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%I #3 Mar 30 2012 18:49:10
%S 1,3,5,7,9,11,13,15,17,19,21,23,27,29,31,37,41,43,45,47,51,53,59,61,
%T 63,67,71,73,79,83,85,89,93,95,97,101,103,107,109,111,113,119,123,127,
%U 131,137,139,149,151,153,157,163,167,173,179,181,187,189,191,193,197,199
%N Fixed points of A161955.
%C Fixed points of the TITO2 operation (the TITO operation in binary): numbers a(n) such that A161955(a(n)) = a(n).
%C All numbers in the sequence are odd. All odd primes A065091 belong to the sequence.
%H T. Khovanova, <a href="http://blog.tanyakhovanova.com/?p=144">Turning Numbers Inside Out</a> [From _Tanya Khovanova_, Jul 07 2009]
%e 95 is in this sequence because 95 = 5*19. Prime factors in binary are: 101 and 10011.
%e Reversing them we get 101 and 11001. The product of the last two numbers is 1111101, which is
%e the reverse of the binary representation of 95 (1011111).
%t reverseBinPower[{n_, k_}] := FromDigits[Reverse[IntegerDigits[n, 2]], 2]^k fBin[n_] := FromDigits[ Reverse[IntegerDigits[ Times @@ Map[reverseBinPower, FactorInteger[n]], 2]], 2] Select[Range[300], fBin[ # ] == # &]
%Y Cf. A161955, A161594, A161597.
%K base,nonn
%O 1,2
%A _Tanya Khovanova_, Jun 22 2009
%E Comments condensed by _R. J. Mathar_, Aug 14 2009
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