A373469 -id:A373469 - cp's OEIS frontend

A086748 Odd numbers m such that when C(2k, k) == 1 (mod m) then k is necessarily even.

3, 5, 9, 15, 21, 25, 27, 33, 35, 39, 45, 51, 55, 57, 63, 65, 69, 75, 81, 85, 87, 93, 95, 99, 105, 111, 115, 117, 123, 125, 129, 135, 141, 145, 147, 153, 155, 159, 165, 171, 175, 177, 183, 185, 189, 195, 201, 205, 207, 213, 215, 219, 225, 231, 235, 237, 243, 245

Offset: 1

Benoit Cloitre, Jul 30 2003

nonn

From Jinyuan Wang, Apr 05 2020: (Start)
All terms are odd, because C(2k, k) is always divisible by 2.
If m is a term, then m*t is also a term for odd numbers t.
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.
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!
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.
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)
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
Numbers m such that A099976(k) = 1 (mod m) has no solutions k. - R. J. Mathar, Jul 11 2024
Is this  A005408 \ A007775 ? - Antti Karttunen, Jul 11 2024
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

			m = 7 is not a term because C(2k,k) = 1 (mod 7) is solvable by the odd k=17.
m = 11 is not a term because C(2k,k) = 1 (mod 11) is solvable by the odd k=13.
m = 13 is not a term because C(2k,k) = 1 (mod 13) is solvable by the odd k=2383.
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
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

Cf. A000984.

Cf. A373469 (least k such that C(2k,k)=1 mod A007775(n)).

13 removed and offset changed by Jinyuan Wang, Apr 04 2020
23 removed and more terms added by Giovanni Resta, Apr 05 2020
Definition corrected by Max Alekseyev, Jul 12 2024

cp's OEIS Frontend

A086748 Odd numbers m such that when C(2k, k) == 1 (mod m) then k is necessarily even.

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