

A046022


Primes together with 1 and 4.


29



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
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OFFSET

1,2


COMMENTS

Also the values of n which are incrementally largest values of A002034.  validated by Franklin T. AdamsWatters, Jul 13 2012
Solutions to A000005[x]+A000010[x]x1=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 (n1)!.  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(a_i,i=1..m) >= sum(a_i, i=1..m) 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 (x1)*(y1)>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: prod(a_i,i=1..m+1) = a_(m+1)* prod(a_i,i=1..m) >= a_(m+1) * sum(a_i,i=1..m).
Since a_(m+1) >= 2 and the sum >= 4, the lemma applies, and we find a_(m+1) * sum(a_i,i=1..m) > a_(m+1) + sum(a_i,i=1..m) = sum(a_i,i=1..m+1) and thus prod(a_i,i=1..m+1) > sum(a_i,i=1..m+1) 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 n such that the nth Fibonacci number is relatively prime to all smaller Fibonacci numbers.  Charles R Greathouse IV, Jul 13 2012
Numbers n such that (1)^n*floor(d(n)*(1)^n/2) = 1, where d(n) is the number of divisors of n.  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 n such that sigma(n!) is divisible by sigma((n1)!).  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 !! (n1)
a046022_list = [1..4] ++ drop 2 a000040_list
 Reinhard Zumkeller, Apr 06 2014
(PARI) a(n)=if(n<6, n, prime(n2)) \\ Charles R Greathouse IV, Apr 28 2015


CROSSREFS

Cf. A002034, A046021, A001751, A178156, A174460, A000040.
Sequence in context: A284696 A033070 A211781 * A175787 A073019 A174291
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



