

A105033


Read binary numbers downwards to the right.


5



0, 1, 0, 3, 2, 1, 4, 7, 6, 5, 0, 11, 10, 9, 12, 15, 14, 13, 8, 3, 18, 17, 20, 23, 22, 21, 16, 27, 26, 25, 28, 31, 30, 29, 24, 19, 2, 33, 36, 39, 38, 37, 32, 43, 42, 41, 44, 47, 46, 45, 40, 35, 50, 49, 52, 55, 54, 53, 48, 59, 58, 57, 60, 63, 62, 61, 56, 51, 34, 1, 68, 71, 70, 69, 64, 75
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OFFSET

0,4


COMMENTS

Equals A103530(n+2)  1.  Philippe Deléham, Apr 06 2005


REFERENCES

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.


LINKS

Table of n, a(n) for n=0..75.
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.


FORMULA

a(n) = n  Sum_{ k >= 0, 2^{k+1} <= n, n == k mod 2^(k+1) } 2^(k+1).
Structure: blocks of size 2^k taken from A105025, interspersed with terms a(n) itself! Thus a(2^k + k  1 ) = a(k1) for k >= 1.
From David Applegate, Apr 06 2005: (Start)
"a(n) = 2^k + a(n2^k) if k >= 1 and 0 <= n  2^k  k < 2^k, = a(n2^k) if k >= 1 and n  2^k  k = 1, or = 0 if n = 0 (and exactly one of the three conditions is true for any n >= 0).
"Equivalently, a(2^k + k + x) = 2^k + a(k+x) if 0 <= x < 2^k, = a(k+x) if x = 1 (for each n >= 0, there is a unique k, x such that 2^k + k + x = n, k >= 0, 1 <= x < 2^k). This recurrence follows immediately from the definition.
"The recurrence captures three observed facts about a: a(2^k + k  1) = a(k1); a consists of blocks of length 2^k of A105025 interspersed with terms of a; a(n) = n  Sum_{ k >= 0, 2^{k+1} <= n, n = k mod 2^(k+1) } 2^(k+1)." (End)
a(n) = sum_{k=0..n} A103589(n,k)*2^(nk).  L. Edson Jeffery, Dec 01 2013


EXAMPLE

Start with the binary numbers:
........0
........1
.......10
.......11
......100
......101
......110
......111
.....1000
.........
and read downwards to the right, getting 0, 1, 0, 11, 10, 1, 100, 111, ...


MAPLE

f:= proc (n) local t1, l; t1 := n; for l from 0 to n do if `mod`(nl, 2^(l+1)) = 0 and n >= 2^(l+1) then t1 := t12^(l+1) fi; od; t1; end proc;


MATHEMATICA

f[n_] := Block[{k = 0, s = 0}, While[2^(k + 1) < n + 1, If[ Mod[n, 2^(k + 1)] == k, s = s + 2^(k + 1)]; k++ ]; n  s]; Table[ f[n], {n, 0, 75}] (* Robert G. Wilson v, Apr 06 2005 *)


CROSSREFS

Analog of A102370. Cf. A105034, A105025.
Cf. triangular array in A103589.
Sequence in context: A271513 A117212 A208153 * A092486 A159966 A119263
Adjacent sequences: A105030 A105031 A105032 * A105034 A105035 A105036


KEYWORD

nonn,base


AUTHOR

N. J. A. Sloane, Apr 04 2005


STATUS

approved



