login
Moduli n for which A248218(n) = 2 (length of the terminating cycle of 0 under x -> x^2+1 modulo n).
9

%I #15 Feb 01 2023 15:00:24

%S 2,4,6,8,11,12,14,16,22,23,24,28,29,32,33,38,42,44,46,48,53,56,58,62,

%T 64,66,67,69,74,76,77,84,86,87,88,92,96,106,107,109,112,114,116,124,

%U 127,128,132,134,138,148,152,154,159,161,163,168,172,174,176,184,186,192

%N Moduli n for which A248218(n) = 2 (length of the terminating cycle of 0 under x -> x^2+1 modulo n).

%C If x is a member and y is a member of this sequence or A248219, then LCM(x,y) is a member. - _Robert Israel_, Mar 09 2021

%H Robert Israel, <a href="/A256342/b256342.txt">Table of n, a(n) for n = 1..10000</a>

%e In Z/mZ with m = 2, the iteration of x -> x^2+1 starting at x = 0 yields (0, 1, 0, ...), and m = 2 is the least positive number for which there is such a cycle of length 2, here [0, 1], therefore a(1) = 2.

%e For m = 3, the iteration yields (0, 1, 2, 2, ...), i.e., a cycle [2] of length 1, therefore 3 is not in this sequence.

%e For m = 4, the iterations yield (0, 1, 2, 1, ...), and since there is again a cycle [1, 2] of length 2, a(2)=4.

%p filter:= proc(n) local x, k, R,p;

%p x:= 0; R[0]:= 0;

%p for k from 1 do

%p x:= x^2+1 mod n;

%p if assigned(R[x]) then return evalb(k-R[x] = 2)

%p else R[x]:= k

%p fi

%p od;

%p end proc:

%p select(filter, [$1..1000]); # _Robert Israel_, Mar 09 2021

%t filterQ[n_] := Module[{x, k, R}, x = 0; R[0] = 0; For[k = 1, True, k++, x = Mod[x^2 + 1, n]; If[IntegerQ[R[x]], Return[k - R[x] == 2], R[x] = k]]];

%t Select[Range[1000], filterQ] (* _Jean-François Alcover_, Feb 01 2023, after _Robert Israel_ *)

%o (PARI) for(i=1,200,A248218(i)==2&&print1(i","))

%Y Cf. A248218, A248219, A256343 - A256349, A003095, A247981.

%K nonn

%O 1,1

%A _M. F. Hasler_, Mar 25 2015