A014689 - OEIS

A014689

a(n) = prime(n)-n, the number of nonprimes less than prime(n).

1, 1, 2, 3, 6, 7, 10, 11, 14, 19, 20, 25, 28, 29, 32, 37, 42, 43, 48, 51, 52, 57, 60, 65, 72, 75, 76, 79, 80, 83, 96, 99, 104, 105, 114, 115, 120, 125, 128, 133, 138, 139, 148, 149, 152, 153, 164, 175, 178, 179, 182, 187, 188, 197, 202, 207, 212, 213, 218, 221, 222

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OFFSET

1,3

COMMENTS

a(n) = A048864(A000040(n)) = number of nonprimes in RRS of n-th prime. - Labos Elemer, Oct 10 2002

A000040 - A014689 = A000027; in other words, the sequence of natural numbers subtracted from the prime sequence produces A014689. - Enoch Haga, May 25 2009

a(n) = A000040(n) - n. a(n) = inverse (frequency distribution) sequence of A073425(n), i.e., number of terms of sequence A073425(n) less than n. a(n) = A065890(n) + 1, for n >= 1. a(n) - 1 = A065890(n) = the number of composite numbers, i.e., (A002808) less than n-th primes, (i.e., < A000040(n)). - Jaroslav Krizek, Jun 27 2009

a(n) = A162177(n+1) + 1, for n >= 1. a(n) - 1 = A162177(n+1) = the number of composite numbers, i.e., (A002808) less than (n+1)-th number of set {1, primes}, (i.e., < A008578(n+1)). - Jaroslav Krizek, Jun 28 2009

Conjecture: Each residue class contains infinitely many terms of this sequence. Similarly, for any integers m > 0 and r, we have prime(n) + n == r (mod m) for infinitely many positive integers n. - Zhi-Wei Sun, Nov 25 2013

First differences are A046933 = differences minus one between successive primes. - Gus Wiseman, Jan 18 2020

LINKS

T. D. Noe, Table of n, a(n) for n=1..1000

FORMULA

G.f: b(x) - x/((1-x)^2), where b(x) is the g.f. of A000040. - Mario C. Enriquez, Dec 13 2016

MATHEMATICA

Table[Prime[n] - n, {n, 61}] (* Alonso del Arte *)

PROG

(PARI) a(n) = prime(n)-n \\ Charles R Greathouse IV, Sep 05 2011

(Haskell)

a014689 n = a000040 n - fromIntegral n

-- Reinhard Zumkeller, Apr 09 2012