OFFSET
0,2
COMMENTS
Interpretation of A109812(n) written in binary instead as if written in "multiplicity notation", that is, as if we write 1 if divisible by prime(k+1), otherwise 0 in the k-th place. Example, decimal 12 is written in binary as 1100 = 2^2 + 2^3, and take exponents 2 and 3 and instead construe them as prime(2+1) * prime(3+1) = 5*7 = 35.
Permutation of squarefree numbers A005117.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..10000
Michael De Vlieger, Annotated log log scatterplot of a(n), n = 1..2^14, with records in red and local minima in blue, highlighting primes in green and fixed points in gold.
FORMULA
a(0) = 1; a(n) = Product p_k where A109812(n) = Sum 2^(k-1) for n > 0.
EXAMPLE
Table showing n, A109812(n), and b(n), the binary expansion of A109812(n) writing "." for zeros for clarity. a(n) interprets 1's in the k-th place of b(n) as prime(k+1) and thereafter takes the product. We find a(n) = A005117(j). Note that A109812(0) is not defined.
n A109812(n) b(n) a(n) j
----------------------------
0 - . 1 1
1 1 1 2 2
2 2 1. 3 3
3 4 1.. 5 4
4 3 11 6 5
5 8 1... 7 6
6 5 1.1 10 7
7 10 1.1. 21 14
8 16 1.... 11 8
9 6 11. 15 11
10 9 1..1 14 10
11 18 1..1. 33 21
12 12 11.. 35 23
13 17 1...1 22 15
14 14 111. 105 65
15 32 1..... 13 9
16 7 111 30 19
...
MATHEMATICA
nn = 60; c[_] = 0; a[0] = c[1] = j = 1; a[1] = u = 2; Do[k = u; While[Nand[c[k] == 0, BitAnd[j, k] == 0], k++]; If[k == u, While[c[u] > 0, u++]]; j = k; Set[{a[i], c[k]}, {Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, Reverse@ IntegerDigits[k, 2]], i}], {i, 2, nn}]; Array[a, nn + 1, 0]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael De Vlieger, May 12 2022
STATUS
approved