A087119 -id:A087119 - cp's OEIS frontend

A087116 Number of maximal groups of consecutive zeros in binary representation of n.

1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2

Offset: 0

Reinhard Zumkeller, Aug 14 2003

nonn

The following four statements are equivalent: a(n) = 0; n = 2^k - 1 for some k; A087117(n) = 0; A023416(n) = 0.

			G.f. = 1 + x^2 + x^4 + x^5 + x^6 + x^8 + x^9 + 2*x^10 + x^11 + x^12 + x^13 + x^14 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A087118, A087119, A087120, A023416, A007088.
Cf. A023416, A087117.
Cf. A033264, A003817, A069010.
Essentially the same as A033264.

a087116 0 = 1
a087116 n = f 0 n where
   f y 0 = y
   f y x = if r == 0 then g x' else f y x'
           where (x', r) = divMod x 2
                 g z = if r == 0 then g z' else f (y + 1) z'
                       where (z', r) = divMod z 2
-- Reinhard Zumkeller, Mar 31 2015

a[n_] := SequenceCount[IntegerDigits[n, 2], {Longest[0..]}];
Table[a[n], {n, 0, 101}] (* Jean-François Alcover, Oct 18 2021 *)

a(n) = if (n == 0, 1, hammingweight(bitxor(n, n>>1)) >> 1);
vector(102, i, a(i-1))  \\ Gheorghe Coserea, Sep 17 2015

def A087116(n):
    return sum(1 for d in bin(n)[2:].split('1') if len(d)) # Chai Wah Wu, Nov 04 2016

a(n) = A033264(n) for n > 0 since strings of 0's alternate with strings of 1's. - Jonathan Sondow, Jan 17 2016
a(n) = a(2*n + 1) = a(4*n + 2) - 1, if n > 0. - Michael Somos, Nov 04 2016
a(n) = A069010(A003817(n)-n) for n > 0. - Chai Wah Wu, Nov 04 2016

A087118 Numbers having exactly one maximal group of consecutive zeros in binary representation of n.

Original entry on oeis.org

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

Reinhard Zumkeller, Aug 14 2003

nonn

A087116(a(n)) = 1.

a(n) = A043687(n-1) for 1 < n < 1000. - Georg Fischer, Oct 19 2018

Gheorghe Coserea, Table of n, a(n) for n = 1..12343
Index entries for sequences related to binary expansion of n

Cf. A007088, A023416, A043687, A087119.

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

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 *)

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

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)

cp's OEIS Frontend

A087116 Number of maximal groups of consecutive zeros in binary representation of n.

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