A360573 Odd numbers with exactly three zeros in their binary expansion.

17, 35, 37, 41, 49, 71, 75, 77, 83, 85, 89, 99, 101, 105, 113, 143, 151, 155, 157, 167, 171, 173, 179, 181, 185, 199, 203, 205, 211, 213, 217, 227, 229, 233, 241, 287, 303, 311, 315, 317, 335, 343, 347, 349, 359, 363, 365, 371, 373, 377, 399, 407, 411, 413

Offset: 1

Author

Bernard Schott, Feb 12 2023

Keywords

• nonn
• base

Comments

If m is a term then 2*m+1 is another term, since if M is the binary expansion of m, then M.1 where . stands for concatenation is the binary expansion of 2*m+1.

A052996 \ {1,3,8} is a subsequence, since for m >= 3, A052996(m) = 9*2^(m-2) - 1 has 100011..11 with m-2 trailing 1 for binary expansion.

A171389 \ {20} is a subsequence, since for m >= 1, A171389(m) = 21*2^m - 1 has 1010011..11 with m trailing 1 for binary expansion.

A198276 \ {18} is a subsequence, since for m >= 1, A198276(m) = 19*2^m - 1 has 1001011..11 with m trailing 1 for binary expansion.

Binary expansion of a(n) is A360574(n).

{8*a(n), n>0} form a subsequence of A353654 (numbers with three trailing 0 bits and three other 0 bits).

Numbers of the form 2^(a+1) - 2^b - 2^c - 2^d - 1 where a > b > c > d > 0. - Robert Israel, Feb 13 2023

Examples

			35_10 = 100011_2, so 35 is a term.
		

Crossrefs

Cf. A005408, A023416, A056557, A333516, A353654, A360010, A360574.

Subsequences: A052996 \ {1,3,8}, A171389 \ {20}, A198276 \ {18}.

Odd numbers with k zeros in their binary expansion: A000225 (k=0), A190620 (k=1), A357773 (k=2), this sequence (k=3).

Programs

Maple

q:= n-> n::odd and add(1-i, i=Bits[Split](n))=3:
select(q, [$1..575])[];  # Alois P. Heinz, Feb 12 2023
# Alternative:
[seq(seq(seq(seq(2^(a+1) - 2^b - 2^c - 2^d - 1, d = c-1..1,-1), c=b-1..2,-1),b=a-1..3,-1),a=4..12)]; # Robert Israel, Feb 13 2023

Mathematica

Select[Range[1, 500, 2], DigitCount[#, 2, 0] == 3 &] (* Amiram Eldar, Feb 12 2023 *)

PARI

isok(m) = (m%2) && #select(x->(x==0), binary(m)) == 3; \\ Michel Marcus, Feb 13 2023

Python

def ok(n): return n&1 and bin(n)[2:].count("0") == 3
print([k for k in range(414) if ok(k)]) # Michael S. Branicky, Feb 12 2023

Python

from itertools import count, islice
from sympy.utilities.iterables import multiset_permutations
def A360573_gen(): # generator of terms
    yield from (int('1'+''.join(d)+'1',2) for l in count(0) for d in  multiset_permutations('000'+'1'*l))
A360573_list = list(islice(A360573_gen(),54)) # Chai Wah Wu, Feb 18 2023

Python

from itertools import combinations, count, islice
def agen(): yield from ((1<Michael S. Branicky, Feb 18 2023

Python

from math import comb, isqrt
from sympy import integer_nthroot
def A056557(n): return (k:=isqrt(r:=n+1-comb((m:=integer_nthroot(6*(n+1), 3)[0])-(nA333516(n): return (r:=n-1-comb((m:=integer_nthroot(6*n, 3)[0])+(n>comb(m+2, 3))+1, 3))-comb((k:=isqrt(m:=r+1<<1))+(m>k*(k+1)), 2)+1
def A360010(n): return (m:=integer_nthroot(6*n, 3)[0])+(n>comb(m+2, 3))
def A360573(n):
    a = (a2:=integer_nthroot(24*n, 4)[0])+(n>comb(a2+2, 4))+3
    j = comb(a-1,4)-n
    b, c, d = A360010(j+1)+2, A056557(j)+2, A333516(j+1)
    return (1<Chai Wah Wu, Dec 18 2024

Formula

A023416(a(n)) = 3.

Let a = floor((24n)^(1/4))+4 if n>binomial(floor((24n)^(1/4))+2,4) and a = floor((24n)^(1/4))+3 otherwise. Let j = binomial(a-1,4)-n. Then a(n) = 2^a-1-2^(A360010(j+1)+2)-2^(A056557(j)+2)-2^(A333516(j+1)). - Chai Wah Wu, Dec 18 2024