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
Examples
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.
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.
Crossrefs
Programs
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Haskell
a008477 n = product $ zipWith (^) (a124010_row n) (a027748_row n) -- Reinhard Zumkeller, Feb 17 2012
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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
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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 *) -
PARI
A008477(n)=factorback(factor(n)*[0,1;1,0]) \\ M. F. Hasler, May 20 2012
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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
Formula
Multiplicative with a(p^e) = e^p. - David W. Wilson, Aug 01 2001
Comments