A247477 Largest m such that n/m + n divides (n/m)^(n/m) + n, (n/m)^n + n/m and n^(n/m) + n/m, or 0 if no such m exists.
1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 4, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 4, 1, 6, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2
Offset: 1
Keywords
Examples
a(10) = 2 because 10/2 + 10 = 15 divides (10/2)^(10/2) + 10 = 3135, (10/2)^10 + 10/2 = 9765630, 10^(10/2) + 10/2 = 100005, i.e., 3135/15 = 209, 9765630/15 = 651042, 100005/15 = 6667.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
superdivisors[n_] := Select[Range@ n, And[Mod[(n/#)^(n/#) + n, n/# + n] == 0, Mod[(n/#)^n + n/#, n/# + n] == 0, Mod[n^(n/#) + n/#, n/# + n] == 0] &] /. {} -> 0; Min /@ Array[superdivisors, 94] (* Michael De Vlieger, Feb 09 2015 *)
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PARI
a(n)=fordiv(n,d,my(m=n/d,k=d+n); if(Mod(d,k)^d==-n && Mod(d,k)^n==-d && Mod(n,k)^d==-d, return(m))); 0 \\ Charles R Greathouse IV, Feb 19 2015
Comments