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A352782
The binary expansion of a(n) encodes the runs of consecutive 1's in the binary expansion of n (see Comments section for precise definition).
2
0, 1, 2, 4, 8, 3, 16, 32, 64, 5, 6, 12, 128, 9, 256, 512, 1024, 17, 10, 20, 24, 7, 48, 96, 2048, 33, 18, 36, 4096, 65, 8192, 16384, 32768, 129, 34, 68, 40, 11, 80, 160, 192, 13, 14, 28, 384, 25, 768, 1536, 65536, 257, 66, 132, 72, 19, 144, 288, 131072, 513
OFFSET
0,3
COMMENTS
For any nonnegative integer n:
- the binary expansion of n can be uniquely expressed as the concatenation of k = A069010(n) positive terms of A023758 separated by 0's:
n = A023758(m_k+1) | 0 | A023758(m_{k-1}+1) | 0 | ... | 0 | A023758(m_1+1)
(where | denotes binary concatenation)
- a(n) = ( Sum_{i = 1..k} 2^Sum_{j = 1..i} m_j ) / 2.
This sequence is a permutation of the nonnegative integers, with inverse A352783.
FORMULA
a(4*n+1) = 2*a(n)+1.
A000120(a(n)) = A069010(n).
a(A023758(k+1)) = 2^k for any k >= 0.
a(2^k) = A006125(k+1) for any k >= 0.
a(2^k-1) = A036442(k+1) for any k >= 0.
a(n) = n iff n = 0 or n belongs to A131865 or n/2 belongs to A131865.
EXAMPLE
For n = 89:
- the binary expansion of 89 is "1011001",
- "1011001" = "1" | 0 | "110" | 0 | "1"
= A023758(1+1) | 0 | A023758(5+1) | 0 | A023758(1+1)
- so 2*a(89) = 2^(1+5+1) + 2^(5+1) + 2^1 = 194,
- and a(89) = 97.
PROG
(PARI) a(n) = { my (v=0, s=-1, z, o, i); while (n, n\=2^z=valuation(n, 2); n\=2^o=valuation(n+1, 2); n\=2; i=(o+z)*(o+z-1)/2 + o; v+=2^s+=i); v }
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Apr 02 2022
STATUS
approved