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%I #15 Jun 08 2017 23:50:29
%S 2,3,2,2,3,2,2,3,2,2,2,2,2,2
%N a(n) = length of the repeating part of row n of A288097.
%C a(n) + A268479(n) = total number of different terms in the trajectory of p.
%C a(15) is unknown, since there is no known Wieferich prime in base 47 (cf. Fischer link).
%C Obviously, a(n) != 1 for all n.
%C Period length of the repeating part of prime(n)-th row of A281001. - _Felix Fröhlich_, Jan 14 2017
%H R. Fischer, <a href="http://www.fermatquotient.com/FermatQuotienten/FermQ_Sort.txt">Thema: Fermatquotient B^(P-1) == 1 (mod P^2)</a>
%e The trajectory of 31 starts 31, 7, 5, 2, 1093, 2, 1093, 2, 1093, ...., entering a repeating cycle of length 2, so a(11) = 2.
%t Table[Length@ DeleteCases[Values@ PositionIndex@ NestList[Function[n, Block[{p = 2}, While[! Divisible[n^(p - 1) - 1, p^2], p = NextPrime@ p]; p]], Prime@ n, 12], _?(Length@ # == 1 &)], {n, 12}] (* _Michael De Vlieger_, Jun 06 2017, Version 10 *)
%o (PARI) a039951(n) = forprime(p=1, , if(Mod(n, p^2)^(p-1)==1, return(p)))
%o trajectory(n, terms) = my(v=[n]); while(#v < terms, v=concat(v, a039951(v[#v]))); v
%o a(n) = my(p=prime(n), i=0, len=2, t=trajectory(p, len), k=#t); while(1, while(k > 1, k--; if(t[k]==t[#t], return(#t-k))); len++; t=trajectory(p, len); k=#t) \\ _Felix Fröhlich_, Jan 14 2017
%Y Cf. A039951, A244550, A252801, A252802, A252812, A268479, A281001, A288097.
%K nonn,hard,more
%O 1,1
%A _Felix Fröhlich_, Feb 19 2016
%E Definition simplified by _Felix Fröhlich_, Jun 05 2017
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