OFFSET
2,1
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
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Alexander Adamchuk, Jul 04 2006
EXTENSIONS
a(16)-a(28) from Robert G. Wilson v, Nov 14 2015
a(29)-a(86) from Robert Israel, Nov 18 2015
STATUS
approved