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A120303
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Largest prime factor of Catalan number A000108(n).
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4
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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
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OFFSET
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2,1
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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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 + ...
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MATHEMATICA
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Table[Max[FactorInteger[(2n)!/n!/(n+1)! ]], {n, 2, 100}]
FactorInteger[CatalanNumber[#]][[-1, 1]]&/@Range[2, 70] (* Harvey P. Dale, May 02 2017 *)
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PROG
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(PARI) a(n) = vecmax(factor(binomial(2*n, n)/(n+1))[, 1]); \\ Michel Marcus, Nov 14 2015
(Python)
from gmpy2 import is_prime
for n in range(3, 801):
for k in range(2*n-1, n, -2):
if is_prime(k, n):
break
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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