Search: a036993 -id:a036993
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A036994
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Numbers k with the property that reading from right to left in the binary expansion of k, the number of 1's always stays ahead of the number of 0's.
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+10
3
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1, 3, 7, 11, 15, 23, 27, 31, 39, 43, 47, 55, 59, 63, 79, 87, 91, 95, 103, 107, 111, 119, 123, 127, 143, 151, 155, 159, 167, 171, 175, 183, 187, 191, 207, 215, 219, 223, 231, 235, 239, 247, 251, 255, 287, 303, 311, 315, 319, 335, 343, 347, 351, 359, 363, 367
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OFFSET
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1,2
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COMMENTS
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LINKS
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MATHEMATICA
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aheadOnesRLQ[n_Integer] := Module[{digits, len, flag = True, iter = 1, ones = 0, zeros = 0}, digits = Reverse[IntegerDigits[n, 2]]; len = Length[digits]; While[flag && iter < len, If[digits[[iter]] == 1, ones++, zeros++]; flag = ones > zeros; iter++]; flag]; Select[Range[1, 201, 2], aheadOnesRLQ] (* Alonso del Arte, Sep 21 2011 *)
Select[Range[400], Min[Accumulate[Reverse[IntegerDigits[#, 2]/. (0->-1)]]]> 0&] (* Harvey P. Dale, Apr 23 2016 *)
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PROG
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(Haskell)
a036994 n = a036994_list !! (n-1)
a036994_list = filter ((p 0) . a030308_row) [1, 3 ..] where
p ones [] = ones > 0
p ones (0:bs) = ones > 1 && p (ones - 1) bs
p ones (1:bs) = p (ones + 1) bs
(Python)
from itertools import count, islice
def A036994_gen(startvalue=0): # generator of terms >= startvalue
for n in count(max(startvalue, 0)):
s = bin(n)[2:]
c, l = 0, len(s)
for i in range(l):
c += int(s[l-i-1])
if 2*c <= i + 1:
break
else:
yield n
(PARI) ok(x)={if(x<1, return(0)); my(c=logint(x, 2), c0=0, c1=0); for(i=0, c, if(bittest(x, i), c1++, c0++); if(c1<=c0, return(0))); 1} \\ for(n=1, 367, if(ok(n), print1(n, ", "))) - Ruud H.G. van Tol, Sep 14 2022
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A343049
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The k-th binary digit of a(n) is the most frequent digit among the first k binary digits of n (in case of a tie, take the k-th binary digit of n).
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+10
2
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0, 1, 2, 7, 0, 5, 6, 31, 0, 9, 10, 31, 8, 29, 30, 127, 0, 1, 2, 23, 0, 21, 22, 127, 0, 25, 26, 127, 24, 125, 126, 511, 0, 1, 2, 39, 0, 37, 38, 127, 0, 41, 42, 127, 40, 125, 126, 511, 0, 33, 34, 119, 32, 117, 118, 511, 32, 121, 122, 511, 120, 509, 510, 2047, 0
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listen;
history;
text;
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OFFSET
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0,3
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COMMENTS
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Leading zeros are taken into account up to the point the number of zeros exceeds the total number of ones.
We scan the binary representation of a number starting from the least significant digit. See A343271 for the other way.
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LINKS
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FORMULA
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a(n) = n iff n = 0 or n belongs to A032925.
a(2^k-1) = 2^(2*k-1)-1 for any k > 1.
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EXAMPLE
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The first terms, in decimal and in binary, are:
n a(n) bin(n) bin(a(n))
-- ---- ------ ---------
0 0 0 0
1 1 1 1
2 2 10 10
3 7 11 111
4 0 100 0
5 5 101 101
6 6 110 110
7 31 111 11111
8 0 1000 0
9 9 1001 1001
10 10 1010 1010
11 31 1011 11111
12 8 1100 1000
13 29 1101 11101
14 30 1110 11110
15 127 1111 1111111
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PROG
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(PARI) a(n, base=2) = { my (d=digits(n, base), t, f=vector(base)); d=concat(vector(#d), d); forstep (k=#d, 1, -1, f[1+d[k]]++; if (vecmax(f)==f[1+d[k]], t=d[k]; ); d[k]=t); fromdigits(d, base) }
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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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