OFFSET
1,1
COMMENTS
The polynomial x^n - n is reducible over the integers for n in this sequence.
A result of Vahlen shows that the polynomial x^n - n is reducible over the integers for n in this sequence and no other n.
The representation (k*p)^p is generally not unique, e.g. a(120) = 46656 = (108*2)^2 = (12*3)^3. - Reinhard Zumkeller, Feb 14 2015
This is also numbers of the form (km)^m for any m > 1, not just primes. Let m be > 1; then m has a prime factor, so let m=pj, p a prime and j an integer > 0. Then (km)^m = (kpj)^pj = (k^j p^j j^j)^p = ((k^j p^(j-1) j^j) p) ^ p. - Franklin T. Adams-Watters, Sep 13 2015
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
A. Schinzel, Problems and results on polynomials.
MATHEMATICA
nMax=10000; lst={}; n=1; While[p=Prime[n]; p^p<=nMax, k=1; While[(k*p)^p<=nMax, AppendTo[lst, (k*p)^p]; k++ ]; n++ ]; Union[lst]
PROG
(Haskell)
import Data.Set (singleton, deleteFindMin, insert)
a097764 n = a097764_list !! (n-1)
a097764_list = f 0 (singleton (4, 2, 2)) $
tail $ zip a051674_list a000040_list where
f m s ppps'@((pp, p) : ppps)
| pp < qq = f m (insert (pp, p, 2) s) ppps
| qq == m = f m (insert ((k * q) ^ q, q, k + 1) s') ppps'
| otherwise = qq : f qq (insert ((k * q) ^ q, q, k + 1) s') ppps'
where ((qq, q, k), s') = deleteFindMin s
-- Reinhard Zumkeller, Feb 14 2015
(PARI) is(n)=my(b, e=ispower(n, , &b), f); if(e==0, return(0)); f=factor(e)[, 1]; for(i=1, #f, if(b%f[i]==0, return(1))); 0 \\ Charles R Greathouse IV, Aug 29 2016
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
T. D. Noe, Aug 24 2004
STATUS
approved