A159918 -id:A159918 - cp's OEIS frontend

A340956 Integers n such that s(n^2) = s(n)*(s(n)+1)/2, where s(n) is the sum of binary digits of n (A000120).

0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37, 40, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 84, 128, 129, 130, 132, 133, 136, 137, 138, 144, 146, 148, 160, 161, 162, 168, 256, 257, 258, 260, 261, 264, 265, 266, 272, 273, 274, 276, 277, 288, 289, 292, 293, 296, 320, 321, 322, 324, 336, 337, 512, 513, 514, 516

Offset: 1

Max Alekseyev, Feb 01 2021

Also, integers n such that A159918(n) = A000217(A000120(n)).

For any n, s(n^2) <= s(n)*(s(n)+1)/2. This sequence gives n for which the equality is achieved.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A000120, A000217, A000290, A159918.

q:= n-> (s-> is(s(n^2)=(t->t(s(n)))(h->h*(h+1)/2)))(
         k-> add(i, i=Bits[Split](k))):
select(q, [$0..555])[];  # Alois P. Heinz, Feb 01 2021

s[n_] := DigitCount[n, 2, 1]; t[n_] := n*(n + 1)/2; Select[Range[0, 500], s[#^2] == t[s[#]] &] (* Amiram Eldar, Feb 01 2021 *)

isok(n) = my(hn=hammingweight(n)); hammingweight(n^2) == hn*(hn+1)/2; \\ Michel Marcus, Feb 01 2021

A355505 a(n) is the number of distinct cycles when iterating the function f_n(x), where f_n(x) is the sum of the digits in base n of x^2.

Original entry on oeis.org

2, 5, 3, 4, 4, 7, 4, 3, 4, 6, 4, 7, 4, 8, 6, 3, 3, 7, 5, 7, 9, 7, 4, 6, 4, 7, 5, 9, 5, 12, 7, 3, 9, 5, 8, 9, 5, 10, 9, 6, 4, 16, 8, 9, 8, 7, 5, 7, 9, 7, 7, 8, 4, 9, 8, 8, 11, 9, 4, 14, 7, 13, 11, 3, 8, 16, 7, 6, 9, 16, 8, 8, 5, 9, 9, 11, 13, 17, 7, 6, 6, 7, 5, 17, 6, 15, 11, 9, 4

Offset: 2

Wouter Zandsteeg, Jul 04 2022

The trajectory from every starting point will enter a cycle given a sufficient number of iterations of f_n.
To determine a(n), only starting points 0, 1, 2, ..., n^2 have to be checked for cycles. Larger starting points will always lead to a cycle reached from one (or more) of 0, 1, 2, ..., n^2.
From Iain Fox, Jul 09 2022: (Start)
Since f_n(x) <= (n-1)*(1 + floor(2*log_n(x))), only numbers less than the largest zero of (n-1)*(1 + floor(2*log_n(x))) - x need to be checked.
The value mentioned above is less than or equal to (2-2n)*W_{-1}(log(n)/((2-2n)*sqrt(n)))/log(n) where W_k(x) is the k-th branch of the Lambert W function. (End)

			a(8) = 4 because there are 4 cycles for n = 8:
  f_8(0) = 0 (since 0^2 = 0 = 0_8, with digit sum 0),
  f_8(1) = 1 (1^2 = 1 = 1_8, with digit sum 1),
  f_8(4) = 2 (4^2 = 20_8) and f_8(2) = 4 (2^2 = 4_8), and
  f_8(7) = 7 (7^2 = 61_8, with digit sum 7).

Iain Fox, Table of n, a(n) for n = 2..10000

Cf. A004159, A061903, A159918.

a(n) = my(d=1); while(d<=logint(((n-1)*d)^2, n)+1, d++); my(l=(n-1)*(d-1)+1, x=vector(l, i, l-i), y=[], z=[], j, c=0); for(i=1, #x, y=setunion(y, z); j=x[i]; z=[]; while(!setsearch(z, j), if(setsearch(y, j), next(2)); z=setunion(z, [j]); j=sumdigits(j^2, n)); c++); c \\ Iain Fox, Jul 09 2022

A357656 a(n) is a lower bound for the largest Hamming weight of squares with exactly n binary zeros.

Original entry on oeis.org

1, 0, 13, 8, 13, 16, 37, 38, 44

Offset: 0

Karl-Heinz Hofmann and Hugo Pfoertner, Oct 07 2022

The terms from a(2) onwards must be regarded as lower bounds, because no proof for the non-existence of squares with a very small number of binary zeros in the range k^2 > 2^90 (see b-file of A230097) is known.

a(9) >= 63, a(10) >= 57.

