OFFSET
0,3
COMMENTS
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.
EXAMPLE
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.
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
...
MATHEMATICA
Select[Range[0, 300], AllTrue[Differences@ MapIndexed[Prime[First[#2]]^#1 &, Length[#] - Position[#, 1][[All, 1]] &@ IntegerDigits[#, 2] + 1], # < 0 &] &]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael De Vlieger, Jun 09 2023
STATUS
approved