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

Author

Alexander Adamchuk, Jul 31 2006, Oct 11 2007

Keywords

• nonn

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

• Eric Weisstein's World of Mathematics, Catalan Number.

Crossrefs

Cf. A038003, A000108, A120274, A120275.

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

Programs

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}]

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

Formula

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

Extensions

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

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