|
|
A003159
|
|
Numbers whose binary representation ends in an even number of zeros.
(Formerly M2306)
|
|
106
|
|
|
1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 23, 25, 27, 28, 29, 31, 33, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 57, 59, 60, 61, 63, 64, 65, 67, 68, 69, 71, 73, 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 89, 91, 92, 93, 95, 97, 99, 100, 101, 103, 105
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Fraenkel (2010) called these the "vile" numbers.
Minimal with respect to property that parity of number of 1's in binary expansion alternates.
Minimal with respect to property that sequence is half its complement. [Corrected by Aviezri S. Fraenkel, Jan 29 2010]
If k appears then 2k does not.
Increasing sequence of positive integers k such that A035263(k)=1 (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
A different permutation of the same terms may be found in A141290 and the accompanying array. - Gary W. Adamson, Jun 14 2008
a(n) = n-th clockwise Tower of Hanoi move; counterclockwise if not in the sequence. - Gary W. Adamson, Jun 14 2008
Indices of terms of Thue-Morse sequence A010060 which are different from the previous term. - Tanya Khovanova, Jan 06 2009
The sequence has the following fractal property. Remove the odd numbers from the sequence, leaving 4,12,16,20,28,36,44,48,52,... Dividing these terms by 4 we get 1,3,4,5,7,9,11,12,..., which is the original sequence back again. - Benoit Cloitre, Apr 06 2010
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeff A174065 = the Euler transform of A035263 = 1/((1-x)*(1-x^3)*(1-x^4)*(1-x^5)*...) = 1 + x + x^2 + 2*x^3 + 3*x^4 + 4*x^5 + ... and the aerated variant = the Euler transform of the complement of A035263: 1/((1-x^2)*(1-x^6)*(1-x^8)*...) = 1 + x^2 + x^4 + 2*x^6 + 3*x^8 + 4*x^10 + ....
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013. - Gary W. Adamson, Jan 22 2014
If the lower s-Wythoff sequence of s is s, then s=A003159. (See A184117 for the definition of lower and upper s-Wythoff sequences.) Starting with any nondecreasing sequence s of positive integers, A003159 is the limit when the lower s-Wythoff operation is iterated. For example, starting with s=(1,4,9,16,...)=(n^2), we obtain lower and upper s-Wythoff sequences
a=(1,3,4,5,6,8,9,10,11,12,14,...)=A184427;
b=(2,7,12,21,31,44,58,74,...)=A184428.
Then putting s=a and repeating the operation gives a'=(1,3,4,5,7,9,11,12,14,...), which has the same first eight terms as A003159. - Clark Kimberling, Jan 14 2011
|
|
REFERENCES
|
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
LINKS
|
J.-P. Allouche, Andre Arnold, Jean Berstel, Srecko Brlek, William Jockusch, Simon Plouffe, and Bruce E. Sagan, A sequence related to that of Thue-Morse, Discrete Math., 139 (1995), 455-461.
J.-P. Allouche and Jeffrey Shallit, The Ubiquitous Prouhet-Thue-Morse Sequence, in C. Ding. T. Helleseth and H. Niederreiter, eds., Sequences and Their Applications: Proceedings of SETA '98, Springer-Verlag, 1999, pp. 1-16.
C. Kimberling, Problem E2850, Amer. Math. Monthly, 87 (1980), 671.
|
|
FORMULA
|
a(0) = 1; for n >= 0, a(n+1) = a(n) + 1 if (a(n) + 1)/2 is not already in the sequence, = a(n) + 2 otherwise.
a(n+1) = (if a(n) mod 4 = 3 then A007814(a(n) + 1) mod 2 else a(n) mod 2) + a(n) + 1; a(1) = 1. - Reinhard Zumkeller, Aug 03 2003
Sequence consists of numbers of the form 4^i*(2*j + 1), i>=0, j>=0. - Jon Perry, Jun 06 2004
a(1) = 1, a(2) = 3, and for n >= 2 we get a(n+1) = 4 + a(n) + a(n-1) - a(a(n)-n+1) - a(a(n-1)-n+2). - Benoit Cloitre, Apr 08 2010
If A(x) is the counting function for a(n) <= x, then A(2^n) = (2^(n+1) + (-1)^n)/3. - Vladimir Shevelev, Apr 15 2010
|
|
EXAMPLE
|
1=1, 3=11, 5=101 and 7=111 have no (0 = even) trailing zeros, 4=100 has 2 (= even) trailing zeros in the base-2 representation.
2=10 and 6=110 end in one (=odd number) of trailing zeros in their base-2 representation, therefore are not terms of this sequence. - M. F. Hasler, Oct 29 2013
|
|
MAPLE
|
filter:= n -> type(padic:-ordp(n, 2), even):
|
|
MATHEMATICA
|
f[n_Integer] := Block[{k = n, c = 0}, While[ EvenQ[k], c++; k /= 2]; c]; Select[ Range[105], EvenQ[ f[ # ]] & ]
Select[Range[150], EvenQ[IntegerExponent[#, 2]]&] (* Harvey P. Dale, Oct 19 2011 *)
|
|
PROG
|
(PARI) a(n)=if(n<2, n>0, n=a(n-1); until(valuation(n, 2)%2==0, n++); n)
(Haskell)
import Data.List (delete)
a003159 n = a003159_list !! (n-1)
a003159_list = f [1..] where f (x:xs) = x : f (delete (2*x) xs)
(Python)
from itertools import count, islice
def A003159_gen(startvalue=1): # generator of terms >= startvalue
return filter(lambda n:(n&-n).bit_length()&1, count(max(startvalue, 1)))
|
|
CROSSREFS
|
For the actual binary numbers see A280049.
Indices of even numbers in A007814.
|
|
KEYWORD
|
nonn,nice,easy,eigen,base
|
|
AUTHOR
|
|
|
EXTENSIONS
|
Incorrect formula removed by Peter Munn, Dec 04 2020
|
|
STATUS
|
approved
|
|
|
|