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n appears A006519(n) times.
1

%I #48 Jan 24 2021 05:53:36

%S 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,

%T 14,15,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,17,18,18,19,20,

%U 20,20,20,21,22,22,23,24,24,24,24,24,24,24,24,25,26,26

%N n appears A006519(n) times.

%C This sequence has similarities with the Cantor staircase function.

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

%C 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:

%C a(n)

%C ^

%C | *

%C | **

%C | *

%C | ****

%C | *

%C | **

%C |*

%C +-------------> n

%H Rémy Sigrist, <a href="/A340619/b340619.txt">Table of n, a(n) for n = 1..11264</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cantor_function">Cantor function</a>

%F a(A006520(n)) = n.

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

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

%F 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

%e The first rows, alongside A006519(n), are:

%e n | n-th row | A006519(n)

%e ---+------------------------+-----------

%e 1 | 1 | 1

%e 2 | 2, 2 | 2

%e 3 | 3 | 1

%e 4 | 4, 4, 4, 4 | 4

%e 5 | 5 | 1

%e 6 | 6, 6 | 2

%e 7 | 7 | 1

%e 8 | 8, 8, 8, 8, 8, 8, 8, 8 | 8

%e 9 | 9 | 1

%e 10 | 10, 10 | 2

%t A340619[n_] := Array[n &, Table[BitAnd[BitNot[i - 1], i], {i, 1, n}][[n]]];

%t Table[A340619[n], {n, 1, 26}] // Flatten (* _Robert P. P. McKone_, Jan 19 2021 *)

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

%o (PARI) a(n) = my(ret=0); forstep(k=logint(n,2),0,-1, if(n > k<<(k-1), ret+=1<<k; n-=(k+2)<<(k-1))); ret; \\ _Kevin Ryde_, Jan 18 2021

%Y See A061392 and A340500 for similar sequences.

%Y Cf. A006519, A006520, A046699.

%K nonn,easy,tabf

%O 1,2

%A _Rémy Sigrist_, Jan 13 2021