			                A357657(n)
   n  a(n) bits     k       k^2          k^2 in binary
   0    1    1      1         1                      1
   1    0    1      0         0                      0
   2   13   15    181     32761        111111111111001
   3    8   11     45      2025            11111101001
   4   13   17    362    131044      11111111111100100
   5   16   21   1241   1540081  101110111111111110001

A357657 are the corresponding square roots of the record-setting squares.

Cf. A000120, A000290, A159918, A230097, A356878, A357304, A357305.

A357657 a(n) is a lower bound for the square root of the maximum square with exactly n zeros in its binary representation.

Original entry on oeis.org

1, 0, 181, 45, 362, 1241, 2965685, 5931189, 57804981

Offset: 0

Karl-Heinz Hofmann and Hugo Pfoertner, Oct 08 2022

See A357656 for more information.

a(9) >= 66537313397, a(10) >= 10520476455.

A357656 gives the Hamming weight of the squared terms.

Cf. A000120, A000290, A159918, A230097, A356838, A357304, A357305, A357753, A357754.

A357742 a(n) is the maximum binary weight of the squares of n-bit numbers.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 13, 13, 15, 16, 18, 20, 22, 24, 25, 27, 29, 31, 34, 34, 37, 38, 39, 41, 44, 44, 47, 49, 51, 52, 54, 55, 57, 59, 63, 63, 64, 66, 68, 69, 72, 73, 76, 77, 78, 80, 82, 85, 87

Offset: 1

Karl-Heinz Hofmann and Hugo Pfoertner , Oct 11 2022

			   bit   |
  length |          possible binary weight of k^2
   of k  | 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
   = n   |          the rightmost value is a(n)
  -------+--------------------------------------------------------------
     1   | 0  1
     2   |    1  2  -  -
     3   |    1  2  3  -  -  -
     4   |    1  2  3  4  5  -  -  -
     5   |    1  2  3  4  5  6  -  -  -  -
     6   |    1  2  3  4  5  6  7  8  -  -  -  -
     7   |    1  2  3  4  5  6  7  8  9  -  -  -  -  -
     8   |    1  2  3  4  5  6  7  8  9 10 11  - 13  -  -  -
     9   |    1  2  3  4  5  6  7  8  9 10 11 12 13  -  -  -  -  -
    10   |    1  2  3  4  5  6  7  8  9 10 11 12 13 14 15  -  -  -  -  -

Cf. A000290, A159918, A357304, A357658.

a(n) = max(A357658(2*n-2), A357658(2*n-1)).

a(47)-a(50) from Martin Ehrenstein, Dec 26 2023

A358701 a(n) is the least number > 1 that needs n toggles in the trailing bits of its binary representation to become a square.

Original entry on oeis.org

4, 5, 7, 14, 79, 831, 6495, 247614, 7361278, 743300286, 121387475838

Offset: 0

Hugo Pfoertner, Dec 16 2022

			a(0) = 4 = 100 in binary, 0 toggled bits needed;
a(1) = 5 = 101_2, 1 toggled bit -> 100_2 = 4;
a(2) = 7 = 111_2, 2 toggled bits -> 100_2 = 4;
a(3) = 14 = 1110_2, 3 toggled bits -> 1001_2 = 9;
a(4) = 79 = 1001111_2, 4 toggled bits -> 1000000_2 = 64;
a(5) = 831 = 1100111111_2, 5 toggled bits -> 1100010000_2 = 784 = 28^2.

Michael S. Branicky, Go program.
Code Golf StackExchange, Toggle some bits and get a square, coding challenge started by user Arnauld, Aug 18 2018.
Delbert L. Johnson, C# program

Cf. A000120, A000290, A001737, A159918.

// see linked program
(C#) // see linked program

a(9) from Michael S. Branicky, Dec 19 2022

a(10) from Delbert L. Johnson, Apr 13 2024

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

A340956 Integers n such that s(n^2) = s(n)*(s(n)+1)/2, where s(n) is the sum of binary digits of n (A000120).

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A355505 a(n) is the number of distinct cycles when iterating the function f_n(x), where f_n(x) is the sum of the digits in base n of x^2.

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A357656 a(n) is a lower bound for the largest Hamming weight of squares with exactly n binary zeros.

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A357657 a(n) is a lower bound for the square root of the maximum square with exactly n zeros in its binary representation.

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A357742 a(n) is the maximum binary weight of the squares of n-bit numbers.

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A358701 a(n) is the least number > 1 that needs n toggles in the trailing bits of its binary representation to become a square.

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

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