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A057521
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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.
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49
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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
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OFFSET
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1,4
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LINKS
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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
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FORMULA
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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
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EXAMPLE
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a(40) = 8 since 40 = 2^3 * 5 so the powerful part is 2^3 = 8.
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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Cf. A001694, A003415, A003557, A055231, A064549.
Sequence in context: A212173 A274006 A203025 * A084885 A112538 A008477
Adjacent sequences: A057518 A057519 A057520 * A057522 A057523 A057524
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KEYWORD
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nonn,mult,easy
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AUTHOR
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Henry Bottomley, Sep 01 2000
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STATUS
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approved
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