A046022 Primes together with 1 and 4.

1, 2, 3, 4, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy

Comments

Also the numbers which are incrementally largest values of A002034. - validated by Franklin T. Adams-Watters, Jul 13 2012

Solutions to A000005(x) + A000010(x) - x - 1 = 0. - Labos Elemer, Aug 23 2001

Also numbers m such that m, phi(m) and tau(m) form an integer triangle, where phi=A000010 is the totient and tau=A000005 the number of divisors (see also A084820). - Reinhard Zumkeller, Jun 04 2003

Terms > 1 are n such that n does not divide (n-1)!. - Benoit Cloitre, Nov 12 2003

Terms > 1 are the sum of their prime factors; 4 (= 2+2) is the only such composite number. - Stuart Orford (sjorford(AT)yahoo.co.uk), Aug 04 2005

From Jonathan Vos Post, Aug 23 2010, Robert G. Wilson v, Aug 25 2010, proof by D. S. McNeil, Aug 29 2010: (Start)

Also the numbers n which divide A001414(n), or equivalently divide A075254(n). Proof:

Theorem: for a multiset of m >= 2 integers a_i, each a_i >= 2, Product_{i=1..m} a_i >= Sum_{i=1..m} a_i, with equality only at (a_1,a_2) = (2,2).

Lemma: For integers x,y >= 2, if x > 2 or y > 2, x*y > x + y. This follows from distributing (x-1)*(y-1) > 1.

[Proof of the theorem by induction on m:

first consider m=2. We have equality at (2,2) and for any product(a_i) > 4 there is some a_i > 2, so the lemma gives a_1*a_2 > a_1+a_2.

Then the induction m->m+1: Product_{i=1..m+1} a_i = a_(m+1)*Product_{i=1..m} a_i >= a_(m+1) * Sum_{i=1..m} a_i.

Since a_(m+1) >= 2 and the sum >= 4, the lemma applies, and we find a_(m+1) * Sum+{i=1..m} a_i > a_(m+1) + Sum_{i=1..m} a_i = Sum_{i=1..m+1} a_i and thus Product_{i=1..m+1} a_i > Sum_{i=1..m+1} a_i, QED.]

For composite n > 4, applying the theorem to the multiset of prime factors with multiplicity yields n > sopfr(n), so there are no composite numbers greater than 4 such that they divide sopfr(n).

(End)

Numbers k such that the k-th Fibonacci number is relatively prime to all smaller Fibonacci numbers. - Charles R Greathouse IV, Jul 13 2012

Numbers k such that (-1)^k*floor(d(k)*(-1)^k/2) = 1, where d(k) is the number of divisors of k. - Wesley Ivan Hurt, Oct 11 2013

Also, union of odd primes (A065091) and the divisors of 4. Also, union of A008578 and 4. - Omar E. Pol, Nov 04 2013

Numbers k such that sigma(k!) is divisible by sigma((k-1)!). - Altug Alkan, Jul 18 2016

Links

• J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function
• Eric Weisstein's World of Mathematics, Sum of Prime Factors

Crossrefs

Cf. A002034, A046021, A001751, A178156, A174460, A000040.

Programs

Haskell

a046022 n = a046022_list !! (n-1)
a046022_list = [1..4] ++ drop 2 a000040_list
-- Reinhard Zumkeller, Apr 06 2014

Maple

A046022:=n-> `if`((-1)^n*floor(numtheory[tau](n)*(-1)^n/2) = 1, n, NULL); seq(A046022(j), j=1..260); # Wesley Ivan Hurt, Oct 11 2013

Mathematica

max = 0; a = {}; Do[m = FactorInteger[n]; w = Sum[m[[k]][[1]]*m[[k]][[2]], {k, 1, Length[m]}]; If[w > max, AppendTo[a, w]; max = w], {n, 1, 1000}]; a (* Artur Jasinski, Apr 06 2008 *)

PARI

a(n)=if(n<6,n,prime(n-2)) \\ Charles R Greathouse IV, Apr 28 2015

Python

from sympy import prime
def A046022(n): return prime(n-2) if n>4 else n # Chai Wah Wu, Oct 17 2024

Formula

A141295(a(n)) = a(n). - Reinhard Zumkeller, Jun 23 2008

A018194(a(n)) = 1. - Reinhard Zumkeller, Mar 09 2012

A240471(a(n)) = 1. - Reinhard Zumkeller, Apr 06 2014

Extensions

Better description from Frank Ellermann, Jun 15 2001