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If n = Product_{i>0} prime(i)^e(i), then a(n) = Product_{i>0} prime(i)^(XOR_{j>=i} e(j)), where XOR is bitwise XOR.
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%I #20 Jan 03 2021 12:29:40

%S 1,2,6,4,30,3,210,8,36,15,2310,24,30030,105,5,16,510510,72,9699690,

%T 120,35,1155,223092870,12,900,15015,216,840,6469693230,10,

%U 200560490130,32,385,255255,7,9,7420738134810,4849845,5005,60,304250263527210,70,13082761331670030,9240,1080,111546435,614889782588491410

%N If n = Product_{i>0} prime(i)^e(i), then a(n) = Product_{i>0} prime(i)^(XOR_{j>=i} e(j)), where XOR is bitwise XOR.

%C This sequence is a permutation of the natural numbers.

%C The powers of 2 are the fixed points of this sequence.

%C a(prime(i)) = A002110(i) for any i > 0.

%C a(i^2) = a(i)^2 for any i > 0.

%C a(A019565(n)) = A019565(A006068(n)) for any n >= 0.

%H Paul Tek, <a href="/A226532/b226532.txt">Table of n, a(n) for n = 1..2370</a>

%H Paul Tek, <a href="/A226532/a226532.txt">PERL program for this sequence</a>

%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>

%e a(50) = a(2^1 * 3^0 * 5^2)

%e = 2^xor(1,0,2) * 3^xor(0,2) * 5^xor(2)

%e = 2^3 * 3^2 * 5^2

%e = 1800.

%o (Perl) See link.

%o (Haskell)

%o import Data.Bits (xor)

%o a226532 n = product $ zipWith (^)

%o a000040_list (scanr1 xor $ a067255_row n :: [Integer])

%o -- _Reinhard Zumkeller_, Jun 11 2013

%Y Cf. A006068.

%Y Cf. A226569 (inverse), A067255, A000040.

%K nonn

%O 1,2

%A _Paul Tek_, Jun 10 2013