A161594 a(n) = R(f(n)), where R = A004086 = reverse (decimal) digits, f = A071786 = reverse digits of prime factors.
1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 21, 13, 41, 51, 61, 17, 81, 19, 2, 12, 22, 23, 42, 52, 26, 72, 82, 29, 3, 31, 23, 33, 241, 53, 63, 37, 281, 39, 4, 41, 24, 43, 44, 54, 46, 47, 84, 94, 5, 312, 421, 53, 45, 55, 65, 372, 481, 59, 6, 61, 62, 36, 46, 551, 66, 67, 482, 69, 7, 71, 27
Offset: 1
Examples
a(34) = 241, because 34 = 2*17, f(34) = 2*71 = 142, and reversing gives 241.
Links
- M. F. Hasler, Table of n, a(n) for n=1..5000. [From _M. F. Hasler_, Jun 24 2009]
- T. Khovanova, Turning Numbers Inside Out [From _Tanya Khovanova_, Jul 07 2009]
Programs
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Haskell
a161594 = a004086 . a071786 -- Reinhard Zumkeller, Oct 14 2011
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Maple
read("transforms") ; A071786 := proc(n) local ifs,a,d ; ifs := ifactors(n)[2] ; a := 1 ; for d in ifs do a := a*digrev(op(1,d))^op(2,d) ; od: a ; end: A161594 := proc(n) digrev(A071786(n)) ; end: seq(A161594(n),n=1..80) ; # R. J. Mathar, Jun 16 2009 # second Maple program: r:= n-> (s-> parse(cat(seq(s[-i], i=1..length(s)))))(""||n): a:= n-> r(mul(r(i[1])^i[2], i=ifactors(n)[2])): seq(a(n), n=1..100); # Alois P. Heinz, Jun 19 2017
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Mathematica
reversepower[{n_, k_}] := FromDigits[Reverse[IntegerDigits[n]]]^k f[n_] := FromDigits[ Reverse[IntegerDigits[Times @@ Map[reversepower, FactorInteger[n]]]]] Table[f[n], {n, 100}] Table[IntegerReverse[Times@@Flatten[Table[IntegerReverse[#[[1]]],#[[2]]]& /@FactorInteger[n]]],{n,100}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 21 2016 *)
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PARI
R=A004086; A161594(n)={n=factor(n);n[,1]=apply(R,n[,1]);R(factorback(n))} \\ M. F. Hasler, Jun 24 2009. Removed code for R here, see A004086 for most recent & efficient version. - M. F. Hasler, May 11 2015
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Python
from math import prod from sympy import factorint def f(n): return prod(int(str(p)[::-1])**e for p, e in factorint(n).items()) def R(n): return int(str(n)[::-1]) def a(n): return 1 if n == 1 else R(f(n)) print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Mar 28 2022
Formula
a(p) = p, for prime p.
From M. F. Hasler, Jun 25 2009: (Start)
a( p*10^k ) = p for any prime p.
Proof: if gcd( p, 2*5) = 1, then a( p * 10^k ) = R( R(p) * R(2)^k * R(5)^k ) = R( R(p) * 10^k ) = R(R(p)) = p;
if gcd(p, 2*5) = 2, then p=2 and a( p * 10^k ) = R( R(2)^(k+1) * R(5)^k ) = R( 2 * 10^k ) = 2 = p and mutatis mutandis for gcd(p, 2*5) = 5. (End)
Extensions
Simpler definition from R. J. Mathar, Jun 16 2009
Edited by N. J. A. Sloane, Jun 23 2009
Comments