OFFSET
0,3
COMMENTS
The next term has 150 digits. - Harvey P. Dale, Feb 22 2016
Silberger (1968 and 1969) was apparently the first to prove that a(n) = Catalan(2^n-1). - Amiram Eldar, Jan 09 2026
REFERENCES
Thomas Koshy, Catalan Numbers with Applications, Oxford University Press, 2008. See pp. 329-330.
LINKS
Amiram Eldar, Table of n, a(n) for n = 0..10 (terms 0..9 from David Wasserman, May 07 2007)
Thomas Koshy and Mohammad Salmassi, Parity and Primality of Catalan Numbers, The College Mathematics Journal, Vol. 37, No. 1 (2006), pp. 52-53.
Hsueh-Yung Lin, Odd Catalan Numbers modulo 2^k, Integers 11 (2011), #A55.
D. M. Silberger, The parity of the integer (2n-2)!/n!(n-1)!, Notices of Amer. Math. Soc., Vol. 15 (1968), p. 615, Abstract #657-7; entire issue.
D. M. Silberger, Occurrences of the integer (2n-2)!/n!(n-1)!, Roczniki Polskiego Towarzystwa Math., Vol. 13, No. 1 (1969), pp. 91-96.
Eric Weisstein's World of Mathematics, Catalan Number.
FORMULA
a(n) = binomial(2^(n+1)-2, 2^n-1)/(2^n).
a(n-1) = C(2^n,2^(n-1))/(2^n - 1)/2. - Benoit Cloitre, Aug 17 2002
a(n) = A000108(2^n-1). - David Wasserman, May 07 2007
MATHEMATICA
Select[CatalanNumber[Range[0, 300]], OddQ] (* Harvey P. Dale, Feb 22 2016 *)
a[n_] := CatalanNumber[2^n-1]; Array[a, 8, 0] (* Amiram Eldar, Jan 09 2026 *)
PROG
(Python)
from __future__ import division
A038003_list, c, s = [1, 1], 1, 3
for n in range(2, 10**5+1):
c = (c*(4*n-2))//(n+1)
if n == s:
A038003_list.append(c)
s = 2*s+1 # Chai Wah Wu, Feb 12 2015
(PARI) a(n) = binomial(2^(n+1)-2, 2^n-1)/(2^n); \\ Joerg Arndt, Nov 05 2015
(Magma) [Binomial(2^(n+1)-2, 2^n-1)/(2^n): n in [0..10]]; // Vincenzo Librandi, Nov 01 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved