A087118 Numbers having exactly one maximal group of consecutive zeros in binary representation of n.
0, 2, 4, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 23, 24, 25, 27, 28, 29, 30, 32, 33, 35, 39, 47, 48, 49, 51, 55, 56, 57, 59, 60, 61, 62, 64, 65, 67, 71, 79, 95, 96, 97, 99, 103, 111, 112, 113, 115, 119, 120, 121, 123, 124, 125, 126, 128, 129, 131, 135, 143, 159, 191
Offset: 1
Keywords
Links
Programs
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Maple
0, seq(seq(seq(2^n - 2^b + 2^a - 1, a=0..b-1),b=n-1..1,-1),n=0..10); # Robert Israel, Oct 01 2015
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Mathematica
Table[2^n - 2^b + 2^a - 1, {n, 0, 10}, {b, n-1, 1, -1}, {a, 0, b-1}] // Flatten // Prepend[#, 0]& (* Jean-François Alcover, Apr 11 2019, after Robert Israel *) -
PARI
num(a,b,c) = (1 << (a+b+c)) - (1 << (b+c)) + (1 << c) - 1; succ(a,b,c) = { if (b > 1, return([a, b-1, c+1])); if (c > 0, return([a+1, c, 0])); return([1, a+1, 0]); }; seq(n) = { my(a = 1, b = 1, c = 0, v = vector(n)); for (i = 2, n, v[i] = num(a,b,c); my(x = succ(a,b,c)); a = x[1]; b = x[2]; c = x[3]); return(v); }; seq(64) \\ Gheorghe Coserea, Sep 28 2015
Formula
From Gheorghe Coserea, Sep 28-30 2015: (Start)
a((n^3 - n)/6 + 2) = 2^n for n >= 1.
a((n^3 - n)/6 + 2 + n) = 2^n + 2^(n-1) for n >= 2.
a((n^3 - n)/6 + 2 + n + n-1) = 2^n + 2^(n-1) + 2^(n-2) for n >= 3.
a(n) < 2*2^((6*n)^(1/3)) and limsup a(n)/2^((6*n)^(1/3)) = 2.
a(n) > 1/2 * 2^((6*n)^(1/3)) for n>=3 and 1/2 <= liminf a(n)/(2^((6*n)^(1/3))) <= 1.
(End)
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