A363235 a(0) = 1; let e be the largest multiplicity such that p^e | a(n); for n>0, a(n) = Sum_{j=1..k} 2^(e(j)-1) where k is the index of the greatest power factor p(k)^e(k) such that p(k-1)^e(k-1) > p(k)^(e(k)+1).

0, 1, 2, 3, 4, 5, 8, 9, 10, 11, 16, 17, 18, 19, 20, 21, 22, 23, 32, 33, 34, 35, 36, 37, 38, 39, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 144, 145, 146, 147, 148, 149, 150, 151, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267

Offset: 0

Author

Michael De Vlieger, Jun 09 2023

Keywords

• nonn

Comments

A binary compactification of A363250, this sequence rewrites A363250(n) = Product_{i=1..omega(a(n))} p(i)^e(i) instead as Sum_{i=1..omega(a(n))} e(i)-1.

Not a permutation of nonnegative integers.

Examples

			a(1) = 1 since 2^1 is a product of the smallest primes p(i) whose prime power factors decrease as i increases; Hence a(1) = 2^(e(i)-1) = 1.
a(2) = 2 since we can find no power 3^e with e>=1 that is smaller than 2^1, we increment the exponent of 2 and have 2^2, hence a(2) = 2^(e(i)-1) = 2.
a(3) = 3 since indeed we may multiply 2^2 by 3^1; 2^2 > 3^1, hence Sum_{i=1..2} 2^(e(i)-1) = 2^1 + 2^0 = 2+1 = 3.
Table relating this sequence to A363250.
b(n) = A363250(n), f(n) = A067255(n), g(n) = A272011(n), with the latter two
   n      b(n)  f(b(n))  a(n)  g(a(n))
  ------------------------------------
   1        1   0          0   -
   2        2   1          1   0
   3        4   2          2   1
   4       12   2,1        3   1,0
   5        8   3          4   2
   6       24   3,1        5   2,0
   7       16   4          8   3
   8       48   4,1        9   3,0
   9      144   4,2       10   3,1
  10      720   4,2,1     11   3,1,0
  11       32   5         16   4
  12       96   5,1       17   4,0
  13      288   5,2       18   4,1
  14     1440   5,2,1     19   4,1,0
  15      864   5,3       20   4,2
  16     4320   5,3,1     21   4,2,0
  17    21600   5,3,2     22   4,2,1
  18   151200   5,3,2,1   23   4,2,1,0
  19       64   6         32   5
  ...
Therefore, a(18) = 23 = 2^4 + 2^2 + 2^1 + 2^0 since b(18) = 151200 = 2^5 * 3^3 * 5^2 * 7^1.
The sequence is a series of intervals, organized so as to begin with 2^k, that begin as follows:
     0
     1
     2..3
     4..5
     8..11
    16..23
    32..39
    64..75
   128..139     144..151
   256..267     272..279
   512..523     528..535     544..559
  1024..1035   1040..1047   1056..1071
  2048..2059   2064..2071   2080..2095   2112..2127
  ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..11210 (a(11210) = 2^28.)
• Michael De Vlieger, Binary tree indicating natural numbers k in red that appear in this sequence for k = 1..16383.

Crossrefs

Cf. A000079, A067255, A272011, A347284, A363063, A363250.

Programs

Mathematica

Select[Range[0, 300], AllTrue[Differences@ MapIndexed[Prime[First[#2]]^#1 &, Length[#] - Position[#, 1][[All, 1]] &@ IntegerDigits[#, 2] + 1], # < 0 &] &]