A256342 Moduli n for which A248218(n) = 2 (length of the terminating cycle of 0 under x -> x^2+1 modulo n).
2, 4, 6, 8, 11, 12, 14, 16, 22, 23, 24, 28, 29, 32, 33, 38, 42, 44, 46, 48, 53, 56, 58, 62, 64, 66, 67, 69, 74, 76, 77, 84, 86, 87, 88, 92, 96, 106, 107, 109, 112, 114, 116, 124, 127, 128, 132, 134, 138, 148, 152, 154, 159, 161, 163, 168, 172, 174, 176, 184, 186, 192
Offset: 1
Keywords
Examples
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. 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. 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.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
filter:= proc(n) local x, k, R,p; x:= 0; R[0]:= 0; for k from 1 do x:= x^2+1 mod n; if assigned(R[x]) then return evalb(k-R[x] = 2) else R[x]:= k fi od; end proc: select(filter, [$1..1000]); # Robert Israel, Mar 09 2021
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Mathematica
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]]]; Select[Range[1000], filterQ] (* Jean-François Alcover, Feb 01 2023, after Robert Israel *)
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PARI
for(i=1,200,A248218(i)==2&&print1(i","))
Comments