|
| |
|
|
A023758
|
|
Numbers of the form 2^i - 2^j with i >= j.
|
|
42
|
|
|
|
0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 28, 30, 31, 32, 48, 56, 60, 62, 63, 64, 96, 112, 120, 124, 126, 127, 128, 192, 224, 240, 248, 252, 254, 255, 256, 384, 448, 480, 496, 504, 508, 510, 511, 512, 768, 896, 960, 992, 1008, 1016, 1020, 1022, 1023
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
Numbers whose digits in base 2 are in nonincreasing order.
Might be called "nialpdromes".
Subset of A077436. Proof: Since a(n) is of form (2^i-1)2^j, i,j>=0, a(n)^2 = [2^(2i)-2^(i+1)]2^(2j) + 2^(2j) where the first sum term has i-1 one bits and its 2j-th bit is zero, while the second sum term switches the 2j-th bit to one, giving i one bits, as in a(n). - Ralf Stephan, Mar 08 2004
Numbers n such that binary representation contains no "01". - Benoit Cloitre, May 23 2004
Every polynomial with coefficients equal to 1 for the leading terms and 0 after that, evaluated at 2. For instance a(13) = x^4 + x^3 + x^2 at 2, a(14) = x^4 + x^3 + x^2 + x at 2. - Ben Paul Thurston, Jan 11 2008
From Gary W. Adamson, Jul 18 2008: (Start)
As a triangle by rows starting:
1;
2, 3;
4, 6, 7;
8, 12, 14, 15;
16, 24, 28, 30, 31;
...,
equals A000012 * A130123 * A000012, where A130123 = (1, 0,2; 0,0,4; 0,0,0,8;...). Row sums of this triangle = A000337 starting (1, 5, 17, 49, 129,...). (End)
First differences are A057728 = 1; 1; 1; 1; 2,1; 1; 4,2,1; 1; 8,4,2,1; 1;... i.e., decreasing powers of 2, separated by another "1". - M. F. Hasler, May 06 2009
Apart from first term, numbers that are powers of 2 or the sum of some consecutive powers of 2. - Omar E. Pol, Feb 14 2013
A049502(a(n)) = 0. - Reinhard Zumkeller, Jun 17 2015
From Andres Cicuttin, Apr 29 2016: (Start)
Numbers that can be digitally generated with twisted ring (Johnson) counters. This is, the binary digits of a(n) correspond to those stored in a shift register where the input bit of the first bit storage element is the inverted output of the last storage element. After starting with all 0’s, each new state is obtained by rotating the stored bits but inverting at each state transition the last bit that goes to the first position (see link).
Examples: for a(n) represented by three bits
Binary
a(5)= 4 -> 100 last bit = 0
a(6)= 6 -> 110 first bit = 1 (inverted last bit of previous number)
a(7)= 7 -> 111
and for a(n) represented by four bits
Binary
a(8) = 8 -> 1000
a(9) = 12 -> 1100 last bit = 0
a(10)= 14 -> 1110 first bit = 1 (inverted last bit of previous number)
a(11)= 15 -> 1111
(End)
|
|
|
LINKS
|
T. D. Noe and R. Zumkeller, Table of n, a(n) for n = 1..10000, First 5051 terms by T. D. Noe
S. M. Shabab Hossain, Md. Mahmudur Rahman and M. Sohel Rahman, Solving a Generalized Version of the Exact Cover Problem with a Light-Based Device, Optical Supercomputing, Lecture Notes in Computer Science, 2011, Volume 6748/2011, 23-31, DOI: 10.1007/978-3-642-22494-2_4.
