A038003 - OEIS (original) (raw)

A038003

Odd Catalan numbers: a(n) = A000108(2^n-1).

23

1, 1, 5, 429, 9694845, 14544636039226909, 94295850558771979787935384946380125, 11311095732253345760960290897769189975961199415637572612957718759342193629

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

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.

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

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:

(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