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A080359
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The smallest integer x > 0 such that the number of primes in (x/2, x] equals n.
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44
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2, 3, 13, 19, 31, 43, 53, 61, 71, 73, 101, 103, 109, 113, 139, 157, 173, 181, 191, 193, 199, 239, 241, 251, 269, 271, 283, 293, 313, 349, 353, 373, 379, 409, 419, 421, 433, 439, 443, 463, 491, 499, 509, 523, 577, 593, 599, 601, 607, 613, 619, 647, 653, 659
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OFFSET
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1,1
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COMMENTS
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a(n) is the same as: Smallest integer x > 0 such that the number of unitary-prime-divisors of x! equals n.
Let p_n be the n-th prime. If p_n>3 is in the sequence, then all integers (p_n-1)/2, (p_n-3)/2, ..., (p_(n-1)+1)/2 are composite numbers. - Vladimir Shevelev, Aug 12 2009
For n >= 3, denote by q(n) the prime which is the nearest from the left to a(n)/2. Then there exists a prime between 2q(n) and a(n). The converse, generally speaking, is not true; i.e., there exist primes that are outside the sequence, but possess such property (e.g., 131). - Vladimir Shevelev, Aug 14 2009
a(n) is the n-th Labos prime.
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LINKS
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N. Amersi, O. Beckwith, S. J. Miller, R. Ronan, and J. Sondow, Generalized Ramanujan primes, Combinatorial and Additive Number Theory, Springer Proc. in Math. & Stat., CANT 2011 and 2012, Vol. 101 (2014), 1-13.
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FORMULA
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EXAMPLE
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n=5: in 31! five unitary-prime-divisors appear (firstly): {17,19,23,29,31}, while other primes {2,3,5,7,11,13} are at least squared. Thus a(5)=31.
Consider a(9)=71. Then the nearest prime < 71/2 is q(9)=31, and between 2q(9) and a(9), i.e., between 62 and 71 there exists a prime (67). - Vladimir Shevelev, Aug 14 2009
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MATHEMATICA
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nn=1000; t=Table[0, {nn+1}]; s=0; Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--]; If[s<=nn && t[[s+1]]==0, t[[s+1]]=k], {k, Prime[3*nn]}]; Rest[t]
(* Second program: *)
a[1] = 2; a[n_] := a[n] = Module[{x = a[n-1]}, While[(PrimePi[x]-PrimePi[Quotient[x, 2]]) != n, x++ ]; x]; Array[a, 54] (* Jean-François Alcover, Sep 14 2018 *)
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PROG
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(PARI) a(n) = {my(x = 1); while ((primepi(x) - primepi(x\2)) != n, x++; ); x; } \\ Michel Marcus, Jan 15 2014
(Sage)
def A():
i = 0; n = 1
while True:
p = prime_pi(i) - prime_pi(i//2)
if p == n:
yield i
n += 1
i += 1
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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