%I
%S 0,1,2,3,4,5,6,7,8,9,11,10,13,12,14,15,16,17,18,19,22,23,20,21,26,27,
%T 24,25,28,29,30,31,32,33,35,34,36,37,39,38,44,45,47,46,40,41,43,42,53,
%U 52,54,55,49,48,50,51,57,56,58,59,61,60,62,63,64,65,66,67
%N Binary interpretation of the left diagonal of the EQtriangle with first row generated from the binary expansion of n, with most significant bit given by first row.
%C For any nonnegative number n, the EQtriangle for n is built by taking as first row the binary expansion of n (without leading zeros), having each entry in the subsequent rows be the EQ of the two values above it (a "1" indicates that these two values are equal, a "0" indicates that these values are different).
%C This sequence is a selfinverse permutation of the nonnegative numbers.
%H Rémy Sigrist, <a href="/A335133/b335133.txt">Table of n, a(n) for n = 0..8192</a> (n = 0..2^13)
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F a(floor(n/2)) = floor(a(n)/2).
%F abs(a(2*n+1)  a(2*n)) = 1.
%F a(2^k) = 2^k for any k >= 0.
%F a(2^k+1) = 2^k+1 for any k >= 0.
%F a(2^k1) = 2^k1 for any k >= 0.
%F Apparently, a(n) + A334727(n) = A055010(A070939(n)) for any n > 0.
%e For n = 42:
%e  the binary representation of 42 is "101010",
%e  the corresponding EQtriangle is:
%e 1 0 1 0 1 0
%e 0 0 0 0 0
%e 1 1 1 1
%e 1 1 1
%e 1 1
%e 1
%e  the bits on the left diagonal are: 1, 0, 1, 1, 1, 1,
%e  so a(42) = 2^5 + 2^3 + 2^2 + 2^1 + 2^0 = 47.
%o (PARI) a(n) = {
%o my (b=binary(n), v=0);
%o forstep (x=#b1, 0, 1,
%o if (b[1], v+=2^x);
%o b=vector(#b1, k, b[k]==b[k+1])
%o );
%o return (v)
%o }
%Y Cf. A055010, A070939, A279645, A334727 (XOR variant).
%K nonn,look,base
%O 0,3
%A _Rémy Sigrist_, May 24 2020
