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A057521 Powerful (1) part of n: if n = Product_i (pi^ei) then a(n) = Product_{i : ei > 1} (pi^ei); if n=b*c^2*d^3 then a(n)=c^2*d^3 when b is minimized. 51
1, 1, 1, 4, 1, 1, 1, 8, 9, 1, 1, 4, 1, 1, 1, 16, 1, 9, 1, 4, 1, 1, 1, 8, 25, 1, 27, 4, 1, 1, 1, 32, 1, 1, 1, 36, 1, 1, 1, 8, 1, 1, 1, 4, 9, 1, 1, 16, 49, 25, 1, 4, 1, 27, 1, 8, 1, 1, 1, 4, 1, 1, 9, 64, 1, 1, 1, 4, 1, 1, 1, 72, 1, 1, 25, 4, 1, 1, 1, 16, 81, 1, 1, 4, 1, 1, 1, 8, 1, 9, 1, 4, 1, 1, 1, 32, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..16383 (first 1000 terms from T. D. Noe)

Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Christian Krause, LODA, an assembly language, a computational model and a tool for mining integer sequences

Victor Ufnarovski and Bo Åhlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003, #03.3.4.

Index entries for sequences related to powerful numbers

FORMULA

a(n) = n / A055231(n).

Multiplicative with a(p)=1 and a(p^e)=p^e for e>1. - Vladeta Jovovic, Nov 01 2001

From Antti Karttunen, Nov 22 2017: (Start)

a(n) = A064549(A003557(n)).

A003557(a(n)) = A003557(n).

(End)

a(n) = gcd(n, A003415(n)^k), for all k >= 2. [This formula was found in the form k=3 by Christian Krause's LODA miner. See Ufnarovski and Åhlander paper, Theorem 5 on p. 4 for why this holds] - Antti Karttunen, Mar 09 2021

EXAMPLE

a(40) = 8 since 40 = 2^3 * 5 so the powerful part is 2^3 = 8.

MAPLE

A057521 := proc(n)

local a, d, e, p;

a := 1;

for d in ifactors(n)[2] do

e := d[1] ;

p := d[2] ;

if e > 1 then

a := a*p^e ;

end if;

end do:

return a;

end proc: # R. J. Mathar, Jun 09 2016

MATHEMATICA

rad[n_] := Times @@ First /@ FactorInteger[n]; a[n_] := n/Denominator[n/rad[n]^2]; Table[a[n], {n, 1, 97}] (* Jean-François Alcover, Jun 20 2013 *)

f[p_, e_] := If[e > 1, p^e, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 21 2020 *)

PROG

(PARI) a(n)=my(f=factor(n)); prod(i=1, #f~, if(f[i, 2]>1, f[i, 1]^f[i, 2], 1)) \\ Charles R Greathouse IV, Aug 13 2013

(PARI) a(n) = my(f=factor(n)); for (i=1, #f~, if (f[i, 2]==1, f[i, 1]=1)); factorback(f); \\ Michel Marcus, Jan 29 2021

(Python)

from sympy import factorint, prod

def a(n): return 1 if n==1 else prod(1 if e==1 else p**e for p, e in factorint(n).items())

print([a(n) for n in range(1, 51)]) # Indranil Ghosh, Jul 19 2017

(Python)

from math import prod

from sympy import factorint

def A057521(n): return n//prod(p for p, e in factorint(n).items() if e == 1) # Chai Wah Wu, Nov 14 2022

CROSSREFS

Cf. A001694, A003415, A003557, A055231, A064549.

Sequence in context: A212173 A274006 A203025 * A084885 A112538 A008477

Adjacent sequences: A057518 A057519 A057520 * A057522 A057523 A057524

KEYWORD

nonn,mult,easy

AUTHOR

Henry Bottomley, Sep 01 2000

STATUS

approved

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Last modified December 5 17:04 EST 2022. Contains 358588 sequences. (Running on oeis4.)