A353954 a(0) = 1; a(n) = A019565(A109812(n)).

1, 2, 3, 5, 6, 7, 10, 21, 11, 15, 14, 33, 35, 22, 105, 13, 30, 77, 26, 55, 42, 65, 66, 91, 110, 39, 70, 143, 210, 17, 165, 182, 51, 154, 195, 34, 231, 130, 119, 330, 221, 462, 85, 78, 385, 102, 455, 187, 390, 1309, 19, 770, 663, 38, 1155, 442, 57, 910, 561, 95, 273

Offset: 0

Author

Michael De Vlieger, May 12 2022

Keywords

• nonn

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.

Examples

			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
  ...
		

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.

Crossrefs

Cf. A005117, A019565, A109812.

Programs

Mathematica

Clear[c, a]; 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]

Formula

a(0) = 1; a(n) = Product p_k where A109812(n) = Sum 2^(k-1) for n > 0.