OFFSET
1,4
COMMENTS
a(n) < log(4)*n/log(n) < 7*n/(5*log(n)) for n > 1. - Reinhard Zumkeller, Mar 04 2008
Bertrand's postulate is equivalent to the formula a(n) >= 1 for all positive integers n. - Jonathan Vos Post, Jul 30 2008
Number of distinct prime factors > n of binomial(2*n,n). - T. D. Noe, Aug 18 2011
f(2, 2n) - f(3, n) < a(n) < f(3, 2n) - f(2, n) for n > 5889 where f(k, x) = x/log x * (1 + 1/log x + k/(log x)^2). The constant 3 can be improved. - Charles R Greathouse IV, May 02 2012
For n >= 2, a(n) is the number of primes appearing in the 2nd row of an n X n square array whose elements are the numbers from 1..n^2, listed in increasing order by rows. - Wesley Ivan Hurt, May 17 2021
REFERENCES
F. Irschebeck, Einladung zur Zahlentheorie, BI Wissenschaftsverlag 1992, p. 40
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
Tsutomu Hashimoto, On a certain relation between Legendre's conjecture and Bertrand's postulate, arXiv:0807.3690 [math.GM], 2008.
FORMULA
For n > 1, a(n) = A060715(n). - David Wasserman, Nov 04 2005
Conjecture: G.f.: Sum_{i>0} Sum_{j>=i|i+j is prime} x^j. - Benedict W. J. Irwin, Mar 31 2017
From Wesley Ivan Hurt, Sep 20 2021: (Start)
a(n) = Sum_{k=1..n} A010051(2*n-k+1).
MAPLE
MATHEMATICA
Table[Length[Select[Transpose[FactorInteger[Binomial[2 n, n]]][[1]], # > n &]], {n, 100}] (* T. D. Noe, Aug 18 2011 *)
f[n_] := Length@ Select[ Range[n + 1, 2n], PrimeQ]; Array[f, 100] (* Robert G. Wilson v, Mar 20 2012 *)
Table[PrimePi[2n]-PrimePi[n], {n, 90}] (* Harvey P. Dale, Mar 11 2013 *)
PROG
(PARI) g(n) = for(x=1, n, y=primepi(2*x)-primepi(x); print1(y", "))
(Python)
from sympy import primepi
def A108954(n): return primepi(n<<1)-primepi(n) # Chai Wah Wu, Aug 19 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Cino Hilliard, Jul 22 2005
STATUS
approved