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A008835 Largest 4th power dividing n.

%I A008835 #41 Sep 01 2024 09:36:22
%S A008835 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,16,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,16,1,
%T A008835 1,1,1,1,1,1,1,1,1,1,1,1,1,1,16,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,16,1,1,
%U A008835 1,1,1,1,1,1,1,1,1,1,1,1,1,16,81
%N A008835 Largest 4th power dividing n.
%H A008835 Michael De Vlieger, <a href="/A008835/b008835.txt">Table of n, a(n) for n = 1..10000</a>
%H A008835 Henry Bottomley, <a href="http://fs.gallup.unm.edu/Bottomley-Sm-Mult-Functions.htm">Some Smarandache-type multiplicative sequences</a>.
%F A008835 a(n) = A000188(A000188(n))^4.
%F A008835 Multiplicative with a(p^e) = p^(4[e/4]). - _Mitch Harris_, Apr 19 2005
%F A008835 Dirichlet g.f.: zeta(s) * zeta(4s-4) / zeta(4s). - _Álvar Ibeas_, Feb 12 2015
%F A008835 Sum_{k=1..n} a(k) ~ zeta(5/4) * n^(5/4) / (5*zeta(5)) - 45*n/Pi^4. - _Vaclav Kotesovec_, Feb 03 2019
%F A008835 a(n) = n/A053165(n). - _Amiram Eldar_, Aug 15 2023
%F A008835 a(n) = A053164(n)^4. - _Amiram Eldar_, Sep 01 2024
%p A008835 with(numtheory): [ seq( expand(nthpow(i,4)),i=1..200) ];
%t A008835 Max@ Select[Divisors@ #, IntegerQ@ Power[#, 1/4] &] & /@ Range@ 81 (* _Michael De Vlieger_, Mar 18 2015 *)
%t A008835 f[p_, e_] := p^(e - Mod[e, 4]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Aug 15 2023 *)
%o A008835 (PARI) a(n) = {f = factor(n); for (i=1, #f~, f[i,2] = 4*(f[i,2]\4);); factorback(f);} \\ _Michel Marcus_, Mar 16 2015
%o A008835 (Python)
%o A008835 from math import prod
%o A008835 from sympy import factorint
%o A008835 def A008835(n): return prod(p**(e&-4) for p, e in factorint(n).items()) # _Chai Wah Wu_, Aug 08 2024
%Y A008835 Cf. A000188, A000190, A008833, A008834, A053164, A053165.
%K A008835 nonn,easy,mult
%O A008835 1,16
%A A008835 _N. J. A. Sloane_
%E A008835 Entry improved by comments from _Henry Bottomley_, Feb 29 2000