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%I #87 Feb 16 2024 18:36:59
%S 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,
%T 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,
%U 36,1,1,1,4,1,1,1,72,1
%N If n = Product (p_j^k_j) then a(n) = Product (k_j^p_j).
%C 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
%C a(A005117(n)) = 1; a(A013929(n)) > 1; A010052(a(A122132(n))) = 1. - _Reinhard Zumkeller_, Feb 17 2012
%C From _Bernard Schott_, Mar 26 2021: (Start)
%C The study of some properties of this sequence was proposed in the 1st problem of Concours Général in 2012 in France (see links).
%C Terms are precisely the powerful numbers in A001694.
%C If m is a term, there is a term q such that a(q) = m.
%C a(a(n)) <= n (see examples). (End)
%H M. F. Hasler, <a href="/A008477/b008477.txt">Table of n, a(n) for n = 1..10000</a>
%H Annales Concours Général, <a href="https://www.freemaths.fr/annales-composition-mathematiques-concours-general/concours-general-mathematiques-sujet-serie-s-2012.pdf">Sujet Concours Général 2012</a> (in French, problems).
%H Annales Concours Général, <a href="https://www.freemaths.fr/annales-composition-mathematiques-concours-general/concours-general-mathematiques-corrige-serie-s-2012.pdf">Corrigé Concours Général 2012</a> (in French, solutions).
%H M. Farrokhi, <a href="https://www.jstor.org/stable/40391138">The Prime Exponentiation of an Integer: Problem 11315</a>, Amer. Math. Monthly, 116 (2009), 470. - from _N. J. A. Sloane_, Apr 25 2009
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical Competitions</a>.
%F Multiplicative with a(p^e) = e^p. - _David W. Wilson_, Aug 01 2001
%e 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.
%e For n = 243 = 3^5, a(243) = 5^3 = 125, so a(125) = 3^5 = 243 and a(a(243)) = 243.
%p A008477 := proc(n) local e,j; e := ifactors(n)[2]:
%p mul (e[j][2]^e[j][1], j=1..nops(e)) end:
%p seq (A008477(n), n=1..60);
%p # _Peter Luschny_, Jan 17 2010
%t Prepend[ Array[ Times @@ Map[ Power @@ RotateLeft[ #1, 1 ]&, FactorInteger[ # ] ]&, 100, 2 ], 1 ]
%t Table[Times@@(First[#]^Last[#]&/@Transpose[Reverse[ Transpose[ FactorInteger[ n]]]]),{n,80}] (* _Harvey P. Dale_, Jul 22 2014 *)
%o (Haskell)
%o a008477 n = product $ zipWith (^) (a124010_row n) (a027748_row n)
%o -- _Reinhard Zumkeller_, Feb 17 2012
%o (PARI) A008477(n)=factorback(factor(n)*[0,1;1,0]) \\ _M. F. Hasler_, May 20 2012
%o (Python)
%o from sympy import factorint, prod
%o a = lambda n: prod([pk[1]**pk[0] for pk in factorint(n).items()])
%o print([a(n) for n in range(1,61)]) # _Darío Clavijo_, Nov 06 2023
%o (APL, Dyalog dialect) A008477 ← {×/{⍺*⍨≢⍵}⌸factors(⍵)} ⍝ Needs also factors function from https://dfns.dyalog.com/c_factors.htm - _Antti Karttunen_, Feb 16 2024
%Y Cf. A001694, A027748, A124010.
%Y Cf. A005117, A013929, A010052, A062307, A122132, A342551.
%Y Cf. A008478 (fixed points).
%K nonn,mult
%O 1,4
%A _Olivier Gérard_
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