A159918 -id:A159918 - cp's OEIS frontend

A083567 Let B(k) be the number of binary digits in k equal to 1. This is the sequence of positive integers k such that 2B(k)=B(k^2).

21, 37, 42, 45, 53, 69, 73, 74, 81, 83, 84, 90, 106, 133, 137, 138, 141, 146, 148, 155, 161, 162, 165, 166, 168, 177, 180, 211, 212, 261, 265, 266, 269, 273, 274, 276, 281, 282, 289, 291, 292, 295, 296, 299, 310, 321, 322, 324, 330, 332, 336, 354, 359, 360

Offset: 1

Giuseppe Melfi, Jun 13 2003

This includes all k > 1 such that the average of ones in the binary expansion of k is the same of the average of ones in binary expansion of k^2; these are the values in the sequence with sqrt(2)*2^j < a(k) < 2^(j+1) for some j. - Corrected by Franklin T. Adams-Watters, Aug 23 2012

Conjecture: The counting function p(n) satisfies p(n)=c n/log n + o(n/log n).

			21 is in the sequence because 21=10101_2 (3 1's) and 441=110111001_2 (6 1's).

Robert Israel, Table of n, a(n) for n = 1..10000
Giuseppe Melfi, Su alcune successioni di interi, 2° Incontro Italiano di Teoria dei Numeri, Parma, 13-15 novembre 2003.
Giuseppe Melfi, On certain positive integer sequences, arXiv:math/0404555 [math.NT], 30 Apr 2004.

Cf. A000120, A077436, A159918.

select(t -> 2*convert(convert(t,base,2),`+`) = convert(convert(t^2,base,2),`+`), [$1..1000]); # Robert Israel, Aug 27 2015

f[n_] := Total@ IntegerDigits[n, 2]; Select[Range@ 360, 2 f@ # == f[#^2] &] (* Michael De Vlieger, Aug 27 2015 *)

isok(n) =  norml2(binary(n^2)) == 2*norml2(binary(n)) \\ Michel Marcus, Jun 20 2013

A192085 Number of ones in the binary expansion of n^3.

Original entry on oeis.org

0, 1, 1, 4, 1, 6, 4, 6, 1, 6, 6, 6, 4, 5, 6, 8, 1, 6, 6, 8, 6, 6, 6, 9, 4, 7, 5, 8, 6, 9, 8, 10, 1, 6, 6, 11, 6, 10, 8, 12, 6, 8, 6, 9, 6, 11, 9, 10, 4, 9, 7, 7, 5, 8, 8, 10, 6, 10, 9, 7, 8, 11, 10, 12, 1, 6, 6, 11, 6, 9, 11, 11, 6, 13, 10, 14, 8, 13, 12, 13

Offset: 0

Carl R. White, Jun 23 2011

The binary weight of n^3.

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

Cf. A000120, A000578, A007088, A159918.

[&+Intseq(n^3,2): n in [0..79]];  // Bruno Berselli, Jun 24 2011

A192085 := proc(n) return add(b,b=convert(n^3,base,2)): end: seq(A192085(n),n=0..100); # Nathaniel Johnston, Jun 23 2011

a(n)=hammingweight(n^3) \\ Charles R Greathouse IV, Sep 27 2016

def A192085(n):
    return bin(n**3).count('1') # Chai Wah Wu, Sep 03 2014

a(n) = A000120(A000578(n)).

a(n) = [x^(n^3)] (1/(1 - x))*Sum_{k>=0} x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Mar 27 2018

A235001 Odious squares.

Original entry on oeis.org

1, 4, 16, 25, 49, 64, 81, 100, 121, 196, 256, 289, 324, 361, 400, 484, 529, 625, 784, 841, 961, 1024, 1089, 1156, 1225, 1296, 1444, 1521, 1600, 1681, 1849, 1936, 2116, 2401, 2500, 2601, 3025, 3136, 3364, 3481, 3844, 4096, 4225, 4356, 4489, 4624, 4900, 5041, 5184, 5625

Offset: 1

Gerasimov Sergey, Jan 02 2014

nonn

This sequence is the intersection of A000069 and A000290. Numbers n^2 such that A159918 is odd.

