A038179 Result of second stage of sieve of Eratosthenes (after eliminating multiples of 2 and 3).
2, 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151
Offset: 1
References
- Fred S. Roberts, Applied Combinatorics, Prentice-Hall, 1984, p. 256.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..2500
- Ahmed Hamdy A. Diab, Sequence eliminating law (SEL) and the interval formulas of prime numbers, arXiv:2012.03052 [math.NT], 2020.
- H. B. Meyer, Eratosthenes' sieve.
- Index entries for sequences generated by sieves.
Programs
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Mathematica
max = 200; Complement[Range[2, max], 2Range[2, Ceiling[max/2]], 6Range[2, Ceiling[max/6]] + 3] (* Alonso del Arte, May 16 2014 *) Prepend[Table[3*n - Mod[ Mod[n, 2] + 1, n], {n, 1, 999}], 2] (* Mikk Heidemaa, Nov 02 2017 *)
Formula
O.g.f.: x*(2 + x + x^3 + 2x^4)/((1+x)*(1-x)^2). - R. J. Mathar, May 23 2008
a(n) = (1/9)*(4*n^3 + 3*n^2 + 1 - Kronecker(-3,n+1)). - Ralf Stephan, Jun 01 2014
From Mikk Heidemaa, Oct 28 2017: (Start)
a(n) = floor((41/21 - (3 mod n))^(-3*n+5)) + 3*n - 4 (n > 0).
a(n+1) = 3*n - ((n mod 2)+1) mod n (n > 0). (End)
a(n+2) = 2*floor((3*n+1)/2) + 1 for n>=1; see (17) in Diab link. - Michel Marcus, Dec 14 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = (7-sqrt(3)*Pi)/6. - Amiram Eldar, Sep 22 2022
Extensions
Name edited by Michel Marcus, Dec 14 2020
Comments