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A086748
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Numbers m such that when C(2k, k) == 1 (mod m) then k is necessarily even.
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0
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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
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1,1
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COMMENTS
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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*(2t-1) or 5*(2t-1) (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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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