%I
%S 1,1,3,1,5,3,7,1,9,5,11,3,13,7,15,1,17,9,19,5,21,11,23,3,19,13,27,7,
%T 29,15,31,1,57,17,49,9,37,19,33,5,41,21,43,11,45,23,47,3,35,19,51,13,
%U 53,27,65,7,105,29,59,15,61,31,63,1,59,57,67,17,117,49,71,9,73,37,105,19,109
%N TITO2(n): The operation A161594 in binary, digitreversals carried out in base 2.
%C The TITO function in binary: Represent n as a product of its prime factors in binary.
%C Revert the binary digits of each of these factors, then multiply them with the same multiplicities as in nso the base2 representation does not effect the exponents in the canonical prime factorization. Reverse the product in binary to get a(n).
%H Alois P. Heinz, <a href="/A161955/b161955.txt">Table of n, a(n) for n = 1..40000</a>
%H T. Khovanova, <a href="http://blog.tanyakhovanova.com/?p=144">Turning Numbers Inside Out</a> [From _Tanya Khovanova_, Jul 07 2009]
%F a(n) = A030101(A162742(n))  _R. J. Mathar_, Aug 03 2009
%e To calculate TITO2(n=99): 99 = 3^3*11. Prime factors 3 and 11 in binary are 11 and 1011 correspondingly. Reversing those numbers we get 11 and 1101. The product with multiplicities is the binary product of 11*11*1101 = 1110101. Reversing that we get 1010111, which corresponds to 87. Hence a(99) = 87.
%p r:= proc(n) local m, t; m, t:=n, 0; while m>0
%p do t:=2*t+irem(m, 2, 'm') od; t end:
%p a:= n> r(mul(r(i[1])^i[2], i=ifactors(n)[2])):
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Jun 29 2017
%t reverseBinPower[{n_, k_}] := FromDigits[Reverse[IntegerDigits[n, 2]], 2]^k fBin[n_] := FromDigits[ Reverse[IntegerDigits[ Times @@ Map[reverseBinPower, FactorInteger[n]], 2]], 2] Table[fBin[n], {n, 200}]
%Y Cf. A161594.
%K base,nonn
%O 1,3
%A _Tanya Khovanova_, Jun 22 2009
%E Edited by _R. J. Mathar_, Aug 03 2009
