A120303 Largest prime factor of Catalan number A000108(n).

2, 5, 7, 7, 11, 13, 13, 17, 19, 19, 23, 23, 23, 29, 31, 31, 31, 37, 37, 41, 43, 43, 47, 47, 47, 53, 53, 53, 59, 61, 61, 61, 67, 67, 71, 73, 73, 73, 79, 79, 83, 83, 83, 89, 89, 89, 89, 97, 97, 101, 103, 103, 107, 109, 109, 113, 113, 113, 113, 113, 113, 113, 127, 127, 131, 131

Offset: 2

Author

Alexander Adamchuk, Jul 13 2006

Keywords

• nonn

Comments

All prime numbers (except 3) are present in this sequence in their natural order with repetition. The number of repetitions is equal to A028334(n): differences between consecutive primes, divided by 2. - Alexander Adamchuk, Jul 30 2006

For p>3 a((p+1)/2) = p and all a(n) = p for n >= (p+1)/2 until the first occurrence of the next prime q = NextPrime(p) at a((q+1)/2) = q. - Alexander Adamchuk, Dec 27 2013

For n>2, a(n) is the largest prime less than 2*n. - Gennady Eremin, Mar 02 2021

Examples

			G.f. = 2*x^2 + 5*x^3 + 7*x^4 + 7*x^5 + 11*x^6 + 13*x^7 + 13*x^8 + 17*x^9 + ...
		

Links

• Gennady Eremin, Table of n, a(n) for n = 2..800
• Gennady Eremin, Factoring Catalan numbers, arXiv:1908.03752 [math.NT], 2019.

Crossrefs

Cf. A000108, A006530, A020482, A028334, A060265, A060308, A152765.

Programs

Mathematica

Table[Max[FactorInteger[(2n)!/n!/(n+1)! ]],{n,2,100}]
FactorInteger[CatalanNumber[#]][[-1,1]]&/@Range[2,70] (* Harvey P. Dale, May 02 2017 *)

PARI

a(n) = vecmax(factor(binomial(2*n, n)/(n+1))[,1]); \\ Michel Marcus, Nov 14 2015

PARI

a(n)=if(n>2,precprime(2*n),2) \\ Charles R Greathouse IV, Nov 17 2015

Python

from gmpy2 import is_prime
A120303 = [2]
for n in range(3, 801):
    for k in range(2*n-1, n, -2):
        if is_prime(k, n):
            A120303.append(k)
            break
for n in range(len(A120303)):
    print(n+2, A120303[n])  # Gennady Eremin, Mar 17 2021

Formula

a(n) = A060308(n) = A060265(n) for n>2.

a(n) = A006530(A000108(n)). - Michel Marcus, Nov 14 2015

G.f.: A(x) - x^2, where A(x) is the g.f. of A060265. - Gennady Eremin, Mar 02 2021