A086748 - OEIS

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

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OFFSET

1,1

COMMENTS

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

LINKS

Table of n, a(n) for n=1..58.

EXAMPLE

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

CROSSREFS

Cf. A000984.

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

Sequence in context: A350166 A018685 A107994 * A364561 A014957 A014876