A128059 a(n) = numerator((2*n-1)^2/(2*(2*n)!)).
1, 1, 3, 5, 7, 1, 11, 13, 1, 17, 19, 1, 23, 1, 1, 29, 31, 1, 1, 37, 1, 41, 43, 1, 47, 1, 1, 53, 1, 1, 59, 61, 1, 1, 67, 1, 71, 73, 1, 1, 79, 1, 83, 1, 1, 89, 1, 1, 1, 97, 1, 101, 103, 1, 107, 109, 1, 113, 1, 1, 1, 1, 1, 1, 127
Offset: 0
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Programs
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Haskell
a128059 0 = 1 a128059 n = f n n where f 1 _ = 1 f x q = if a010051' q' == 1 then q' else f x' q' where x' = x - 1; q' = q + x' -- Reinhard Zumkeller, Jun 14 2015 -
Maple
A128059 := proc(n): numer(((2*n-1)^2)/(2*(2*n)!)) end: seq(A128059(n), n=0..64); # Artur Jasinski, Nov 29 2008 A128059 := proc(n): if isprime(2*n-1) then 2*n-1 else 1 fi: end: seq(A128059(n), n=0..64); # Johannes W. Meijer, Oct 25 2012, Jun 01 2016
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Mathematica
Table[Numerator[(2 n - 1)^2/(2 (2 n)!)], {n, 0, 64}] (* Michael De Vlieger, Jun 01 2016 *) -
Python
from sympy import isprime def A128059(n): return a if isprime(a:=(n<<1)-1) else 1 # Chai Wah Wu, Feb 26 2024
Formula
Conjecture: a(n) = denominator(f(n-1)) with f(n) = lcm(2,3,4,5,...,n)*(Sum_{k=0..n} frac(Bernoulli(2*k))*binomial(n+k,k)). - Yalcin Aktar, Jul 23 2008
a(n) = 2*n-3 if 2*n-3 is prime and a(n) = 1 otherwise. a(n+4) = A145737(n+2), for n >= 1. - Artur Jasinski, Nov 29 2008
a(n+1) = denominator( (2n)!/(2n+1) ), n > 0. - Wesley Ivan Hurt, Jun 19 2013
a(n+1) = abs(2n*(pi(2n) - pi(2n-2)) - 1) where abs is the absolute value function and pi is the prime counting function (A000720). - Anthony Browne, Jun 28 2016
a(n+1) = denominator(Bernoulli(2*n)*(2*n)!) = numerator(Clausen(2*n,1)/(2*n)!) with Clausen defined in A160014. - Peter Luschny, Sep 25 2016
Comments