A261586 -id:A261586 - cp's OEIS frontend

A077436 Let B(n) be the sum of binary digits of n. This sequence contains n such that B(n) = B(n^2).

0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 28, 30, 31, 32, 48, 56, 60, 62, 63, 64, 79, 91, 96, 112, 120, 124, 126, 127, 128, 157, 158, 159, 182, 183, 187, 192, 224, 240, 248, 252, 254, 255, 256, 279, 287, 314, 316, 317, 318, 319, 351, 364, 365, 366, 374, 375, 379, 384

Offset: 1

Giuseppe Melfi, Nov 30 2002

Superset of A023758.

Hare, Laishram, & Stoll show that this sequence contains infinitely many odd numbers. In particular for each k in {12, 13, 16, 17, 18, 19, 20, ...} there are infinitely many terms in this sequence with binary digit sum k. - Charles R Greathouse IV, Aug 25 2015

			The element 79 belongs to the sequence because 79=(1001111) and 79^2=(1100001100001), so B(79)=B(79^2)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10476, all terms <= 2^20
Karam Aloui, Damien Jamet, Hajime Kaneko, Steffen Kopecki, Pierre Popoli, and Thomas Stoll, On the binary digits of n and n^2, arXiv:2203.05451 [math.NT], 2022.
K. G. Hare, S. Laishram, and T. Stoll, The sum of digits of n and n^2, International Journal of Number Theory 7:7 (2011), pp. 1737-1752.
Giuseppe Melfi, On simultaneous binary expansions of n and n^2, arXiv:math/0402458 [math.NT], 2004.
Giuseppe Melfi, Su alcune successioni di interi (English with an Italian title)

Cf. A058369, A000120, A000290, A083567.
Cf. A211676 (number of n-bit numbers in this sequence).
A261586 is a subsequence. Subsequence of A352084.

import Data.List (elemIndices)
import Data.Function (on)
a077436 n = a077436_list !! (n-1)
a077436_list = elemIndices 0
   $ zipWith ((-) `on` a000120) [0..] a000290_list
-- Reinhard Zumkeller, Apr 12 2011

[n: n in [0..400] | &+Intseq(n, 2) eq &+Intseq(n^2, 2)]; // Vincenzo Librandi, Aug 30 2015

select(t -> convert(convert(t,base,2),`+`) = convert(convert(t^2,base,2),`+`), [$0..1000]); # Robert Israel, Aug 27 2015

t={}; Do[If[DigitCount[n, 2, 1] == DigitCount[n^2, 2, 1], AppendTo[t, n]], {n, 0, 364}]; t
f[n_] := Total@ IntegerDigits[n, 2]; Select[Range[0, 384], f@ # == f[#^2] &] (* Michael De Vlieger, Aug 27 2015 *)

is(n)=hammingweight(n)==hammingweight(n^2) \\ Charles R Greathouse IV, Aug 25 2015

def ok(n): return bin(n).count('1') == bin(n**2).count('1')
print([m for m in range(400) if ok(m)]) # Michael S. Branicky, Mar 11 2022

A159918(a(n)) = A000120(a(n)). - Reinhard Zumkeller, Apr 25 2009

Initial 0 added by Reinhard Zumkeller, Apr 28 2012, Apr 12 2011

A261593 Odd numbers n such that the sum of the binary digits of n and n^2 both equal 12.

Original entry on oeis.org

4095, 10239, 11263, 12159, 12223, 12255, 12271, 12279, 12283, 14333, 15351, 15355, 15743, 15807, 18431, 19455, 19967, 20351, 20477, 22015, 22495, 22511, 24031, 24303, 24431, 24445, 25599, 26615, 26621, 27519, 27631, 27639, 28095, 28411, 28413, 28511, 28541, 28575

Offset: 1

Charles R Greathouse IV, Aug 25 2015

Hare, Laishram, & Stoll show that this sequence is infinite.
12 is the least constant for which the associated sequence is known to be infinite. 1 through 8 make finite sequences, the sequences with 9 and 10 are conjectured finite, and the sequence with 11 is unknown. (See Hare-Laishram-Stoll Theorem 1.3 and 1.4 plus section 5.)
Odd numbers n such that the sum of the binary digits of n-1 and n^2-1 both equal 11. - Chai Wah Wu, Aug 26 2015

			4095 = 111111111111_2 and 4095^2 = 111111111110000000000001_2 both have 12 1s in binary.

Charles R Greathouse IV, Table of n, a(n) for n = 1..1329
K. G. Hare, S. Laishram, and T. Stoll, The sum of digits of n and n^2, International Journal of Number Theory 7:7 (2011), pp. 1737-1752.

Subsequence of A261586 and hence of A077436.

Select[2 Range@ 15000 - 1, Total@ IntegerDigits[#, 2] == 12 && Total@ IntegerDigits[#^2, 2] == 12 &] (* Michael De Vlieger, Aug 27 2015 *)

is(n)=n%2 && hammingweight(n)==12 && hammingweight(n^2)==12

\\ List the elements below 2^(N+1).
go(N)=my(v=List(),n); forvec(u=vector(11,i,[1,N-11+i]), n=sum(i=1,11,2^u[i])+1; if(hammingweight(n^2)==12, listput(v,n)), 2); Set(v)

from itertools import combinations
A261593_list = []
for c in combinations((2**x for x in range(15)),11):
    n = sum(c)
    if sum(int(d) for d in format(n*(n+1),'b')) == 11:
        A261593_list.append(2*n+1)
A261593_list = sorted(A261593_list) # Chai Wah Wu, Aug 26 2015

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A077436 Let B(n) be the sum of binary digits of n. This sequence contains n such that B(n) = B(n^2).

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A261593 Odd numbers n such that the sum of the binary digits of n and n^2 both equal 12.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python