cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

Views

Author

Antti Karttunen, May 12 2015

Keywords

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

Examples

			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.
		

Crossrefs

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

Programs

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).

A257259 a(n) = A218542(n) - A218543(n).

Original entry on oeis.org

1, -1, 0, -1, -1, -3, -1, -6, -8, -10, -7, -4, 2, 2, -14, -90, -318, -896, -2166, -4691, -9298, -17175, -30007, -50261, -81664, -129637, -200973, -301205, -426500, -545183, -561511, -252122, 836543, 3542930, 9312475, 20508985, 40858087, 76119037, 135203839, 232236417, 390709345, 652711050, 1101492173, 1921013671, 3558812217, 7170401227, 15800043812, 37400877416
Offset: 0

Views

Author

Antti Karttunen, May 13 2015

Keywords

Comments

a(n) = the difference between the number of even and odd numbers encountered when traversing from 2^(n+1)-1 to (2^n)-1 by iterating the map A011371: x -> x - (number of 1's in binary representation of x).

Crossrefs

Partial sums: A257805.

Programs

Formula

a(n) = A218542(n) - A218543(n).
Showing 1-2 of 2 results.