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Smallest primitive prime p such that 1/p has duodecimal period factor of 12^n-1.
a(n) = the Also, smallest primitive prime factor of 12^n-p such that 1/p has duodecimal period n.
a(n) is known up to n = 310.
Max Alekseyev, <a href="/A252170/b252170.txt">Table of n, a(n) for n = 1..310</a>
Edited by Max Alekseyev, Aug 26 2021
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(PARI) listap(nn) = {prf = []; for (n=1, nn, vp = (factor(12^n-1)[, 1])~; f = setminus(Set(vp), Set(prf)); prf = concat(prf, f); print1(vecmin(Vec(f)), ", "); ); } \\ Michel Marcus, Dec 15 2014; after A007138
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(PARI) listap(nn) = {prf = []; for (n=1, nn, vp = (factor(12^n-1)[, 1])~; f = setminus(Set(vp), Set(prf)); prf = concat(prf, f); print1(vecmin(Vec(f)), ", "); ); } \\ Michel Marcus, Dec 15 2014
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a(4) = 5 because 1/5 = 0.249724972497... and 5 is the smallest prime with period 4 in base 12.
a(5) = 22621 because 1/22621 = 0.0000100001... and 22621 is the smallest (in fact, the only one) prime with period 5 in base 12.
prms={}; Table[f=First/@FactorInteger[12^n-1]; p=Complement[f, prms]; prms=Join[prms, p]; If[p=={}, 1, First[p]], {n, 72}]