

A086748


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


0



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 values are either terms, because of the form 3*(2t1) or 5*(2t1) (as proved by Jinyuan Wang), or there exist an odd k <= 7412629 such that C(2k, k) == 1 (mod m).  Giovanni Resta, Apr 05 2020


LINKS

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


CROSSREFS

Cf. A000984.
Sequence in context: A029533 A018685 A107994 * A014957 A014876 A045602
Adjacent sequences: A086745 A086746 A086747 * A086749 A086750 A086751


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Jul 30 2003


EXTENSIONS

13 removed and offset changed by Jinyuan Wang, Apr 04 2020
23 removed and more terms added by Giovanni Resta, Apr 05 2020


STATUS

approved



