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A161955 TITO2(n): The operation A161594 in binary, digit-reversals carried out in base 2.

%I A161955 #15 Sep 23 2021 02:24:38
%S A161955 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 A161955 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 A161955 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 A161955 TITO2(n): The operation A161594 in binary, digit-reversals carried out in base 2.
%C A161955 The TITO function in binary: Represent n as a product of its prime factors in binary.
%C A161955 Revert the binary digits of each of these factors, then multiply them with the same multiplicities as in n--so the base-2 representation does not affect the exponents in the canonical prime factorization. Reverse the product in binary to get a(n).
%H A161955 Alois P. Heinz, <a href="/A161955/b161955.txt">Table of n, a(n) for n = 1..40000</a>
%H A161955 Tanya Khovanova, <a href="http://blog.tanyakhovanova.com/?p=144">Turning Numbers Inside Out</a>
%F A161955 a(n) = A030101(A162742(n)) - _R. J. Mathar_, Aug 03 2009
%e A161955 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 A161955 r:= proc(n) local m, t; m, t:=n, 0; while m>0
%p A161955       do t:=2*t+irem(m, 2, 'm') od; t end:
%p A161955 a:= n-> r(mul(r(i[1])^i[2], i=ifactors(n)[2])):
%p A161955 seq(a(n), n=1..100);  # _Alois P. Heinz_, Jun 29 2017
%t A161955 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 A161955 Cf. A161594.
%K A161955 base,nonn
%O A161955 1,3
%A A161955 _Tanya Khovanova_, Jun 22 2009
%E A161955 Edited by _R. J. Mathar_, Aug 03 2009