%I
%S 0,1,0,1,0,1,2,1,2,1,2,3,2,3,4,5,6,5,4,5,6,5,6,5,4,3,4,3,4,5,4,5,6,7,
%T 6,5,6,7,6,7,8,7,6,7,8,9,10,11,12,11,12,13,12,11,10,9,10,9,10,11,10,
%U 11,12,13,12,11,12,13,12,13,12,13,14,13,12,11,10,9,10,11,12,11,10,9,10,11,12,13,14,15,14,15,16,15,16,15,14
%N a(n) = A257808(n)  A257807(n).
%C Alternative description: Start with a(0) = 0, and then to obtain each a(n), look at each successive term in the infinite trunk of inverted binary beanstalk, from A233271(1) onward, subtracting one from a(n1) if A233271(n) is odd, and adding one to a(n1) if A233271(n) is even.
%C In other words, starting from zero, iterate the map x > {x + 1 + number of nonleading zeros in the binary representation of x}, and note each time whether the result is odd or even: With odd results go one step down, and even results go one step up.
%C After the zeros at a(0), a(2) and a(4) and 1 at a(1), the terms stay strictly positive for a long time, although from the terms of A257805 it can be seen that the sequence must again fall to the negative side somewhere between n = 541110611 and n = 1051158027 (i.e., A218600(33) .. A218600(34)). Indeed the fourth zero occurs at n = 671605896, and the second negative term right after that as a(671605897) = 1.
%C The maximum positive value reached prior to the slide into negative territory is 2614822 for a(278998626) and a(278998628).  _Hans Havermann_, May 23 2015
%H Antti Karttunen, <a href="/A257806/b257806.txt">Table of n, a(n) for n = 0..8727</a>
%H Hans Havermann, <a href="http://chesswanks.com/num/a257806.png">Graph of 2*10^9 terms</a>
%H Hans Havermann, <a href="http://chesswanks.com/num/a257806zeros.png">Detail graph of the eventual crossover to negative terms and a listing of its associated zeros</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Blancmange_curve">Blancmange curve</a>
%F a(n) = A257808(n)  A257807(n).
%F a(0) = 0; and for n >= 1, a(n) = a(n1) + (1)^A233271(n).
%F Other identities. For all n >= 0:
%F a(A218600(n+1)) = A257805(n).
%e We consider 0 to have no nonleading zeros, so first we get to 0 > 0+1+0 = 1, and 1 is odd, so we go one step down from the starting value a(0)=0, and thus a(1) = 1.
%e 1 has no nonleading zeros, so we get 1 > 1+1+0 = 2, and 2 is even, so we go one step up, and thus a(2) = 0.
%e 2 has one nonleading zero in binary "10", so we get 2 > 2+1+1 = 4, and 4 is also even, so we go one step up, and thus a(3) = 1.
%e 4 has two nonleading zeros in binary "100", so we get 4 > 4+2+1 = 7, 7 is odd, so we go one step down, and thus a(4) = 0.
%o (PARI)
%o A070939 = n>#binary(n)+!n; \\ From _M. F. Hasler_
%o A080791 = n>if(n<1,0,(A070939(n)hammingweight(n)));
%o A233272 = n>(n + A080791(n) + 1);
%o A257806_write_bfile(up_to_n) = { my(n,a_n=0,b_n=0); for(n=0, up_to_n, write("b257806.txt", n, " ", a_n); b_n = A233272(b_n); a_n += ((1)^b_n)); };
%o A257806_write_bfile(8727);
%o (Python)
%o def A257806_print_upto(n):
%o a = 0
%o b = 0
%o for n in range(n):
%o print ",",b,
%o ta = a
%o c0 = 0
%o while ta>0:
%o c0 += 1(ta&1)
%o ta >>= 1
%o a += 1 + c0
%o b += ((2*(1(a&1)))  1)
%o # By _Antti Karttunen_ after _Alex Ratushnyak_'s Pythoncode for A216431.
%o (Scheme, two alternatives, the latter using memoizing definecmacro)
%o (define (A257806 n) ( (A257808 n) (A257807 n)))
%o (definec (A257806 n) (if (zero? n) n (+ (expt 1 (A233271 n)) (A257806 ( n 1)))))
%Y Cf. A218600, A233271, A233272, A257807, A257808, A257803, A257804, A257805.
%Y Cf. also A218542, A218543, A218789 and A233270 (compare the scatter plots).
%K sign,base,look
%O 0,7
%A _Antti Karttunen_, May 12 2015
