A379972 Numbers x with 2^(m-1)

1, 2, 3, 4, 5, 7, 8, 13, 15, 16, 25, 29, 31, 32, 57, 61, 63, 64, 113, 121, 125, 127, 128, 241, 249, 253, 255, 256, 481, 497, 505, 509, 511, 512, 993, 1009, 1017, 1021, 1023, 1024, 1985, 2017, 2033, 2041, 2045, 2047, 2048

Offset: 1

Author

Gerhard Kirchner, Jan 08 2025

Keywords

• nonn

Comments

Each circle, except for the last one, intersects the x-axis at the center of the next one. Any x>1 is represented by intersection points: y(1)=2, y(2),..., y(m)=x, but not necessarily in a unique way. The standard representation can be found by a backward algorithm: If y(j) is even, y(j-1)= y(j)/2, otherwise y(j)=(y(j)+1)/2. This way, only circles intersecting the x-axis at 0 or 1 are used. If no other representation exists, x belongs to the sequence, see examples.

Further comments, proof of the formula and images, see link "Construction with circles".

Examples

			Example 1: k=0, x=2^m belongs to the sequence.
 Standard repesentation: (2,..,2^j,..,2^m)
Example 2: m=6, k=3, x=57 belongs to the sequence.
 Standard repesentation: (2,4,8,15,29,57)
Counterexample 3: m=6, k=4, x=49 does not belong to the sequence.
 Standard repesentation: (2,4,7,13,25,49)
 Other repesentation:    (2,4,7,14,28,49)
Counterexample 4: x=48 does not belong to the sequence.
 Standard repesentation: (2,3,6,12,24,48)
 Other repesentation:    (2,4,7,13,25,48)
		

Links

• Gerhard Kirchner, Construction with circles

Programs

Maxima

block(u:[],
for m from 0 thru 11 do
  for k from floor((m+1)/2) thru 0 step -1 do
    if m=0 or k
				

Formula

x=2^m-2^k+1 with m>=0 and 0<=k<=(m+1)/2.

For x=1, only m=0 makes sense. Therefore k=m=1 is excluded.