A165634 -id:A165634 - cp's OEIS frontend

A095117 a(n) = pi(n) + n, where pi(n) = A000720(n) is the number of primes <= n.

0, 1, 3, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 20, 21, 22, 24, 25, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 39, 40, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 54, 55, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 69, 70, 71, 72, 73, 74, 76, 77, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89, 91

Offset: 0

Dean Hickerson, following a suggestion of Leroy Quet, May 28 2004

Positions of first occurrences of n in A165634: A165634(a(n))=n for n>0. - Reinhard Zumkeller, Sep 23 2009

There exists at least one prime number p such that n < p <= a(n) for n >= 2. For example, 2 is in (2, 3], 5 in (3, 5], 5 in (4, 6], ..., and primes 73, 79, 83 and 89 are in (71, 91] (see Corollary 1 in the paper by Ya-Ping Lu attached in the links section). - Ya-Ping Lu, Feb 21 2021

Carmine Suriano, Table of n, a(n) for n = 0..9999
Ya-Ping Lu, Lower Bounds for the Number of Primes in Some Integer Intervals

Complement of A095116.

Cf. A000720, A064427.

a095117 n = a000720 n + toInteger n  -- Reinhard Zumkeller, Apr 17 2012

with(numtheory): seq(n+pi(n),n=1..90); # Emeric Deutsch, May 02 2007

Table[ PrimePi@n + n, {n, 0, 71}] (* Robert G. Wilson v, Apr 22 2007 *)

a(n) = n + primepi(n); \\ Michel Marcus, Feb 21 2021

from sympy import primepi
def a(n): return primepi(n) + n
print([a(n) for n in range(72)]) # Michael S. Branicky, Feb 21 2021

a(0) = 0; for n>0, a(n) = a(n-1) + (if n is prime then 2, else 1). - Robert G. Wilson v, Apr 22 2007; corrected by David James Sycamore, Aug 16 2018

Edited by N. J. A. Sloane, Jul 02 2008 at the suggestion of R. J. Mathar

A095116 a(n) = prime(n) + n - 1.

Original entry on oeis.org

2, 4, 7, 10, 15, 18, 23, 26, 31, 38, 41, 48, 53, 56, 61, 68, 75, 78, 85, 90, 93, 100, 105, 112, 121, 126, 129, 134, 137, 142, 157, 162, 169, 172, 183, 186, 193, 200, 205, 212, 219, 222, 233, 236, 241, 244, 257, 270, 275, 278, 283, 290, 293, 304, 311, 318, 325

Offset: 1

Dean Hickerson, following a suggestion of Leroy Quet, May 28 2004

nonn

Positions of second occurrences of n in A165634: A165634(a(n)) = n. [Reinhard Zumkeller, Sep 23 2009]
a(n) = b(n)-th highest positive integer not equal to any a(k), 1 <= k <= n-1, where b(n) = primes = A000040(n). a(1) = 2, a(n) = a(n-1) + A000040(n) - A000040(n-1) + 1 for n >= 2. a(1) = 2, a(n) = a(n-1) + A001223(n-1) + 1 for n >= 2. a(n) = A014688(n) - 1. [Jaroslav Krizek, Oct 28 2009]
Comment from N. J. A. Sloane, Mar 28 2024 (Start):
On March 23 2024, Davide Rotondo sent me an email with the following conjecture. (I've simplified it a bit.)
For a positive integer n, define a sequence b by b(0) = n; b(i) = n - pi(b(i-1)) for i >= 1, where pi(x) = number of primes <= x.
The conjecture is that after some initial terms, b becomes periodic with period length 1 or 2, and the n for which the period is 2 are 3 together with the present sequence, that is, 2, 3, 4, 7, 10, 15, 18, 23, 26, 31, 38, 41, 48, ... (End)
Proof from Robert Israel, Mar 26 2024 (Start):
This is simply a consequence of the fact that if x < y, 0 <= pi(y) - pi(x) <= y - x and the inequality on the right is strict if y-x > 1 except for the case of 1 and 3.
Thus we start with b(0) - b(1) = pi(n). While |b(i) - b(i+1)| > 2 we get |b(i+1) - b(i+2)| = |pi(b(i+1)) - pi(b(i+2))| < |b(i) - b(i+1)|.
Eventually we must either reach |b(j+1) - b(j)| = 0 or |b(j+1) - b(j)| = 1.
If we reach 0, i.e. b(j+1) = b(j), then clearly b(k) = b(j) for all k > j.
If b(j+1) = b(j) + 1 = n - pi(b(j)), then b(j+2) = n - pi(b(j)+1) = b(j+1) or b(j+1)-1.
If b(j+1) = b(j) - 1, then b(j+2) = n - pi(b(j)-1) = b(j+1) or b(j+1)+1.
Thus from this point on we either get a 2-cycle or a 1-cycle. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Complement of A095117.

Essentially the same sequence as A014690.

a095116 n = a000040 n + toInteger n - 1
-- Reinhard Zumkeller, Apr 17 2012

[n+NthPrime(n)-1: n in [1..60]]; // Vincenzo Librandi, Aug 15 2017

with (numtheory):seq(n+ithprime(n+1),n=0..56); # Zerinvary Lajos, Aug 24 2008

Table[n + Prime[n] - 1, {n, 100}] (* Vincenzo Librandi, Aug 15 2017 *)

a(n) = A014690(n-1), n > 1. [R. J. Mathar, Sep 05 2008]

cp's OEIS Frontend

A095117 a(n) = pi(n) + n, where pi(n) = A000720(n) is the number of primes <= n.

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