A054272 Number of primes in the interval [prime(n), prime(n)^2].

2, 3, 7, 12, 26, 34, 55, 65, 91, 137, 152, 208, 251, 270, 315, 394, 471, 502, 591, 656, 685, 790, 864, 977, 1139, 1227, 1268, 1354, 1395, 1494, 1847, 1945, 2109, 2157, 2455, 2512, 2693, 2878, 3005, 3202, 3396, 3471, 3826, 3902, 4045, 4119, 4581, 5059

Offset: 1

Author

Labos Elemer, May 05 2000

Keywords

• nonn

Comments

These primes are candidates for fortunate numbers (A005235).

These are precisely the primes available for the solution of Aguilar's conjecture or Haga's conjecture in Carlos Rivera's The Prime Puzzles and Problems Connection, (conjecture 26). Aguilar's conjecture states that at least one prime will be available for placement on each row and column of a p X p square array. Haga's conjecture states that just p primes are required for such placement in any p X p array. - Enoch Haga, Jan 23 2002

Also number of times p_n (the n-th prime) occurs as the least prime factor (A020639) among numbers in range [(p_n)+1, ((p_n)^3)-1]. For n=1, p_1 = 2 and there are two even numbers in range [3, 7], namely 4 and 6, so a(1) = 2. See also A250474. - Antti Karttunen, Dec 05 2014

The number of consecutive primes after the leading 1 in the prime(n)-rough numbers. - Benedict W. J. Irwin, Mar 24 2016

Examples

			n=4, the zone in question is [7,49] and encloses a(4)=12 primes, as follows: {7,11,13,17,19,23,29,31,37,41,43,47}.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..5000
• Carlos Rivera, Conjecture 26. The Calendar-like square Conjecture, The Prime Puzzles and Problems Connection.

Crossrefs

One less than A250473, two less than A250474.

First differences: A251723.

Cf. A000040, A000879, A001248, A002110, A005235, A005867, A008683, A008836, A020639, A078898, A249747.

Programs

Mathematica

a[n_] := PrimePi[Prime[n]^2] - n + 1; Array[a, 50] (* Jean-François Alcover, Dec 07 2015 *)

PARI

\\ A fast version:
default(primelimit, 2^31 + 2^30);
A054272(n) = 1 + primepi(prime(n)^2) - n;
for(n=1, 5000, write("b054272.txt", n, " ", A054272(n)));
\\ The following mirrors the given new formula. It is far from an optimal way to compute this sequence:
allocatemem(234567890);
A002110(n) = prod(i=1, n, prime(i));
A054272(n) = { my(p2); p2 = prime(n)^2; sumdiv(A002110(n), d, moebius(d)*floor(p2/d)); };
for(n=1, 22, print1(A054272(n),", ")); \\ Antti Karttunen, Dec 05 2014

Formula

a(n) = A000879(n) - n + 1.

From Antti Karttunen, Dec 05-08 2014: (Start)

a(n) = A250473(n) - 1 = A250474(n) - 2.

a(n) = sum_{d | A002110(n)} moebius(d) * floor((p_n)^2 / d). [Where p_n is the n-th prime (A000040(n)) and A002110(n) gives the product of the first n primes. Because the latter is always squarefree, one could also use Liouville's lambda (A008836) instead of Moebius mu (A008683).]

The ratio (a(n) * A002110(n)) / (A001248(n) * A005867(n)) stays near 1, which follows from the above summation formula. See also A249747.

(End)