A231898 -id:A231898 - cp's OEIS frontend

A159918 Number of ones in binary representation of n^2.

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

Offset: 0

Reinhard Zumkeller, Apr 25 2009

The binary weight (A000120) of n^2.

a(n) = 0 iff n = 0. a(n) = 1 iff n = 2^k for some k >= 0. a(n) = 2 iff n = 3*2^k for some k >= 0. Szalay proves that a(n) = 3 iff n = 7*2^k, 23*2^k, or 2^a + 2^b for k >= 0 and a > b >= 0. It seems that a(n) = 4 iff n = 13*2^k, 15*2^k, 47*2^k, or 111*2^k but this has not been proven! Any other n with a(n) = 4 are greater than 10^50, and there are finitely many odd solutions. - Charles R Greathouse IV, Jan 20 2022

L. Szalay, The equations 2^n ± 2^m ± 2^l = z^2, Indagationes Mathematicae (N.S.) 13, no. 1 (2002), pp. 131-142.

Nathaniel Johnston, Table of n, a(n) for n = 0..10000
Bernt Lindström, On the binary digits of a power, Journal of Number Theory, Volume 65, Issue 2, August 1997, Pages 321-324.
Nick MacKinnon, Problems and Solutions #12140, The American Mathematical Monthly, 126:9 (2019), 850.
K. B. Stolarsky, The binary digits of a power, Proc. Amer. Math. Soc. 71 (1978), 1-5.
Index entries for sequences related to binary expansion of n

Cf. A000120, A007088, A192085, A004159, A214560, A231897, A231898. For records see A230097.

a159918 = a000120 . a000290  -- Reinhard Zumkeller, Oct 12 2013

A159918 := proc(n) return add(b, b=convert(n^2, base, 2)): end: seq(A159918(n), n=0..100); # Nathaniel Johnston, Jun 23 2011

a[n_] := Total[IntegerDigits[n^2, 2]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Oct 27 2021 *)

a(n)=hammingweight(n^2) \\ Charles R Greathouse IV, Aug 06 2015

def A159918(n):
    return bin(n*n).count('1') # Chai Wah Wu, Sep 03 2014

a(n) = A000120(A000290(n)); a(A077436(n)) = A000120(A077436(n)).
Lindström shows that lim sup wt(m^2)/log_2 m = 2. - N. J. A. Sloane, Oct 11 2013
a(n) = [x^(n^2)] (1/(1 - x))*Sum_{k>=0} x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Mar 27 2018

A231897 a(n) = smallest m such that wt(m^2) = n (where wt(i) = A000120(i)), or -1 if no such m exists.

Original entry on oeis.org

0, 1, 3, 5, 13, 11, 21, 39, 45, 75, 155, 217, 331, 181, 627, 923, 1241, 2505, 3915, 5221, 6475, 11309, 15595, 19637, 31595, 44491, 69451, 113447, 185269, 244661, 357081, 453677, 1015143, 908091, 980853, 2960011, 4568757, 2965685, 5931189, 11862197, 20437147

Offset: 0

N. J. A. Sloane, Nov 19 2013

Conjecture: a(n) is never -1. (It seems likely that the arguments of Lindström (1997) could be modified to establish this conjecture.)

a(n) is the smallest m such that A159918(m) = n (or -1 if ...).

Hugo Pfoertner, Table of n, a(n) for n = 0..110 (terms 0..70 from Donovan Johnson, significant extension enabled by programs provided in Code Golf challenge).
Code Golf Stackexchange, Smallest and largest 100-bit square with maximum Hamming weight, fastest code challenge started Dec 15 2022.
Bernt Lindström, On the binary digits of a power, Journal of Number Theory, Volume 65, Issue 2, August 1997, Pages 321-324.

Cf. A000120, A159918, A230097, A211201, A231898, A214560.

