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
Keywords
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.
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.
Comments