A046101 - OEIS

%I #51 Aug 05 2024 16:03:42

%S 16,32,48,64,80,81,96,112,128,144,160,162,176,192,208,224,240,243,256,

%T 272,288,304,320,324,336,352,368,384,400,405,416,432,448,464,480,486,

%U 496,512,528,544,560,567,576,592,608,624,625,640,648,656,672,688,704

%N Biquadrateful numbers.

%C The convention in the OEIS is that squareful, cubeful (A046099), biquadrateful, ... mean the same as "not squarefree" etc., while 2- or square-full, 3- or cube-full (A036966), 4-full (A036967) are used for Golomb's notion of powerful numbers (A001694, see references there), when each prime factor occurs to a power > 1. - _M. F. Hasler_, Feb 12 2008

%C Also solutions to equation tau_{-3}(n)=0, where tau_{-3} is A007428. - _Enrique Pérez Herrero_, Jan 19 2013

%C Sum_{n>0} 1/a(n)^s = Zeta(s) - Zeta(s)/Zeta(4s). - _Enrique Pérez Herrero_, Jan 21 2013

%C A051903(a(n)) > 3. - _Reinhard Zumkeller_, Sep 03 2015

%C The asymptotic density of this sequence is 1 - 1/zeta(4) = 1 - 90/Pi^4 = 0.076061... - _Amiram Eldar_, Jul 09 2020

%H T. D. Noe, <a href="/A046101/b046101.txt">Table of n, a(n) for n = 1..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Biquadratefree.html">Biquadratefree.</a>

%p with(NumberTheory):

%p isBiquadrateful := n -> is(denom(Radical(n) / LargestNthPower(n, 2)) <> 1):

%p select(isBiquadrateful, [`$`(1..704)]);  # _Peter Luschny_, Jul 12 2022

%t lst={};Do[a=0;Do[If[FactorInteger[m][[n, 2]]>3, a=1], {n, Length[FactorInteger[m]]}];If[a==1, AppendTo[lst, m]], {m, 10^3}];lst (* _Vladimir Joseph Stephan Orlovsky_, Aug 15 2008 *)

%t Select[Range[1000],Max[Transpose[FactorInteger[#]][[2]]]>3&] (* _Harvey P. Dale_, May 25 2014 *)

%o (Haskell)

%o a046101 n = a046101_list !! (n-1)

%o a046101_list = filter ((> 3) . a051903) [1..]

%o -- _Reinhard Zumkeller_, Sep 03 2015

%o (PARI) is(n)=n>9 && vecmax(factor(n)[,2])>3 \\ _Charles R Greathouse IV_, Sep 03 2015

%o (Python)

%o from sympy import mobius, integer_nthroot

%o def A046101(n):

%o     def f(x): return n+sum(mobius(k)*(x//k**4) for k in range(1, integer_nthroot(x,4)[0]+1))

%o     m, k = n, f(n)

%o     while m != k:

%o         m, k = k, f(k)

%o     return m # _Chai Wah Wu_, Aug 05 2024

%Y Cf. A046100, A046099, A036967, A001694, A051903, A215267.

%K nonn

%O 1,1

%A _Eric W. Weisstein_