A053164 4th root of largest 4th power dividing n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2

Offset

1,16

Comments

Multiplicative with a(p^e) = p^[e/4]. - Mitch Harris, Apr 19 2005

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

Henry Bottomley, Some Smarandache-type multiplicative sequences.

Formula

a(n) = A000188(A000188(n)) = A008835(n)^(1/4).

Multiplicative with a(p^e) = p^[e/4].

Dirichlet g.f.: zeta(4s-1)*zeta(s)/zeta(4s). - R. J. Mathar, Apr 09 2011

Sum_{k=1..n} a(k) ~ 90*zeta(3)*n/Pi^4 + 3*zeta(1/2)*sqrt(n)/Pi^2. - Vaclav Kotesovec, Dec 01 2020

a(n) = Sum_{d^4|n} phi(d). - Ridouane Oudra, Dec 31 2020

G.f.: Sum_{k>=1} phi(k) * x^(k^4) / (1 - x^(k^4)). - Ilya Gutkovskiy, Aug 20 2021

Example

a(32) = 2 since 2 = 16^(1/4) and 16 is the largest 4th power dividing 32.

Maple

A053164 := proc(n) local a, f, e, p ; for f in ifactors(n)[2] do e:= op(2, f) ; p := op(1, f) ; a := a*p^floor(e/4) ; end do ; a ; end proc: # R. J. Mathar, Jan 11 2012

Mathematica

f[list_] := list[[1]]^Quotient[list[[2]], 4]; Table[Apply[Times, Map[f, FactorInteger[n]]], {n, 1, 81}] (* Geoffrey Critzer, Jan 21 2015 *)

Prog

(Scheme, with memoization macro definec)
(definec (A053164 n) (if (= 1 n) n (* (expt (A020639 n) (A002265 (A067029 n))) (A053164 (A028234 n)))))
(define (A002265 n) (floor->exact (/ n 4))) ;; For MIT/GNU Scheme
;; Antti Karttunen, Sep 13 2017

Crossrefs

Cf. A000188, A000190, A008835, A053150.

Sequence in context: A203640 A043289 A063775 * A365333 A295658 A307427

Adjacent sequences: A053161 A053162 A053163 * A053165 A053166 A053167

Keyword

nonn,mult

Author

Henry Bottomley, Feb 29 2000

Extensions

More terms from Antti Karttunen, Sep 13 2017

Status

approved