cp's OEIS Frontend

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

This sequence is closed under multiplication by odd values, and in particular with multiplication by itself. - Charles R Greathouse IV, Feb 19 2013

For all n, n > tau(n) and n > phi(n) and if n is prime then n-tau(n) = n-2 and phi(n) = n-1. So n = 5 gives the triangle {3, 4, 5} which is a primitive Pythagorean triangle and this is the only one. Other Pythagorean triangles are {30, 16, 34} and {756, 192, 780}, the remainder are only Heronian.

It is not known if this sequence is infinite. Prime numbers in the sequence are 5, 53 and 140453 and generate triangles {3, 4, 5}, {51, 52, 53} and {140451, 140452, 140453}.

If n = 2p where p is prime then n-tau(n) = n-4 and phi(n) = n/2-1. So n = 34 gives the triangle {16, 30, 34}. Similar numbers in this sequence are a(8), a(9) and a(11). See A272365 for generating Heronian triangles with sides n, n-4, n/2-1.

a(12) > 2*10^12. - Giovanni Resta, Apr 14 2016

Next prime value of a(n) after 140453 is > 2*10^5719. See A003500 for generating Heronian triangles with consecutive sides. - Frank M Jackson, Apr 19 2016

A003500(n)+1 is a member of this sequence iff it is prime. Also A272365(n) is a member of this sequence iff A272365(n)/2 is prime. - Frank M Jackson, Apr 29 2016

Some values of m have multiple solutions.

For example, for m = 49, 25^25 == 26^26 (mod 49) and 37^37 == 38^38 (mod 49).

All terms are odd.

First differs from A334420 at a(70) which is 167 for this sequence and 165 for A334420.

First differs from A056911 at a(21) which is 49 for this sequence and 51 for A056911.