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.
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
Offset: 1
Examples
a(40) = 8 since 40 = 2^3 * 5 so the powerful part is 2^3 = 8.
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
- Vaclav Kotesovec, Graph - the asymptotic ratio (100000000 terms).
- 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 Seq., Vol. 6 (2003), Article 03.3.4.
- Index entries for sequences related to powerful numbers.
Programs
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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
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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 *) -
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
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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
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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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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
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)
(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
Dirichlet g.f.: zeta(s-1) * Product_{p prime} (1 + 1/p^s - 1/ p^(s-1) + 1/p^(2*s-2) - 1/p^(2*s-1)). - Amiram Eldar, Sep 18 2023
From Vaclav Kotesovec, Apr 09 2025, simplified May 11 2025: (Start)
Dirichlet g.f.: zeta(2*s-2) * Product_{p prime} (1 - 1/p^(3*s-2) + 1/p^(3*s-3) + 1/p^s).
Sum_{k=1..n} a(k) ~ c * n^(3/2) / 3, where c = Product_{p prime} (1 + 2/p^(3/2) - 1/p^(5/2)) = 3.51955505841710664719752940369857817... = A328013. (End)
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