			16 is a square 4^2 and 16 in base 2 is a 10000, having an odd number of 1's, thus 16 is in this sequence.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A231431 (evil squares).

Select[Range[200]^2,OddQ[DigitCount[#,2,1]]&] (* Harvey P. Dale, Jan 14 2014 *)

A231431 Evil squares.

Original entry on oeis.org

0, 9, 36, 144, 169, 225, 441, 576, 676, 729, 900, 1369, 1764, 2025, 2209, 2304, 2704, 2809, 2916, 3249, 3600, 3721, 3969, 4761, 5329, 5476, 6561, 6889, 7056, 8100, 8649, 8836, 9216, 9801, 10816, 11025, 11236, 11449, 11664, 11881, 12321, 12996, 13225, 14161, 14400, 14884, 15129, 15876, 17689, 18769, 19044, 19881, 21316, 21904

Offset: 1

Juri-Stepan Gerasimov, Nov 20 2013

Numbers n^2 such that A159918(n) is even.

Intersection of A000290 and A001969.

			36 is in the sequence because 36 = 6^2 and 36 in base 2 is 100100, having an even number of 1's.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000290, A125498.

Select[Range[0,150]^2,EvenQ[DigitCount[#,2,1]]&] (* Harvey P. Dale, Nov 23 2015 *)

is(n)=hammingweight(n)%2==0 && issquare(n) \\ Charles R Greathouse IV, Nov 20 2013

Conjecture: a(n) ~ 4n^2. - Charles R Greathouse IV, Nov 20 2013

Corrected and extended by Harvey P. Dale, Nov 23 2015

A214560 Number of 0's in binary expansion of n^2.

Original entry on oeis.org

1, 0, 2, 2, 4, 2, 4, 3, 6, 4, 4, 2, 6, 4, 5, 4, 8, 6, 6, 4, 6, 3, 4, 7, 8, 5, 6, 4, 7, 5, 6, 5, 10, 8, 8, 6, 8, 5, 6, 4, 8, 6, 5, 4, 6, 3, 9, 8, 10, 7, 7, 7, 8, 4, 6, 5, 9, 6, 7, 5, 8, 6, 7, 6, 12, 10, 10, 8, 10, 7, 8, 6, 10, 7, 7, 4, 8, 6, 6, 8, 10, 7, 8, 5, 7

Offset: 0

Alex Ratushnyak, Jul 21 2012

Conjecture: for every x>=0 there is an i such that a(n)>x for n>i.

Comment from N. J. A. Sloane, Nov 21 2013: See also the conjecture in A231898.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A000120, A000290, A023416, A078565, A159918, A231898.

a214560 = a023416 . a000290  -- Reinhard Zumkeller, Nov 20 2013

A214560 := proc(n)
    A023416(n^2) ;
end proc: # R. J. Mathar, Jul 21 2012
# second Maple program:
a:= n-> `if`(n=0, 1, add(1-i, i=Bits[Split](n^2))):
seq(a(n), n=0..84);  # Alois P. Heinz, Nov 25 2024

Join[{1},Table[DigitCount[n^2,2,0],{n,100}]] (* Harvey P. Dale, Nov 24 2024 *)

vector(66,n,b=binary((n-1)^2);sum(j=1,#b,1-b[j])) /* Joerg Arndt, Jul 21 2012 */

for n in range(300):
    b = n*n
    c = 0
    while b>0:
        c += 1-(b&1)
        b//=2
    print(c+(n==0), end=', ')

def A214560(n):
    return bin(n*n)[2:].count('0') # Chai Wah Wu, Sep 03 2014

a(n) = A023416(A000290(n)).

A231898 a(n) = smallest k with property that for all m >= k, there is a square N^2 whose binary expansion contains exactly n 1's and m 0's; or -1 if no such k exists.

