



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, 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, 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
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OFFSET

0,7


COMMENTS

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.
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.
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.
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


LINKS



FORMULA

a(0) = 0; and for n >= 1, a(n) = a(n1) + (1)^A233271(n).
Other identities. For all n >= 0:


EXAMPLE

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.
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.
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.
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.


PROG

(PARI)
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)); };
(Python)
a = 0
b = 0
for n in range(n):
print(b, end=", ")
ta = a
c0 = 0
while ta>0:
c0 += 1(ta&1)
ta >>= 1
a += 1 + c0
b += ((2*(1(a&1)))  1)
(Scheme, two alternatives, the latter using memoizing definecmacro)


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



