%I
%S 1,2,7,13,14,21,25,26,31,37,41,42,49,50,55,61,62,69,73,74,81,82,87,93,
%T 97,98,103,109,110,117,121,122,127,133,137,138,145,146,151,157,161,
%U 162,167,173,174,181,185,186,193,194,199,205,206,213,217,218,223,229,233,234,241,242,247,253
%N Odious numbers of order 2: a(n) = A000069(A000069(n)).
%C Odious numbers with odious subscripts.
%C From _Antti Karttunen_, Nov 29 2013: (Start)
%C Starting from 4 and iterating A000069(4), A000069(A000069(4)), A000069(A000069(A000069(4))), etc. gives A004119 from its second term onward: 4, 7, 13, 25, 49, 97, 193, ..., which is thus a subsequence of this sequence from the term 7 onward.
%C Proof: All of the terms A004119(n) are odious although A004119(n)1 is evil, and the formula for A000069(n) reduces to a(n) = 2n  1 when n1 is evil, and iterating that formula starting from 4 gives A004119 from 7 onward (cf. _Philippe DelĂ©ham_'s formula there dated Feb 20 2004).
%C (End)
%C These numbers are never multiples of 4. Probably there are infinitely many multiples of m in this sequence for any m not divisible by 4. Equivalently, A233419(n) > 0 for all n.  _Charles R Greathouse IV_, Dec 05 2013
%H Antti Karttunen, <a href="/A232637/b232637.txt">Table of n, a(n) for n = 1..8193</a>
%F a(n) = A000069(A000069(n)).
%F 4n6 <= a(n) <= 4n3, see PARI script.  _Charles R Greathouse IV_, Dec 05 2013
%e The first odious number, A000069(1) = 1, and A000069(1) = 1, so a(1) = 1.
%e The second odious number, A000069(2) = 2, and A000069(2) = 2, so a(2) = 2.
%e Those were the only fixed points of A000069, and after that, we have:
%e The third odious number, A000069(3) = 4, and A000069(4) = 7, thus a(3) = 7.
%e The fourth odious number, A000069(4) = 7, and A000069(7) = 13, thus a(4) = 13.
%o (Scheme)
%o (define (A232637 n) (A000069 (A000069 n)))
%o (PARI) a(n)=4*nif(hammingweight(n1)%2,if(hammingweight(n2)%2,5,6),3) \\ _Charles R Greathouse IV_, Dec 05 2013
%Y Cf. A000069, A102615, A092246, A232667, A233419.
%Y A004119 from term 7 onward is a subsequence.
%Y Subsequence of A042968.
%K nonn,base,easy
%O 1,2
%A _Gerasimov Sergey_, Nov 27 2013
