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%I #28 Sep 20 2019 07:25:42
%S 364,1755
%N Multiplicative order of 2 (mod p^2), where p is the n-th Wieferich prime (A001220).
%C Meissner discovered the congruence 2^364 == 1 (mod 1093^2) and thus proved that 1093 is a Wieferich prime, i.e., a term of A001220 (cf. Meissner, 1913).
%C Later, Beeger discovered the congruence 2^1755 == 1 (mod 3511^2) and proved that 3511 is also a Wieferich prime (cf. Beeger, 1922).
%C Let b(n) = (A001220(n)-1)/a(n). Then b(1) = 3 and b(2) = 2.
%C From the fact that a(1) and a(2) are composite it follows that A001220(1) = 1093 and A001220(2) = 3511 do not divide any terms of A001348 (cf. Dobson).
%C Curiously, both 364 and 1755 are repdigits in some base. 364 = 444 in base 9 and 1755 = 3333 in base 8. Compare this with Dobson's observation that 1092 and 3510 are 444 in base 16 and 6666 in base 8, respectively (cf. Dobson).
%H N. G. W. H. Beeger, <a href="https://archive.org/stream/messengerofmathe5051cambuoft#page/148/mode/2up">On a new case of the congruence 2^p-1 == 1 (mod p^2)</a>, Messenger of Mathematics 51 (1922), 149-150.
%H J. B. Dobson, <a href="https://johnblythedobson.org/mathematics/Wieferich_primes.html">A note on the two known Wieferich primes</a>
%H W. Meissner, <a href="/A001917/a001917.pdf">Über die Teilbarkeit von 2^p-2 durch das Quadrat der Primzahl p = 1093</a>, Sitzungsberichte der Königlich Preussischen Akademie der Wissenschaften Berlin, 35 (1913), 663-667. [Annotated scanned copy]
%F a(n) = A014664(A000720(A001220(n))) = A243905(A000720(A001220(n))). [Corrected by _Jianing Song_, Sep 20 2019]
%o (PARI) forprime(p=1, , if(Mod(2, p^2)^(p-1)==1, print1(znorder(Mod(2, p^2)), ", ")))
%Y Cf. A001220, A001348, A014664, A243905, A282552, A282902.
%K nonn,hard,bref,more
%O 1,1
%A _Felix Fröhlich_, May 30 2018
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