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%I #17 Nov 20 2015 05:11:06
%S 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,
%T 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,
%U 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3
%N Smallest prime factor of the odd Catalan number A038003(n).
%C A038003(n) = binomial(2^(n+1)-2, 2^n-1)/(2^n).
%C 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
%e a(2) = 5 because A038003(2) = 5.
%e a(3) = 3 because A038003(3) = 429 = 3*11*13.
%p f:= proc(n) local m;
%p m:= 2^n-1;
%p if has(convert(m,base,3),2) then return 3 fi;
%p min(numtheory:-factorset(binomial(2*m,m)/(m+1)));
%p end proc:
%p seq(f(n),n=2..1000); # _Robert Israel_, Nov 18 2015
%t 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 *)
%Y Cf. A038003, A000108.
%K nonn
%O 2,1
%A _Alexander Adamchuk_, Jul 04 2006
%E a(16)-a(28) from _Robert G. Wilson v_, Nov 14 2015
%E a(29)-a(86) from _Robert Israel_, Nov 18 2015
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