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A100922
k appears A000120(k) times (appearances equal number of 1-bits).
4
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
0,2
COMMENTS
Clearly every positive integer appears at least once in this sequence.
LINKS
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
EXAMPLE
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).
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 *)
PROG
(Scheme, with Antti Karttunen's IntSeq-library)
(define A100922 (COMPOSE (LEAST-GTE-I 1 1 A000788) 1+))
(Python)
def A000788(n): return (n+1)*n.bit_count()+(sum((m:=1<<j)*((k:=n>>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
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.
Sequence in context: A166493 A005185 A119466 * A047785 A027923 A167743
KEYWORD
base,easy,nonn,tabf
AUTHOR
Rick L. Shepherd, Nov 21 2004
STATUS
approved