

A023758


Numbers of the form 2^i  2^j with i >= j.


69



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
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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 the form (2^i1)*2^j, i,j >= 0, a(n)^2 = (2^(2i)  2^(i+1))*2^(2j) + 2^(2j) where the first sum term has i1 one bits and its 2jth bit is zero, while the second sum term switches the 2jth bit to one, giving i one bits, as in a(n).  Ralf Stephan, Mar 08 2004
Numbers whose 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
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
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)
Powers of 2 represented in bases which are terms of this sequence must always contain at least one digit which is also a power of 2. This is because 2^i mod (2^i  2^j) = 2^j, which means the last digit always cycles through powers of 2 (or if i=j+1 then the first digit is a power of 2 and the rest are trailing zeros). The only known nonmember of this sequence with this property is 5.  Ely Golden, Sep 05 2017


LINKS

Eric Weisstein's World of Mathematics, Digit.


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(nk1) for 1 < n and k = A003056(n2). The rows of T(r, c) = 2^r2^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
a(n+1) = (2^(n  r(r1)/2)  1) 2^(r(r+1)/2  n), where r=round(sqrt(2n)).  M. F. Hasler, May 06 2009
G.f.: (x^2/((2x)*(1x)))*(1 + Sum_{k>=0} x^((k^2+k)/2)*(1 + x*(2^k1))). The sum is related to Jacobi theta functions.  Robert Israel, Feb 24 2015
a(n) = a(n1) + a(nd)/a(d*(d+1)/2 + 2) if n > 1, d > 0, where d = A002262(n2).  Yuchun Ji, May 11 2020


EXAMPLE

a(22) = 64 = 32 + 32 = 2^5 + a(16) = 2^A003056(20) + a(2251).
a(23) = 96 = 64 + 32 = 2^6 + a(16) = 2^A003056(21) + a(2361).
a(24) = 112 = 64 + 48 = 2^6 + a(17) = 2^A003056(22) + a(2461).


MAPLE

a:=proc(n) local n2, d: n2:=convert(n, base, 2): d:={seq(n2[j]n2[j1], 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 *)
Select[Range[0, 2^10], NoneTrue[Differences@ IntegerDigits[#, 2], # > 0 &] &] (* Michael De Vlieger, Sep 05 2017 *)


PROG

(PARI) for(n=0, 2500, if(prod(k=1, length(binary(n))1, component(binary(n), k)+1component(binary(n), k+1))>0, print1(n, ", ")))
(PARI) A023758(n)= my(r=round(sqrt(2*n))); (1<<(nr*(r1)/2)1)<<(r*(r+1)/2n)
/* 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^(j1))~); a} \\ M. F. Hasler, May 06 2009
(PARI) list(lim)=my(v=List([0]), t); for(i=1, logint(lim\1+1, 2), t=2^i1; 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 !! (n1)
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
(Python)
def a_next(a_n): return (a_n  (a_n >> 1)) + (a_n & 1)
a_n = 1; a = [0]
for i in range(55): a.append(a_n); a_n = a_next(a_n) # Falk Hüffner, Feb 19 2022


CROSSREFS

Positions of nonzero terms in A341509 (apart from the initial zero).
Positions of squarefree terms in A260443.


KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



