derek Orr, May 24 2014

nonn

a(n) is even for all n > 2. 1 and 3 are members of this sequence because (n^n-2^2)/(n-2) becomes (2^2-n^n) and (n^n-2^2), respectively, which are both integers.
Given the term (n^n-k^k)/(n-k) (here, k=2), whenever k = 2^m for some m, there are significantly fewer data values within a given range of numbers. See A242871 for k=3.
These are also numbers n such that (2^n-n^2)/(n-2) is an integer.

			(4^4-2^2)/(4-2) = 252/2 = 126 is an integer. Thus, 4 is a member of this sequence.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A242871.

filter:= proc(n) (n&^n - 4) mod (n-2) = 0 end proc;
select(filter, [1,$3..1000]); # Robert Israel, May 25 2014

Join[{1},Select[Range[3,2000],IntegerQ[(#^#-4)/(#-2)]&]] (* Harvey P. Dale, Apr 24 2016 *)

for(n=1,2500,if(n!=2,s=(n^n-2^2)/(n-2);if(floor(s)==s,print(n))))

A242872 Least number k > 1 such that (k^k-n^n)/(k-n) is an integer.

Original entry on oeis.org

2, 3, 2, 2, 3, 2, 3, 2, 3, 4, 3, 3, 4, 2, 3, 4, 5, 6, 3, 2, 3, 2, 3, 4, 4, 6, 3, 4, 5, 3, 4, 8, 6, 4, 3, 4, 5, 2, 3, 4, 5, 3, 3, 2, 3, 4, 5, 6, 7, 8, 3, 4, 5, 4, 5, 2, 3, 4, 5, 5, 7, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 9, 10, 3, 4, 5, 6, 7, 8, 3, 4, 3, 4, 4, 2, 3, 4, 5, 6, 7, 8, 9, 4

Offset: 1

Derek Orr, May 24 2014

nonn

a(n) <= n-1 for n > 2 (since k > 1).

This is also the least number k such that (k^n-n^k)/(k-n) is an integer.

			(2^2-5^5)/(2-5) = 3121/3 is not an integer. (3^3-5^5)/(3-5) = 3098/2 = 1549 is an integer. Thus a(5) = 3.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A242870, A242871.

A242872:= proc(n)
   local nn, k;
   nn:= n^n;
   for k from 2 to n-1 do
      if (nn-k^k) mod (n-k) = 0 then return k fi
   od;
   return n+1;
end:
seq(A242872(n),n=1..100); # Robert Israel, May 25 2014

a[n_] := Switch[n, 1, 2, 2, 3, _, With[{nn = n^n}, For[k = 2, True, k++, If[Mod[nn-k^k, n-k] == 0, Return[k]]]]];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, May 15 2023 *)

a(n)=for(k=2,n+1,if(k!=n,s=(k^k-n^n)/(k-n);if(floor(s)==s,return(k))));
n=1;while(n<100,print(a(n));n+=1)

cp's OEIS Frontend

A242870 Numbers n such that (n^n-2^2)/(n-2) is an integer.

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