A100922 k appears A000120(k) times (appearances equal number of 1-bits).

1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 15, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 23, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27, 27, 27, 28, 28, 28, 29, 29, 29, 29, 30, 30, 30, 30

Offset: 0

Author

Rick L. Shepherd, Nov 21 2004

Keywords

• base
• easy
• nonn
• tabf

Comments

Clearly every positive integer appears at least once in this sequence.

Examples

			The binary representation of 16 is 10000, which has one 1-bit (and four 0-bits), hence 16 appears once in this sequence (and four times in A100921).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..11264

Crossrefs

Cf. A100921 (n's appearances equal its number of 0-bits), A030530 (n's appearances equal its total number of bits), A227737 (n's appearances equal its total number of runs), A000069, A000120, A000788, A163510, A243067.

Programs

Maple

T:= n-> n$add(i, i=Bits[Split](n)):
seq(T(n), n=1..30);  # Alois P. Heinz, Nov 11 2024

Mathematica

Table[Table[n,DigitCount[n,2,1]],{n,30}]//Flatten (* Harvey P. Dale, Aug 31 2017 *)

Python

def A000788(n): return (n+1)*n.bit_count()+(sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))>>1)
def A100922(n):
    if n == 0: return 1
    m, k = 1, 1
    while A000788(m)<=n: m<<=1
    while m-k>1:
        r = m+k>>1
        if A000788(r)>n:
            m = r
        else:
            k = r
    return m # Chai Wah Wu, Nov 11 2024

Formula

a(n) = the least k such that A000788(k) > n. - Antti Karttunen, Jun 20 2014

Sum_{n>=1} (-1)^(n+1)/a(n) = Sum_{n>=1} (-1)^(n+1)/A000069(n) = 0.67968268... . - Amiram Eldar, Feb 18 2024