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A120275
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Smallest prime factor of the odd Catalan number A038003(n).
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6
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5, 3, 3, 7, 3, 3, 7, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
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OFFSET
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2,1
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COMMENTS
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A038003(n) = binomial(2^(n+1)-2, 2^n-1)/(2^n).
a(n) <> 3 iff the base-3 representation of 2^n-1 has no 2's. Conjecture: this only occurs for n = 2, 5, 8. I verified it up to n = 10^4. - Robert Israel, Nov 18 2015
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LINKS
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EXAMPLE
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a(3) = 3 because A038003(3) = 429 = 3*11*13.
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MAPLE
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f:= proc(n) local m;
m:= 2^n-1;
if has(convert(m, base, 3), 2) then return 3 fi;
min(numtheory:-factorset(binomial(2*m, m)/(m+1)));
end proc:
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MATHEMATICA
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f[n_] := Block[{p = 2, m = Binomial[2^(n+1)-2, 2^n-1]/(2^n)}, While[Mod[m, p] > 0, p = NextPrime@ p]; p]; Array[f, 27, 2] (* Robert G. Wilson v, Nov 14 2015 *)
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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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