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A067934 Let rep(k) = (10^k - 1)/9 be the k-th repunit number = 11111..1111 with k 1 digits, then sequence gives values of k such that rep(k) == 1 (mod k). 2

%I #32 Jul 25 2018 03:46:10

%S 1,2,5,7,10,11,13,17,19,23,29,31,37,41,43,47,53,55,59,61,67,71,73,79,

%T 83,89,91,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,

%U 173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,259

%N Let rep(k) = (10^k - 1)/9 be the k-th repunit number = 11111..1111 with k 1 digits, then sequence gives values of k such that rep(k) == 1 (mod k).

%C Due to Fermat's little theorem, all prime numbers except 3 are in the sequence. E.g., rep(17) = 1 + 17*653594771241830.

%C Numbers n such that 10^n == 10 (mod 9n). The number (10^n - 1)/9 is a term if and only if n is a term. - _Thomas Ordowski_, Apr 28 2018

%C Generally, the repunit theorem: Let integer b <> 1 and n be a positive integer. Define R_b(n) = (b^n-1)/(b-1) = N. Then R_b(N) == 1 (mod N) if and only if N == 1 (mod n). - _Thomas Ordowski_, Apr 28 2018

%C Proof: (b^N-1)/(b-1)-1 = (b^N-b)/(b-1) is divisible by N if and only if b^N-b is divisible by b^n-1. Since b^N-b == b^(N mod n)-b (mod b^n-1), we have that b^N-b is divisible by b^n-1 if and only if N == 1 (mod n). QED. - _Max Alekseyev_, Apr 28 2018

%C Terms which are not prime are 1 U A303608. - _Robert G. Wilson v_, Jun 13 2018

%C No multiples of 3 are in this sequence. - _Eric Chen_, Jun 13 2018

%C A005939 is subsequence. - _Eric Chen_, Jun 13 2018

%H Michael De Vlieger, <a href="/A067934/b067934.txt">Table of n, a(n) for n = 1..10000</a>

%e (10^11 - 1)/9 = 11111111111 == 1 (mod 11), so 11 is a term.

%e We also have the congruence 10^11 == 10 (mod 9*11).

%t {1}~Join~Select[Range[260], Mod[#2, #1] == 1 & @@ {#, (10^# - 1)/9} &] (* _Michael De Vlieger_, May 06 2018 *)

%t fQ[n_] := PowerMod[10, n, 9 n] == 10; fQ[1] = True; Select[Range@260, fQ] (* _Robert G. Wilson v_, Jun 13 2018 *)

%o (PARI) is(n)=n==1 || ((10^n-1)/9)%n==1 \\ _Eric Chen_, Jun 13 2018

%Y Cf. A000864, A015919, A303608, A005939.

%K nonn

%O 1,2

%A _Benoit Cloitre_, Mar 05 2002

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