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A008477
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If n = Product (p_j^k_j) then a(n) = Product (k_j^p_j).
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22
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1, 1, 1, 4, 1, 1, 1, 9, 8, 1, 1, 4, 1, 1, 1, 16, 1, 8, 1, 4, 1, 1, 1, 9, 32, 1, 27, 4, 1, 1, 1, 25, 1, 1, 1, 32, 1, 1, 1, 9, 1, 1, 1, 4, 8, 1, 1, 16, 128, 32, 1, 4, 1, 27, 1, 9, 1, 1, 1, 4, 1, 1, 8, 36, 1, 1, 1, 4, 1, 1, 1, 72, 1
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OFFSET
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1,4
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COMMENTS
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For any n, the sequence n, a(n), a(a(n)), a(a(a(n))), ... is eventually periodic with period <= 2 [Farrokhi]. - N. J. A. Sloane, Apr 25 2009
The study of some properties of this sequence was proposed in the 1st problem of Concours Général in 2012 in France (see links).
Terms are precisely the powerful numbers in A001694.
If m is a term, there is a term q such that a(q) = m.
a(a(n)) <= n (see examples). (End)
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LINKS
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FORMULA
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EXAMPLE
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For n = 24 = 2^3*3^1, a(24) = 3^2*1^3 = 9, so a(9) = 2^3 = 8 and a(a(24)) = 8 < 24.
For n = 243 = 3^5, a(243) = 5^3 = 125, so a(125) = 3^5 = 243 and a(a(243)) = 243.
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MAPLE
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A008477 := proc(n) local e, j; e := ifactors(n)[2]:
mul (e[j][2]^e[j][1], j=1..nops(e)) end:
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MATHEMATICA
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Prepend[ Array[ Times @@ Map[ Power @@ RotateLeft[ #1, 1 ]&, FactorInteger[ # ] ]&, 100, 2 ], 1 ]
Table[Times@@(First[#]^Last[#]&/@Transpose[Reverse[ Transpose[ FactorInteger[ n]]]]), {n, 80}] (* Harvey P. Dale, Jul 22 2014 *)
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PROG
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(Haskell)
a008477 n = product $ zipWith (^) (a124010_row n) (a027748_row n)
(Python)
from sympy import factorint, prod
a = lambda n: prod([pk[1]**pk[0] for pk in factorint(n).items()])
(APL, Dyalog dialect) A008477 ← {×/{⍺*⍨≢⍵}⌸factors(⍵)} ⍝ Needs also factors function from https://dfns.dyalog.com/c_factors.htm - Antti Karttunen, Feb 16 2024
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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STATUS
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approved
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