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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.

A268289 a(0)=0; thereafter a(n) = a(n-1) - A037861(n).

Original entry on oeis.org

0, 1, 1, 3, 2, 3, 4, 7, 5, 5, 5, 7, 7, 9, 11, 15, 12, 11, 10, 11, 10, 11, 12, 15, 14, 15, 16, 19, 20, 23, 26, 31, 27, 25, 23, 23, 21, 21, 21, 23, 21, 21, 21, 23, 23, 25, 27, 31, 29, 29, 29, 31, 31, 33, 35, 39, 39, 41, 43, 47, 49
Offset: 0

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Author

Mark Moore, Jan 30 2016

Keywords

Comments

The graph of this sequence is related to the Takagi (blancmange) curve: see Lagarias (2012), Section 9, especially Theorem 9.1. [Corrected by Laura Monroe, Oct 21 2020]
Theorem: a(n) is the cardinality of the set { 1<= m <= n, ((n-m) mod 2^floor(log_2(m)+1)) < 2^floor(log_2(m)) }. See links.
From Laura Monroe, Jun 11 2020: (Start)
Consider a full balanced binary tree with n unlabeled leaves such that for each internal node, the number of leaf descendants of the two children differs by at most 1. Call a tree with this even distribution of leaves "pairwise".
Apply labels to the internal nodes, where an internal node is labeled S if its two children have the same number of leaf descendants, and D if its two children have a different number of leaf descendants, and call this an SD-tree. (For a pairwise tree, this is equivalent to saying that a node is an S-node iff it has an even number of leaf descendants.)
a(n) is then the number of S-nodes on a pairwise SD-tree with n+1 leaves.
This is proved in Props. 17 and 18 of the Monroe et al. article in the links.
One example of such a tree is the summation tree generated by a pairwise summation on n+1 summands (see example below). Another example is the tree representing a neutral single-elimination tournament on n+1 teams, as in A096351.
(End)
From Laura Monroe, Oct 23 2020: (Start)
Subtracting a(n) from n gives a sequence of dilations of increasing length on the dyadic rational points of the Takagi function. The number of points in each dilation is 2^k and the scale of each dilation in both the x and y directions is 2^k, where k = floor(log_2(n+1)).
2^(a(n)) is the number of tree automorphisms on the pairwise (i.e., divide-and-conquer) tree with n+1 leaves.
(End)

Examples

			From _Laura Monroe_, Jun 11 2020: (Start)
For n=2, the pairwise summation on 2+1=3 summands takes the form ((a+b)+c). The corresponding summation tree and SD-tree look like:
       +            D
      / \          / \
     +   c        S   c
    / \          / \
   a   b        a   b
and exactly 1 internal node has an even number of leaf descendants, hence is an S-node.
For n=3, the pairwise summation on 3+1=4 summands takes the form ((a+b)+(c+d)). The corresponding summation tree and SD-tree look like:
       +            S
      / \          / \
     +   +        S   S
    /|   |\      /|   |\
   a b   c d    a b   c d
and exactly 3 internal nodes have an even number of leaf descendants, hence are S-nodes.
(End)

Links

Crossrefs

Partial sums of A145037.
Cf. A000120, A000788, A023416, A037861, A096351, A181132, A269735, A233271, A296062.

Programs

  • C
    int a(int n)   {
    int m=n+1;
    int result=0;
    int i=0;
    while (n) {
        int ith_bit_set = m&(1<>= 1;
    }
   return result;
}
/* Laura Monroe, Jun 17 2020 */
  • Julia
    function A268289List(len)
    A = zeros(Int, len)
    for n in 1:len-1
        a, b, c = n, n & 1, 1
        while (a >>= 1) != 0
            b += a & 1
            c += 1
        end
        A[n+1] = A[n] + <<(b, 1) - c
    end
    A
end; println(A268289List(61)) # Peter Luschny, Jun 22 2020
  • Maple
    a000120 := proc(n) add(i, i=convert(n, base, 2)) end:
a023416 := proc(n) if n = 0 then 1; else add(1-e, e=convert(n, base, 2)) ; end if; end proc:
a268289:=proc(n) option remember; global a000120, a023416;
if n=0 then 0 else a268289(n-1)+a000120(n)-a023416(n); fi; end;
[seq(a268289(n),n=0..132)];
# N. J. A. Sloane, Mar 07 2016
# second Maple program:
a:= proc(n) option remember; `if`(n<0, 0,
      a(n-1)+add(2*i-1, i=Bits[Split](n)))
    end:
seq(a(n), n=0..60);  # Alois P. Heinz, Jan 18 2022
  • Mathematica
    Join[{0}, Table[DigitCount[n, 2, 1] - DigitCount[n, 2, 0], {n, 1, 100}] // Accumulate] (* Jean-François Alcover, Oct 24 2016 *)
  • PARI
    a(n) = if (n==0, 0, if (n%2, 2*a((n-1)/2)+1, a(n/2) + a(n/2-1))); \\ Michel Marcus, Jun 16 2020
  • PARI
    a(n) = my(v=binary(n+1),s=-1); for(i=1,#v, v[i]=if(v[i],s++,s--;1)); fromdigits(v,2); \\ Kevin Ryde, Jun 16 2020
  • Python
    def A268289(n): return (sum(i.bit_count() for i in range(1,n+1))<<1)-1-(n+1)*(m:=(n+1).bit_length())+(1<Chai Wah Wu, Mar 01 2023
  • Python
    def A268289(n): return sum((n+1)%m if (n+1)&(m:=1<Chai Wah Wu, Nov 11 2024

Formula

From N. J. A. Sloane, Mar 11 2016: (Start)
a(0)=0; for n > 0, a(n) = a(n-1) + A000120(n) - A023416(n) = A000788(n) - A181132(n).
a(0)=0; thereafter a(2*n) = a(n) + a(n-1), a(2*n+1) = 2*a(n) + 1.
G.f.: (1/(1-x)^2) * Sum_{k >= 0} x^(2^k)*(1-x^(2^k))/(1+x^(2^k)).
a(2^k-1) = 2^k-1, a(3*2^k-1) = 2^(k+1)-1, a(5*2^k-1) = 3*2^k-1, etc.
(End)
From Laura Monroe, Jun 11 2020: (Start)
a(n-1) = Sum_{i=0..floor(log_2(n))} (((floor(n/(2^i))+1) mod 2)*(2^i)+(-1)^((floor(n/(2^i))+1) mod 2)*(n mod (2^i))), for n>=1.
This is an explicit formula for this sequence, and is O(log(n)). This formula is proven in Prop. 18, in the Monroe et al. reference in the links. (End)
From Laura Monroe, Oct 23 2020: (Start)
a(n) = n - A296062(n).
a(n+1) = (n+1) - (2^k)*tau(x/(2^k)), where tau is the Takagi function and n+1 = (2^k)+x with x < 2^k. (End)

Extensions

Simplified definition following a suggestion from Michel Marcus. Corrected start, added more terms. - N. J. A. Sloane, Mar 07 2016

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