Eric Weisstein's World of Mathematics, Digit
Wikipedia, Ring counter
Index entries for 2-automatic sequences.
|
|
|
FORMULA
|
a(n) = 2^s(n) - 2^{[s(n)^2+s(n)-2n]/2} where s(n) = ceiling{[ -1+sqrt(1+8n)]/2}. - Sam Alexander, Jan 08 2005
a(n) = 2^k + a(n-k-1) for 1 < n and k = A003056(n-2). The rows of T(r, c) = 2^r-2^c for 0 <= c < r read from right to left produce this sequence: 1; 2, 3; 4, 6, 7; 8, 12, 14, 15; ... - Frank Ellermann, Dec 06 2001
For n>0, a(n) mod 2 == A010054(n). - Benoit Cloitre, May 23 2004
A140130(a(n)) = 1 and for n>1: A140129(a(n)) = A002262(n-2). - Reinhard Zumkeller, May 14 2008
a(n+1) = (2^(n-r(r-1)/2)-1) 2^(r(r+1)/2-n), where r=round(sqrt(2n)). - M. F. Hasler, May 06 2009
Start with A000225. If n is in sequence, then so is 2n. - Ralf Stephan, Aug 16 2013
G.f.: x^2/((2-x)*(1-x))*(1+sum(k >=0, x^((k^2+k)/2)*(1 + x*(2^k-1)))). The sum is related to Jacobi theta functions. - Robert Israel, Feb 24 2015
|
|
|
EXAMPLE
|
a(22) = 64 = 32 + 32 = 2^5 + a(16) = 2^A003056(20) + a(22-5-1).
a(23) = 96 = 64 + 32 = 2^6 + a(16) = 2^A003056(21) + a(23-6-1).
a(24) = 112 = 64 + 48 = 2^6 + a(17) = 2^A003056(22) + a(24-6-1).
|
|
|
MAPLE
|
a:=proc(n) local n2, d: n2:=convert(n, base, 2): d:={seq(n2[j]-n2[j-1], j=2..nops(n2))}: if n=0 then 0 elif n=1 then 1 elif d={0, 1} or d={0} or d={1} then n else fi end: seq(a(n), n=0..2100); # Emeric Deutsch, Apr 22 2006
|
|
|
MATHEMATICA
|
Union[Flatten[Table[2^i - 2^j, {i, 0, 100}, {j, 0, i}]]] (* T. D. Noe, Mar 15 2011 *)
|
|
|
PROG
|
(PARI) for(n=0, 2500, if(prod(k=1, length(binary(n))-1, component(binary(n), k)+1-component(binary(n), k+1))>0, print1(n, ", ")))
(PARI) A023758(n)= my(r=round(sqrt(2*n--))); (1<<(n-r*(r-1)/2)-1)<<(r*(r+1)/2-n)
/* or, to illustrate the "decreasing digit" property and analogy to A064222: */
A023758(n, show=0)={ my(a=0); while(n--, show & print1(a", "); a=vecsort(binary(a+1)); a*=vector(#a, j, 2^(j-1))~); a} \\ M. F. Hasler, May 06 2009
(PARI) is(n)=if(n<5, 1, n>>=valuation(n, 2); n++; n>>valuation(n, 2)==1) \\ Charles R Greathouse IV, Jan 04 2016
(PARI) list(lim)=my(v=List([0]), t); for(i=1, logint(lim\1+1, 2), t=2^i-1; while(t<=lim, listput(v, t); t*=2)); Set(v) \\ Charles R Greathouse IV, May 03 2016
(Haskell)
import Data.Set (singleton, deleteFindMin, insert)
a023758 n = a023758_list !! (n-1)
a023758_list = 0 : f (singleton 1) where
f s = x : f (if even x then insert z s' else insert z $ insert (z+1) s')
where z = 2*x; (x, s') = deleteFindMin s
-- Reinhard Zumkeller, Sep 24 2014, Dec 19 2012
|
|
|
CROSSREFS
|
A000337(r) = sum of row T(r, c) with 0 <= c < r. See also A003056.
Cf. A130123, A175332, A007088, A049502, A101082 (complement).
This is the base-2 version of A064222. First differences are A057728.
Sequence in context: A077436 A277704 A082752 * A054784 A018585 A018399
Adjacent sequences: A023755 A023756 A023757 * A023759 A023760 A023761
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
Olivier Gérard
|
|
|
EXTENSIONS
|
Definition changed by N. J. A. Sloane, Jan 05 2008
|
|
|
STATUS
|
approved
|
| |
|
|
