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

Offset

1,2

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

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

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)

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

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

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

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

Links

Table of n, a(n) for n=1..57.

J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function

Eric Weisstein's World of Mathematics, Sum of Prime Factors

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 *)

Prog

(Haskell)
a046022 n = a046022_list !! (n-1)
a046022_list = [1..4] ++ drop 2 a000040_list
-- Reinhard Zumkeller, Apr 06 2014
(PARI) a(n)=if(n<6, n, prime(n-2)) \\ Charles R Greathouse IV, Apr 28 2015

Crossrefs

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

Sequence in context: A033070 A211781 A348283 * A175787 A345899 A073019

Adjacent sequences: A046019 A046020 A046021 * A046023 A046024 A046025

Keyword

nonn,easy

Author

Eric W. Weisstein

Extensions

Better description from Frank Ellermann, Jun 15 2001

Status

approved