OFFSET
1,1
COMMENTS
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.
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
A. Schinzel, Problems and results on polynomials, Algorithms Seminar, INRIA, 1992-1993.
K. T. Vahlen, Über reductible Binome, Acta Mathematica 19:1 (December 1895), pp. 195-198.
FORMULA
Is a(n) ~ c * n^3? - David A. Corneth, Jan 12 2019
MATHEMATICA
nMax=500000; lst={}; k=1; While[4k^4<=nMax, AppendTo[lst, 4k^4]; k++ ]; n=2; While[p=Prime[n]; p^p<=nMax, k=1; While[(k*p)^p<=nMax, AppendTo[lst, (k*p)^p]; k++ ]; n++ ]; Union[lst]
PROG
(PARI) upto(n) = {my(res = List()); for(i = 1, sqrtnint(n \ 4, 4), listput(res, 4*i^4) ); forprime(p = 3, log(n), pp = p^p; for(k = 1, sqrtnint(n \ pp, p), listput(res, pp * k ^ p); ) ); listsort(res); res } \\ David A. Corneth, Jan 12 2019
(PARI) select( {is_A097792(n, p=0)= n%4==0 && ispower(n\4, 4) || ((2 < p = ispower(n, , &n)) && if(isprime(p), n%p==0, foreach(factor(p)[, 1], q, q%2 && n%q==0 && return(1))))}, [1..10^4]) \\ M. F. Hasler, Jul 07 2024
(Python)
from sympy import isprime, perfect_power, primefactors
def is_A097792(n):
return n%4==0 and (perfect_power(n//4, [4]) or n==4) or (
p := perfect_power(n)) and p[1] > 2 and (p[0]%p[1]==0 if isprime(p[1])
else any(p[0]%q==0 for q in primefactors(p[1]) if q > 2))
# M. F. Hasler, Jul 07 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Aug 24 2004
STATUS
approved