%I
%S 1068,2057,4193,5182,7318,8307,10443,11432,13568,14557,16693,17682,
%T 19818,20807,22943,23932,26068,27057,29193,30182,32318,33307,35443,
%U 36432,38568,39557,41693,42682,44818,45807,47943,48932,51068,52057,54193,55182,57318,58307
%N Numbers n such that n^2 + 1 is divisible by a 5th power.
%C For each prime p == 1 (mod 4), there are two values of x (mod p^5) that solve x^2 + 1 == 0 (mod p^5), and then x + k*p^5 is in the sequence for every k. Thus the asymptotic density of this sequence should be 1  Product_p (1  2/p^5), where the product is over all primes p == 1 (mod 4).  _Robert Israel_, Sep 04 2018
%H Robert Israel, <a href="/A218564/b218564.txt">Table of n, a(n) for n = 1..10000</a>
%e 1068 is in the sequence because 1068^2+1 = 1140625 = 5^6*73;
%e 143044 is in the sequence because 143044^2+1 = 20461585937 = 13^5*55109;
%e 390112 is in the sequence because 390112^2+1 = 152187372545 = 5*13*17^6*97.
%p N:= 10^5: # to get all terms <= N
%p P:= select(isprime,[seq(i,i=5..floor((N^2+1)^(1/5)),4)]):
%p g:= proc(x,r,N) local t; t:= rhs(op(x)); seq(t+r*k,k=0..(Nt)/r) end proc:
%p R:= `union`(seq(map(g, {msolve(n^2+1,p^5)},p^5,N),p=P)):
%p sort(convert(R,list)); # _Robert Israel_, Sep 04 2018
%t Select[Range[2,20000],Max[Transpose[FactorInteger[#^2+1]][[2]]]>4&]
%o (PARI) isok(n) = vecmax(factor(n^2+1)[,2]) >= 5; \\ _Michel Marcus_, Sep 04 2018
%Y Cf. A002522, A049532, A034939, A218562, A218563.
%K nonn
%O 1,1
%A _Michel Lagneau_, Nov 02 2012
