A185358 The period of the sequence i^i (mod n) starts from i=a(n).
1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 4, 1, 1, 1
Offset: 1
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 1..5000
- R. Hampel, The length of the shortest period of rests of numbers n^n, Ann. Polon. Math. 1 (1955), 360-366.
Crossrefs
Cf. A185359.
Programs
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Mathematica
a[p_,e_]:=1- p*(1+Floor[-e/p]);a[n_]:=Max@Module[{fa=FactorInteger[n]},Table[a[fa[[i,1]],fa[[i,2]]],{i,1,Length[fa]}]];Table[a[n],{n,1,84}]
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Python
from sympy import factorint, floor def a(n): f=factorint(n) return 1 if n==1 else max(1 - i*(1 + (-f[i])//i) for i in f) print([a(n) for n in range(1, 201)]) # Indranil Ghosh, Jun 29 2017
Formula
If n = Product_{pi^ei} then a(n) = Max_{1- pi*(1+floor[-ei/pi])}.