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A109641
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Composite n such that binomial(3n, n) == 3^k (mod n) for some integer k > 0.
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5
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4, 9, 15, 25, 27, 34, 36, 49, 51, 57, 63, 68, 75, 81, 87, 93, 111, 121, 125, 129, 132, 138, 141, 153, 155, 159, 169, 177, 237, 249, 258, 261, 264, 267, 274, 276, 279, 289, 298, 303, 324, 339, 343, 357, 361, 375, 381, 387, 393, 411, 417, 423, 441, 447, 453, 477
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OFFSET
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1,1
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COMMENTS
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Includes p^k for k >= 2 and p > 2 in A019334 but not in A014127, as binomial(3n,n) is coprime to p and 3 is a primitive root mod p^k. - Robert Israel, Nov 12 2017
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LINKS
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EXAMPLE
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Binomial(3*34,34) == 3^6 (mod 34), so 34 is a member.
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MAPLE
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filter:= proc(n) local p, m, k, t;
if isprime(n) then return false fi;
p:= padic:-ordp(n, 3);
p:= p + numtheory:-order(3, n/3^p);
m:= binomial(3*n, n) mod n;
t:= 1;
for k from 1 to p do
t:= t*3 mod n;
if t = m then return true fi;
od:
false
end proc;
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MATHEMATICA
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okQ[n_] := Module[{p, m}, If[PrimeQ[n], Return[False]]; p = IntegerExponent[n, 3]; p = p + MultiplicativeOrder[3, n/3^p]; m = Mod[Binomial[3n, n], n]; AnyTrue[Range[p], m == PowerMod[3, #, n]&]];
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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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