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A286574
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Sum of the binary weights of the lengths of 1-runs in base-2 representation of n: a(n) = A000523(A286575(n)).
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4
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0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 4, 3, 4, 2, 3, 3, 3, 3, 4, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 4, 3, 4, 1
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OFFSET
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0,6
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COMMENTS
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a(0) = 0 (an empty sum).
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LINKS
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FORMULA
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EXAMPLE
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For n = 27, "11011" in binary, there are two 1-runs, both of length 2, thus a(27) = A000120(2) + A000120(2) = 1 + 1 = 2.
For n = 29, "11101" in binary, there are two 1-runs, of lengths 1 and 3, thus a(29) = A000120(1) + A000120(3) = 1 + 2 = 3.
For n = 61, "111101" in binary, there are two 1-runs, of lengths 1 and 4, thus a(61) = A000120(1) + A000120(4) = 1 + 1 = 2.
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PROG
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(Python)
from sympy import factorint, prime, log
import math
def wt(n): return bin(n).count("1")
def a037445(n):
f=factorint(n)
return 2**sum([wt(f[i]) for i in f])
def A(n): return n - 2**int(math.floor(log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a286575(n): return a037445(b(n))
def a(n): return int(math.floor(log(a286575(n), 2))) # Indranil Ghosh, May 30 2017
(Python)
def A286574(n): return len(bin(RLT(n, lambda m: 2**(bin(m).count('1')))))-3 # Chai Wah Wu, Feb 04 2022
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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