A120275 Smallest prime factor of the odd Catalan number A038003(n).

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

Offset: 2

Author

Alexander Adamchuk, Jul 04 2006

Keywords

• nonn

Comments

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

Examples

			a(2) = 5 because A038003(2) = 5.
a(3) = 3 because A038003(3) = 429 = 3*11*13.
		

Crossrefs

Cf. A038003, A000108.

Programs

Maple

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:
seq(f(n),n=2..1000); # Robert Israel, Nov 18 2015

Mathematica

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 *)

Extensions

a(16)-a(28) from Robert G. Wilson v, Nov 14 2015

a(29)-a(86) from Robert Israel, Nov 18 2015