Original entry on oeis.org

-1, -1, 2, -1, 4, 3, 4, 3, 4, 5, 5, 5, 6, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

N. J. A. Sloane, Nov 19 2013

a(n) = -1 for n = 1, 2 and 4, because all squares with exactly 1, 2 or 4 1's in their binary expansion must contain an even number of 0's.
Conjecture: Apart from n=1, 2 and 4, no other a(n) is -1.
See A214560 for a related conjecture.

			Here is a table whose columns give:
N, N^2, number of bits in N^2, number of 1's in N^2, number of 0's in N^2:
0 0 1 0 1
1 1 1 1 0
2 4 3 1 2
3 9 4 2 2
4 16 5 1 4
5 25 5 3 2
6 36 6 2 4
7 49 6 3 3
8 64 7 1 6
9 81 7 3 4
10 100 7 3 4
11 121 7 5 2
12 144 8 2 6
13 169 8 4 4
14 196 8 3 5
15 225 8 4 4
16 256 9 1 8
17 289 9 3 6
18 324 9 3 6
19 361 9 5 4
...
a(n) is defined by the property that for all m >= a(n), the table contains a row ending n m. For example, there are rows ending 3 2, 3 3, 3 4, 3 5, ..., but not 3 1, so a(3) = 2.
a(5)=4: for t>=0, (11*2^t)^2 contains 5 1's and 2t+2 0's and (25*2^t)^2 contains 5 1's and 2t+5 0's, so for m >= 4 there is a number N such that N^2 contains 5 1's and m 0's. Also 4 is the smallest number with this property, so a(5) = 4.

Cf. A000120, A023416, A159918, A214560, A230097, A231897.

Missing word in definition supplied by Jon Perry, Nov 20 2013.

A357304 Records of the Hamming weight of squares.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 8, 9, 10, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 85, 87, 88, 89

Offset: 1

Hugo Pfoertner, Oct 01 2022

			a(70) = 77 corresponds to A230097(70) = 34895284158283. Its square 1217680856487316499797508089 is the smallest and the only 90-bit square with this Hamming weight.

A230097 gives the values of k such that A000120(k^2) sets a new record.

Cf. A000120, A000290, A159918, A357658.

Missing a(68)=75 and a(71)-a(80) from Bert Dobbelaere, Nov 20 2022

A286374 a(n) = A278222(n^2).

Original entry on oeis.org

1, 2, 2, 6, 2, 12, 6, 12, 2, 30, 12, 48, 6, 210, 12, 24, 2, 30, 30, 420, 12, 360, 48, 30, 6, 120, 210, 1260, 12, 420, 24, 48, 2, 30, 30, 420, 30, 4620, 420, 480, 12, 420, 360, 1080, 48, 960, 30, 210, 6, 420, 120, 2310, 210, 3360, 1260, 1680, 12, 1260, 420, 6300, 24, 840, 48, 96, 2, 30, 30, 420, 30, 4620, 420, 2520, 30, 4620, 4620, 6720, 420, 9240, 480, 180

Offset: 0

Antti Karttunen, May 09 2017

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for sequences related to binary expansion of n

Cf. A000290, A278222, A286243, A286375, A286376, A286377.

Cf. A159918 (one of the matched sequences).

from sympy import prime, factorint
import math
def A(n): return n - 2**int(math.floor(math.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 a005940(n): return b(n - 1)
def P(n):
    f = factorint(n)
    return sorted([f[i] for i in f])
def a046523(n):
    x=1
    while True:
        if P(n) == P(x): return x
        else: x+=1
def a278222(n): return a046523(a005940(n + 1))
def a(n): return a278222(n**2) # Indranil Ghosh, May 09 2017

Scheme
```
(define (A286374 n) (A278222 (* n n)))
```

a(n) = A278222(A000290(n)) = A278222(n^2).

