A119861 Number of distinct prime factors of the odd Catalan numbers A038003(n).

0, 1, 3, 6, 11, 20, 36, 64, 117, 209, 381, 699, 1291, 2387, 4445, 8317, 15645, 29494, 55855, 106021, 201778, 384941, 735909, 1409683, 2705277, 5200202

Offset

1,3

Comments

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.

Odd Catalan numbers are listed in A038003[n] = A000108[2^n-1] = binomial(2^(n+1)-2, 2^n-1)/(2^n).

Links

Table of n, a(n) for n=1..26.

Eric Weisstein's World of Mathematics, Catalan Number.

Formula

a(n) = Length[ FactorInteger[ Binomial[ 2^(n+1)-2, 2^n-1] / (2^n) ]].

Maple

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

Mathematica

Table[Length[FactorInteger[Binomial[2^(n+1)-2, 2^n-1]/(2^n)]], {n, 1, 15}]

Prog

(Python)
from sympy import factorint
A119861_list, c, s = [0], {}, 3
for n in range(2, 2**19):
....for p, e in factorint(4*n-2).items():
........if p in c:
............c[p] += e
........else:
............c[p] = e
....for p, e in factorint(n+1).items():
........if c[p] == e:
............del c[p]
........else:
............c[p] -= e
....if n == s:
........A119861_list.append(len(c))
........s = 2*s+1 # Chai Wah Wu, Feb 12 2015

Crossrefs

Cf. A038003, A000108, A120274, A120275.

Cf. A000108 = Catalan Number. Cf. A038003 = Odd Catalan numbers. Cf. A120274, A120275, A119908, A094389.

Sequence in context: A077855 A054887 A019302 * A255061 A018075 A125896

Adjacent sequences: A119858 A119859 A119860 * A119862 A119863 A119864

Keyword

nonn

Author

Alexander Adamchuk, Jul 31 2006, Oct 11 2007

Extensions

a(16)-a(18) from Robert G. Wilson v, May 15 2007

a(19)-a(26) from Chai Wah Wu, Feb 12 2015

Status

approved