A089998 are the corresponding squares.

a231897 n = head [x | x <- [1..], a159918 x == n]
-- Reinhard Zumkeller, Nov 20 2013

a(n)=if(n,my(k); while(hammingweight(k++^2)!=n,); k, 0) \\ Charles R Greathouse IV, Aug 06 2015

def wt(n): return bin(n).count('1')
def a(n):
    m = 2**(n//2) - 1
    while wt(m**2) != n: m += 1
    return m
print([a(n) for n in range(32)]) # Michael S. Branicky, Feb 06 2022

a(n) = 2*A211201(n-1) + 1 for n >= 1. - Hugo Pfoertner, Feb 06 2022

a(26)-a(40) from Reinhard Zumkeller, Nov 20 2013

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)).

A232243 a(n) = wt(n^2) - wt(n), where wt(n) = A000120(n) is the binary weight function.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 2, 0, 1, 0, 0, 0, 1, 1, 2, 1, 3, 2, -1, 0, 2, 1, 2, 0, 1, 0, 0, 0, 1, 1, 2, 1, 3, 2, 3, 1, 2, 3, 3, 2, 4, -1, -1, 0, 2, 2, 1, 1, 4, 2, 2, 0, 2, 1, 2, 0, 1, 0, 0, 0, 1, 1, 2, 1, 3, 2, 3, 1, 3, 3, 5, 2, 3, 3, 0, 1, 3, 2, 4, 3, 3, 3, 2, 2, 5, 4, 0, -1, 1, -1, -1, 0, 2, 2, 2

Offset: 0

Jon Perry, Nov 20 2013

A077436 lists n for which a(n) = 0.

A094694 lists n for which a(n) < 0.

			a(5): 5 = 101_2, 25 = 11001_2, so a(5) = 3 - 2 = 1.
a(23): 23 = 10111_2, 529 = 10001001_2, so a(23) = 3 - 4 = -1.

Dumitru Damian, Table of n, a(n) for n = 0..10000

Cf. A000120, A159918, A077436, A094694, A231898.

function bitCount(n) {
   var i,c,s;
   c=0;
   s=n.toString(2);
   for (i=0;i

a(n) = hammingweight(n^2) - hammingweight(n); \\ Michel Marcus, Mar 05 2023

def A232243(n): return (n**2).bit_count()-n.bit_count()
print(list(A232243(n) for n in range(10**2)))  # Dumitru Damian, Mar 04 2023

a(n) = A159918(n) - A000120(n).

A232245 Sum of the number of ones in binary representation of n and n^2.

Original entry on oeis.org

0, 2, 2, 4, 2, 5, 4, 6, 2, 5, 5, 8, 4, 7, 6, 8, 2, 5, 5, 8, 5, 9, 8, 7, 4, 8, 7, 10, 6, 9, 8, 10, 2, 5, 5, 8, 5, 9, 8, 11, 5, 8, 9, 11, 8, 12, 7, 9, 4, 8, 8, 9, 7, 12, 10, 12, 6, 10, 9, 12, 8, 11, 10, 12, 2, 5, 5, 8, 5, 9, 8, 11, 5, 9, 9, 13, 8, 11, 11, 10, 5, 9

Offset: 0

Jon Perry, Nov 20 2013

The sequence is never 1 or 3, but seems to take on all other values. The fact it is never 3 can be used to prove if n^2 has exactly 4 1's then it must have an even number of 0's (A231898).

			5 is 101 and 25 is 11001, so a(5) = 2 + 3 = 5.

Cf. A000120, A159918, A077436, A094694, A231898, A232243.

function bitCount(n) {
var i,c,s;
c=0;
s=n.toString(2);
for (i=0;i

a(n) = A159918(n) + A000120(n).

cp's OEIS Frontend

A159918 Number of ones in binary representation of n^2.

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A231897 a(n) = smallest m such that wt(m^2) = n (where wt(i) = A000120(i)), or -1 if no such m exists.

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

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A232243 a(n) = wt(n^2) - wt(n), where wt(n) = A000120(n) is the binary weight function.

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A232245 Sum of the number of ones in binary representation of n and n^2.

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