A340619 n appears A006519(n) times.

1, 2, 2, 3, 4, 4, 4, 4, 5, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10, 10, 11, 12, 12, 12, 12, 13, 14, 14, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 18, 18, 19, 20, 20, 20, 20, 21, 22, 22, 23, 24, 24, 24, 24, 24, 24, 24, 24, 25, 26, 26

Offset: 1

Author

Rémy Sigrist, Jan 13 2021

Keywords

• nonn
• easy
• tabf

Comments

This sequence has similarities with the Cantor staircase function.

This sequence can be seen as an irregular table where the n-th row contains A006519(n) times the value n.

For any k > 1, the set of points { (n, a(n)), n = 1..A006520(2^k-1) } is symmetric; for example, for k = 3, we have the following configuration:

a(n)

^

| *

| **

| *

| ****

| *

| **

|*

+-------------> n

Examples

			The first rows, alongside A006519(n), are:
    n | n-th row               | A006519(n)
   ---+------------------------+-----------
    1 | 1                      |          1
    2 | 2, 2                   |          2
    3 | 3                      |          1
    4 | 4, 4, 4, 4             |          4
    5 | 5                      |          1
    6 | 6, 6                   |          2
    7 | 7                      |          1
    8 | 8, 8, 8, 8, 8, 8, 8, 8 |          8
    9 | 9                      |          1
   10 | 10, 10                 |          2
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..11264
• Wikipedia, Cantor function

Crossrefs

See A061392 and A340500 for similar sequences.

Cf. A006519, A006520, A046699.

Programs

Mathematica

A340619[n_] := Array[n &, Table[BitAnd[BitNot[i - 1], i], {i, 1, n}][[n]]];
Table[A340619[n], {n, 1, 26}] // Flatten (* Robert P. P. McKone, Jan 19 2021 *)

PARI

concat(apply(v -> vector(2^valuation(v,2), k, v), [1..26]))

PARI

a(n) = my(ret=0); forstep(k=logint(n,2),0,-1, if(n > k<<(k-1), ret+=1<Kevin Ryde, Jan 18 2021

Formula

a(A006520(n)) = n.

a(A006520(n)+1) = n+1.

a(n) + a(A006520(2^k-1) + 1 - n) = 2^k for any k > 0 and n = 1..A006520(2^k-1).

a(n) = 2^k + (a(r) if r>0), where k such that k*2^(k-1) < n <= (k+1)*2^k and r = n - (k+2)*2^(k-1). - Kevin Ryde, Jan 18 2021