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A046022
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Primes together with 1 and 4.
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15
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| The values of n which are incrementally largest values of the Smarandache function S(n) seem to produce the same sequence.
Solutions to A000005[x]+A000010[x]-x-1=0. - Labos E. (labos(AT)ana.sote.hu), 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 (reinhard.zumkeller(AT)gmail.com), Jun 04 2003
Terms > 1 are n such that n does not divide (n-1)! - Benoit Cloitre (benoit7848c(AT)orange.fr), 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 (reinhard.zumkeller(AT)gmail.com), Jun 23 2008
Contribution from Jonathan Vos Post (jvospost3(AT)gmail.com), Aug 23 2010, Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 25 2010, proof by D. S. McNeil 29 Aug 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 (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: 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)
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LINKS
| Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
Eric Weisstein's World of Mathematics, Sum of Prime Factors
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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 (grafix(AT)csl.pl), Apr 06 2008
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CROSSREFS
| Cf. A002034, A046021.
Cf. A001751, A178156, A174460.
Sequence in context: A062972 A036844 A033070 * A175787 A073019 A174291
Adjacent sequences: A046019 A046020 A046021 * A046023 A046024 A046025
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KEYWORD
| nonn,easy
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AUTHOR
| Eric Weisstein (eric(AT)weisstein.com)
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EXTENSIONS
| Better description from Frank.Ellermann(AT)t-online.de, Jun 15 2001
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