A352084 Integers m such that wt(m) divides wt(m^2) where wt(m) = A000120(m) is the binary weight of m.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 21, 24, 28, 30, 31, 32, 37, 42, 45, 48, 53, 56, 60, 62, 63, 64, 69, 73, 74, 79, 81, 83, 84, 90, 91, 96, 106, 112, 120, 124, 126, 127, 128, 133, 137, 138, 141, 146, 148, 155, 157, 158, 159, 161, 162, 165, 166, 168, 177, 180

Offset: 1

Bernard Schott, Mar 03 2022

Integers m such that A000120(m) divides A159918(m).
This is a problem proposed by the French site Diophante in the links section.
The first 18 terms are the same as A268415, then A268415(19) = 41 while a(19) = 42.
The corresponding quotients are in A352085.
The smallest term k such that the corresponding quotient = n is A352086(n).
Some subsequences:
-> wt(m^2) = wt(m) iff m is in A077436.
-> wt(m^2) / wt(m) = 2 iff m is in A083567.
-> When m is a power of 2 (A000079): wt(2^k) = wt((2^k)^2) = wt(2^(2k)) = 1.

			37_10 = 100101_2, digsum_2(37) = 1+1+1 = 3; then 37^2 = 1369_10 = 10101011001_2, digsum_2(1369) = 1+1+1+1+1+1 = 6; as 3 divides 6, 37 is a term.

Martin Ehrenstein, Table of n, a(n) for n = 1..10000
Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

Cf. A000120, A159918, A268415, A351650, A352085, A352086.
Cf. A351650 (similar for base 10).
Subsequences: A000079, A023758, A077436, A083567.

Select[Range[180], Divisible[Total[IntegerDigits[#^2, 2]], Total[IntegerDigits[#, 2]]] &] (* Amiram Eldar, Mar 03 2022 *)

isok(m) = !(hammingweight(m^2) % hammingweight(m)); \\ Michel Marcus, Mar 03 2022

def ok(n): return n > 0 and bin(n**2).count('1')%bin(n).count('1') == 0
print([m for m in range(1, 200) if ok(m)]) # Michael S. Branicky, Mar 03 2022

More terms from Amiram Eldar, Mar 03 2022

A352085 a(n) is the quotient wt(m^2) / wt(m), where wt = binary weight = A000120 and m = A352084(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2

Offset: 1

Bernard Schott, Mar 05 2022

All positive integers are terms of this sequence and the smallest integer k such that a(k) = n is A352086(n).

a(n) = 1 iff m = A352084(n) is a term of A077436, and a(n) = 2 iff m = A352084(n) is a term of A083567.

			For n=19, A352084(19) = 42_10 = 101010_2 => wt(42) = 3 ones; then 42^2 = 1764_10 = 11011100100_2 => wt(1764) = 6 ones, so that a(19) = wt(42^2) / wt(42) = 6/3 = 2.

Martin Ehrenstein, Table of n, a(n) for n = 1..10000
Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

Cf. A000120, A077436, A083567, A159918, A352084, A352086.

Select[Total[IntegerDigits[#^2, 2]]/Total[IntegerDigits[#, 2]] & /@ Range[300], IntegerQ] (* Amiram Eldar, Mar 05 2022 *)

lista(nn) = my(list = List(), q); for (n=1, nn, if (denominator(q=hammingweight(n^2)/hammingweight(n)) == 1, listput(list, q));); Vec(list); \\ Michel Marcus, Mar 05 2022

from itertools import count, islice
def agen(): # generator of terms
    for m in count(1):
        q, r = divmod(bin(m**2).count('1'), bin(m).count('1'))
        if r == 0:
            yield q
print(list(islice(agen(), 100))) # Michael S. Branicky, Mar 05 2022

a(n) = A159918(A352084(n))/A000120(A352084(n)).

More terms from Michel Marcus, Mar 05 2022

cp's OEIS Frontend

A083567 Let B(k) be the number of binary digits in k equal to 1. This is the sequence of positive integers k such that 2B(k)=B(k^2).

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A192085 Number of ones in the binary expansion of n^3.

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A235001 Odious squares.

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Mathematica

A231431 Evil squares.

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A214560 Number of 0's in binary expansion of n^2.

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A231898 a(n) = smallest k with property that for all m >= k, there is a square N^2 whose binary expansion contains exactly n 1's and m 0's; or -1 if no such k exists.

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A357304 Records of the Hamming weight of squares.

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A286374 a(n) = A278222(n^2).