OFFSET
0,2
COMMENTS
In other words, let S(n) contain place values of 1's in the binary expansion of n, ordered greatest to least, where S(n,1) = floor(log_2(n+1)) = A000523(n+1) and the remaining terms in S strictly decrease. This sequence reads S(n,k)+1 instead as a multiplicity of prime(k) so as to produce a number with strictly decreasing prime exponents.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..16384
EXAMPLE
a(0) = 1 since 1 is the empty product.
a(1) = 2 since 1 = 2^0, s = {0}, hence a(1) = prime(1)^(0+1) = 2^1 = 2.
a(2) = 4 since 2 = 2^1, s = {1}, hence a(2) = 2^(1+1) = 4.
a(3) = 12 since 3 = 2^1+2^0, s = {1,0}, hence a(3) = 2^2*3^1 = 12, etc.
------------------------------------
1 0 2 1
2 1 4 2
3 1,0 12 2,1
4 2 8 3
5 2,0 24 3,1
6 2,1 72 3,2
7 2,1,0 360 3,2,1
8 3 16 4
9 3,0 48 4,1
10 3,1 144 4,2
11 3,1,0 720 4,2,1
12 3,2 432 4,3
13 3,2,0 2160 4,3,1
14 3,2,1 10800 4,3,2
15 3,2,1,0 75600 4,3,2,1
16 4 32 5
...
This sequence appears below, seen as an irregular triangle T(m,j) delimited by 2^m where j = 1..2^(m-1) for m > 0:
1;
2;
4, 12;
8, 24, 72, 360;
16, 48, 144, 720, 432, 2160, 10800, 75600;
...
T(m,1) = 2^m.
T(m,2^(m-1)) = A006939(m) for m > 0.
MATHEMATICA
Array[Times @@ MapIndexed[Prime[First[#2]]^(#1 + 1) &, Length[#] - Position[#, 1][[All, 1]] ] &[IntegerDigits[#, 2]] &, 48, 0]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michael De Vlieger, Jun 08 2023
STATUS
approved