A008477 If n = Product (p_j^k_j) then a(n) = Product (k_j^p_j).

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

Offset

1,4

Comments

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

a(A005117(n)) = 1; a(A013929(n)) > 1; A010052(a(A122132(n))) = 1. - Reinhard Zumkeller, Feb 17 2012

From Bernard Schott, Mar 26 2021: (Start)

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)

Links

M. F. Hasler, Table of n, a(n) for n = 1..10000

Annales Concours Général, Sujet Concours Général 2012 (in French, problems).

Annales Concours Général, Corrigé Concours Général 2012 (in French, solutions).

M. Farrokhi, The Prime Exponentiation of an Integer: Problem 11315, Amer. Math. Monthly, 116 (2009), 470. - from N. J. A. Sloane, Apr 25 2009

Index to sequences related to Olympiads and other Mathematical Competitions.

Formula

Multiplicative with a(p^e) = e^p. - David W. Wilson, Aug 01 2001

Example

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.

Maple

A008477 := proc(n) local e, j; e := ifactors(n)[2]:
mul (e[j][2]^e[j][1], j=1..nops(e)) end:
seq (A008477(n), n=1..60);
# Peter Luschny, Jan 17 2010

Mathematica

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 *)

Prog

(Haskell)
a008477 n = product $ zipWith (^) (a124010_row n) (a027748_row n)
-- Reinhard Zumkeller, Feb 17 2012
(PARI) A008477(n)=factorback(factor(n)*[0, 1; 1, 0]) \\ M. F. Hasler, May 20 2012
(Python)
from sympy import factorint, prod
a = lambda n: prod([pk[1]**pk[0] for pk in factorint(n).items()])
print([a(n) for n in range(1, 61)]) # Darío Clavijo, Nov 06 2023
(APL, Dyalog dialect) A008477 ← {×/{⍺*⍨≢⍵}⌸factors(⍵)} ⍝ Needs also factors function from https://dfns.dyalog.com/c_factors.htm - Antti Karttunen, Feb 16 2024

Crossrefs

Cf. A001694, A027748, A124010.

Cf. A005117, A013929, A010052, A062307, A122132, A342551.

Cf. A008478 (fixed points).

Sequence in context: A084885 A360969 A112538 * A303278 A222639 A184729

Adjacent sequences: A008474 A008475 A008476 * A008478 A008479 A008480

Keyword

nonn,mult

Author

Olivier Gérard

Status

approved