A257806 a(n) = A257808(n) - A257807(n).

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

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(n-1) if A233271(n) is odd, and adding one to a(n-1) 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

Antti Karttunen, Table of n, a(n) for n = 0..8727

Hans Havermann, Graph of 2*10^9 terms

Hans Havermann, Detail graph of the eventual crossover to negative terms and a listing of its associated zeros

Wikipedia, Blancmange curve

Formula

a(n) = A257808(n) - A257807(n).

a(0) = 0; and for n >= 1, a(n) = a(n-1) + (-1)^A233271(n).

Other identities. For all n >= 0:

a(A218600(n+1)) = -A257805(n).

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)
A070939 = n->#binary(n)+!n; \\ From M. F. Hasler
A080791 = n->if(n<1, 0, (A070939(n)-hammingweight(n)));
A233272 = n->(n + A080791(n) + 1);
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)); };
A257806_write_bfile(8727);
(Python)
def A257806_print_upto(n):
  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)
# By Antti Karttunen after Alex Ratushnyak's Python-code for A216431.
(Scheme, two alternatives, the latter using memoizing definec-macro)
(define (A257806 n) (- (A257808 n) (A257807 n)))
(definec (A257806 n) (if (zero? n) n (+  (expt -1 (A233271 n)) (A257806 (- n 1)))))

Crossrefs

Cf. A218600, A233271, A233272, A257807, A257808, A257803, A257804, A257805.

Cf. also A218542, A218543, A218789 and A233270 (compare the scatter plots).

Sequence in context: A325622 A060145 A358997 * A035391 A242552 A249717

Adjacent sequences: A257803 A257804 A257805 * A257807 A257808 A257809

Keyword

sign,base,look

Author

Antti Karttunen, May 12 2015

Status

approved