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A101681
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Numbers k such that gcd (C(2k,k), 2k+1) > 1.
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2
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7, 16, 17, 19, 22, 25, 31, 34, 38, 42, 43, 45, 46, 47, 49, 52, 55, 58, 61, 64, 67, 70, 71, 72, 73, 76, 77, 79, 80, 82, 87, 88, 92, 93, 94, 97, 100, 102, 103, 104, 106, 107, 110, 112, 115, 117, 122, 123, 124, 127, 129, 130, 133, 136, 139, 142, 143, 145, 147, 148
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OFFSET
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1,1
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COMMENTS
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The set seems to have greater cardinality than its complement.
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LINKS
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EXAMPLE
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7 is in the sequence as gcd(binomial(2*7, 7), 2*7 + 1)) = gcd(3432, 15) = gcd(3*1144, 3*5) > 1. - David A. Corneth, Apr 03 2021
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MATHEMATICA
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Select[Range[200], GCD[Binomial[2 #, #], 2 #+1]>1&] (* Harvey P. Dale, May 11 2019 *)
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PROG
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(PARI) is(n) = { my(f = factor(2*n+1)); for(i = 1, #f~, if(val(2*n, f[i, 1])-2*val(n, f[i, 1]) > 0, return(1))); 0 }
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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