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%I #12 Feb 15 2015 00:17:52
%S 0,1,3,6,11,20,36,64,117,209,381,699,1291,2387,4445,8317,15645,29494,
%T 55855,106021,201778,384941,735909,1409683,2705277,5200202
%N Number of distinct prime factors of the odd Catalan numbers A038003(n).
%C A038003[n] = A000108[2^n-1] = binomial(2^(n+1)-2, 2^n-1)/(2^n). a(1) = 0 because A038003[1] = 1. a(2) = 1 because A038003[2] = 5. a(3) = 3 because A038003[3] = 429 = 3*11*13. a(4) = 6 because A038003[4] = 9694845 = 3^2*5*17*19*23*29.
%C Odd Catalan numbers are listed in A038003[n] = A000108[2^n-1] = binomial(2^(n+1)-2, 2^n-1)/(2^n).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CatalanNumber.html">Catalan Number</a>.
%F a(n) = Length[ FactorInteger[ Binomial[ 2^(n+1)-2, 2^n-1] / (2^n) ]].
%p with(numtheory): c:=proc(n) options operator, arrow: binomial(2*n, n)/(n+1) end proc: seq(nops(factorset(c(2^n-1))),n=1..15); # _Emeric Deutsch_, Oct 24 2007
%t Table[Length[FactorInteger[Binomial[2^(n+1)-2, 2^n-1]/(2^n)]],{n,1,15}]
%o (Python)
%o from sympy import factorint
%o A119861_list, c, s = [0], {}, 3
%o for n in range(2,2**19):
%o ....for p,e in factorint(4*n-2).items():
%o ........if p in c:
%o ............c[p] += e
%o ........else:
%o ............c[p] = e
%o ....for p,e in factorint(n+1).items():
%o ........if c[p] == e:
%o ............del c[p]
%o ........else:
%o ............c[p] -= e
%o ....if n == s:
%o ........A119861_list.append(len(c))
%o ........s = 2*s+1 # _Chai Wah Wu_, Feb 12 2015
%Y Cf. A038003, A000108, A120274, A120275.
%Y Cf. A000108 = Catalan Number. Cf. A038003 = Odd Catalan numbers. Cf. A120274, A120275, A119908, A094389.
%K nonn
%O 1,3
%A _Alexander Adamchuk_, Jul 31 2006, Oct 11 2007
%E a(16)-a(18) from _Robert G. Wilson v_, May 15 2007
%E a(19)-a(26) from _Chai Wah Wu_, Feb 12 2015
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