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%I #47 Aug 09 2024 09:32:45
%S 3,5,9,15,21,25,27,33,35,39,45,51,55,57,63,65,69,75,81,85,87,93,95,99,
%T 105,111,115,117,123,125,129,135,141,145,147,153,155,159,165,171,175,
%U 177,183,185,189,195,201,205,207,213,215,219,225,231,235,237,243,245
%N Odd numbers m such that when C(2k, k) == 1 (mod m) then k is necessarily even.
%C From _Jinyuan Wang_, Apr 05 2020: (Start)
%C All terms are odd, because C(2k, k) is always divisible by 2.
%C If m is a term, then m*t is also a term for odd numbers t.
%C Theorem 1: if C(2k, k) == 1 (mod 3) then k is necessarily even. If C(2k, k) == 2 (mod 3) then k is necessarily odd.
%C Proof: for k < 6 it is correct. We have C(6r, 3r) == C(2r, r) (mod 3) and C(6r+4, 3r+2) == C(2r, r)*C(4, 2) == 0 (mod 3). Suppose k is the least value such that theorem 1 is incorrect, then k must be of the form 3r+1. But C(6r+2, 3r+1) == C(2r, r)*C(2, 1) (mod 3), which means that r is a smaller counterexample, a contradiction!
%C Theorem 2: if C(2k, k) == 1 or 4 (mod 5) then k is necessarily even. If C(2k, k) == 2 or 3 (mod 5) then k is necessarily odd.
%C Note that C(10r, 5r) == C(2r, r) (mod 5), C(10r+2, 5r+1) == C(2r, r)*C(2, 1) (mod 5), C(10r+4, 5r+2) == C(2r, r)*C(4, 2) (mod 5), C(10r+6, 5r+3) == C(2r, r)*C(6, 3) (mod 5) and C(10r+8, 5r+4) == C(2r, r)*C(8, 4) (mod 5). The proof is similar to that of theorem 1. (End)
%C Up to m < 1000, all odd m are either of the form 3*(2t-1) or 5*(2t-1) (as proved by _Jinyuan Wang_) and in the sequence, or not in the sequence because an odd k <= 7412629 exists such that C(2k, k) == 1 (mod m). - _Giovanni Resta_, Apr 05 2020
%C Numbers m such that A099976(k) = 1 (mod m) has no solutions k. - _R. J. Mathar_, Jul 11 2024
%C Is this A005408 \ A007775 ? - _Antti Karttunen_, Jul 11 2024
%C It is likely that 1001 = 7*11*13 or other products of at least 3 primes > 5 provide counterexamples, but it is difficult to prove. - _M. F. Hasler_, Jul 13 2024
%e m = 7 is not a term because C(2k,k) = 1 (mod 7) is solvable by the odd k=17.
%e m = 11 is not a term because C(2k,k) = 1 (mod 11) is solvable by the odd k=13.
%e m = 13 is not a term because C(2k,k) = 1 (mod 13) is solvable by the odd k=2383.
%e m = 23 is not a term because C(2k,k) = 1 (mod 23) is solvable by the odd k=3391. - _R. J. Mathar_, Jul 11 2024
%e m=2261 = 7*17*19 is not a term because C(2k,k) = 1 (mod 2261) is solvable by k=57. - _R. J. Mathar_, Aug 09 2024
%Y Cf. A000984.
%Y Cf. A373469 (least k such that C(2k,k)=1 mod A007775(n)).
%K nonn
%O 1,1
%A _Benoit Cloitre_, Jul 30 2003
%E 13 removed and offset changed by _Jinyuan Wang_, Apr 04 2020
%E 23 removed and more terms added by _Giovanni Resta_, Apr 05 2020
%E Definition corrected by _Max Alekseyev_, Jul 12 2024