A242872 Least number k > 1 such that (k^k-n^n)/(k-n) is an integer.
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
Keywords
Examples
(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.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
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
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Mathematica
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 *)
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PARI
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)
Comments