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Numbers k such that k^3 == k (mod 11).
2

%I #39 Jan 01 2024 11:38:57

%S 0,1,10,11,12,21,22,23,32,33,34,43,44,45,54,55,56,65,66,67,76,77,78,

%T 87,88,89,98,99,100,109,110,111,120,121,122,131,132,133,142,143,144,

%U 153,154,155,164,165,166,175,176,177,186,187,188,197,198,199,208,209

%N Numbers k such that k^3 == k (mod 11).

%C Nonnegative integers m such that floor(k*m^2/11) = k*floor(m^2/11), where k can assume the values from 4 to 10. See the second comment in A265187. - _Bruno Berselli_, Dec 03 2015

%H Harvey P. Dale, <a href="/A112654/b112654.txt">Table of n, a(n) for n = 1..1000</a>

%F From _Colin Barker_, Apr 11 2012: (Start)

%F a(n) = a(n-1) + a(n-3) - a(n-4).

%F G.f.: x^2*(1+9*x+x^2)/((1-x)^2*(1+x+x^2)). (End)

%e a(3) = 11 because 11^3 = 1331 == 0 (mod 11) and 11 == 0 (mod 11).

%p m = 11 for n = 1 to 300 if n^3 mod m = n mod m then print n; next n

%t Select[Range@ 209, Mod[#, 11] == Mod[#^3, 11] &] (* _Michael De Vlieger_, Dec 03 2015 *)

%t Select[Range[0,250],PowerMod[#,3,11]==Mod[#,11]&] (* _Harvey P. Dale_, May 15 2016 *)

%K nonn,easy

%O 1,3

%A _Jeremy Gardiner_, Dec 28 2005