A358259 Positions of the first n-bit number to appear in Van Eck's sequence (A181391).

1, 5, 10, 24, 41, 52, 152, 162, 364, 726, 1150, 2451, 4626, 9847, 18131, 36016, 71709, 143848, 276769, 551730, 1086371, 2158296, 4297353, 8607525, 17159741, 34152001, 68194361, 136211839, 271350906, 541199486, 1084811069, 2165421369, 4331203801, 8643518017, 17303787585

Offset: 1

Author

Michael De Vlieger, Nov 05 2022

Keywords

• nonn
• base

Comments

Binary version of the concept behind A358180.

Examples

			First terms written in binary, substituting "." for 0 to enhance the pattern of 1's.
   n      a(n)                   a(n)_2
  -------------------------------------
   1        1                         1
   2        5                       1.1
   3       10                      1.1.
   4       24                     11...
   5       41                    1.1..1
   6       52                    11.1..
   7      152                  1..11...
   8      162                  1.1...1.
   9      364                 1.11.11..
  10      726                1.11.1.11.
  11     1150               1...111111.
  12     2451              1..11..1..11
  13     4626             1..1....1..1.
  14     9847            1..11..111.111
  15    18131           1...11.11.1..11
  16    36016          1...11..1.11....
  17    71709         1...11......111.1
  18   143848        1...11...1111.1...
  19   276769       1....111..1..1....1
  20   551730      1....11.1.11..11..1.
  21  1086371     1....1..1..111.1...11
  22  2158296    1.....111.111.11.11...
  23  4297353   1.....11..1..1.1...1..1
  24  8607525  1.....11.1.1.111..1..1.1
  etc.
		

Crossrefs

Cf. A181391, A358168, A358180, A358258.

Programs

Mathematica

nn = 2^20; q[] = False; q[0] = True; a[] = 0; c[_] = -1; c[0] = 2; m = 1; {1}~Join~Rest@ Reap[Do[j = c[m]; k = m; c[m] = n; m = 0; If[j > 0, m = n - j]; If[! q[#], Sow[n]; q[#] = True] & @ IntegerLength[k, 2], {n, 3, nn}] ][[-1, -1]]

Python

from itertools import count
def A358259(n):
    b, bdict, k = 0, {0:(1,)},1< 1 else 0
    for m in count(2):
        if b >= k:
            return m-1
        if len(l := bdict[b]) > 1:
            b = m-1-l[-2]
            if b in bdict:
                bdict[b] = (bdict[b][-1],m)
            else:
                bdict[b] = (m,)
        else:
            b = 0
            bdict[0] = (bdict[0][-1],m) # Chai Wah Wu, Nov 06 2022

Extensions

a(30)-a(34) from Chai Wah Wu, Nov 06 2022

a(35) from Martin Ehrenstein, Nov 07 2022