A077436 -id:A077436 - cp's OEIS frontend

A114391 Positions where A114390 is a square.

1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 25, 28, 30, 31, 32, 48, 50, 56, 60, 62, 63, 64, 96, 100, 112, 113, 120, 124, 126, 127, 128, 189, 192, 200, 224, 226, 240, 248, 252, 254, 255, 256, 378, 381, 384, 400, 448, 452, 480, 481, 496, 504, 508, 510, 511, 512, 756

Offset: 1

Antti Karttunen, Feb 07 2006

nonn

A114392 gives the corresponding values. Differs from A077436 (and A023758) for the first time at n=13, where a(13)=25, while A077436(13)=28. First differences: A114394.

A118655 Numbers k such that Hamming weight of k equals Hamming weight of k^3.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 56953, 65536, 113906, 131072, 227812, 262144, 455624, 524288, 911248, 1048576, 1822496, 1830651, 2097152, 3644992, 3661302, 4194304, 5502457, 7289984, 7322604

Offset: 1

R. J. Mathar, May 18 2006

A000079(n) is a subsequence because the Hamming weight of any A000079(n) is 1 and cubing such a number just adds binary zeros. A077436 is the sequence when the a(n) is squared, not cubed as here.

Donovan Johnson, Table of n, a(n) for n = 1..177 (terms <= 2^40)

Cf. A077436, A000079, A000120.

Select[Range[10^6], DigitCount[#, 2, 1] == DigitCount[#^3, 2, 1] &] (* Amiram Eldar, Jul 16 2023 *)

bitcnt(n)= { local(bitv,bitl) ; bitv=binary(n) ; bitvl=matsize(bitv) ; return(sum(i=1,bitvl[2],bitv[i])) ; }
{ for(i=1,8000000, if(bitcnt(i)==bitcnt(i^3), print1(i,",") ; ) ; ) ; }

A000120(a(n)) = A000120(a(n)^3).

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

A049343 Numbers m such that 2m and m^2 have same digit sum.

Original entry on oeis.org

0, 2, 9, 11, 18, 20, 29, 38, 45, 47, 90, 99, 101, 110, 119, 144, 146, 180, 182, 189, 198, 200, 245, 290, 299, 335, 344, 351, 362, 369, 380, 398, 450, 452, 459, 461, 468, 470, 479, 488, 495, 497, 639, 729, 794, 839, 848, 900, 929, 954, 990, 999

Offset: 1

R. K. Guy

An easy way to prove that this sequence is infinite is to observe that it contains all numbers of the form 10^k+1. - Stefan Steinerberger, Mar 31 2006

For n>0: digital root (A010888) of 2n or n^2 is either 4 or 9. - Reinhard Zumkeller, Oct 01 2007

Problem 117 in Loren Larson's translation of Paul Vaderlind's book.

Reinhard Zumkeller and Harvey P. Dale, Table of n, a(n) for n = 1..1000 (first 101 terms from Zumkeller)

Cf. A058369, A077436 (binary). - Reinhard Zumkeller, Apr 03 2011

import Data.List (elemIndices)
import Data.Function (on)
a049343 n = a049343_list !! (n-1)
a049343_list = elemIndices 0
   $ zipWith ((-) `on` a007953) a005843_list a000290_list
-- Reinhard Zumkeller, Apr 03 2011

[n: n in [0..1000] | &+Intseq(2*n) eq &+Intseq(n^2)]; // Vincenzo Librandi, Nov 17 2015

Select[Range[0, 1000], Sum[DigitCount[2# ][[i]]*i, {i, 1, 9}] == Sum[DigitCount[ #^2][[i]]*i, {i, 1, 9}] &] (* Stefan Steinerberger, Mar 31 2006 *)
Select[Range[0,1000],Total[IntegerDigits[2#]]==Total[ IntegerDigits[ #^2]]&] (* Harvey P. Dale, Sep 25 2012 *)

A007953(A005843(a(n))) = A007953(A000290(a(n))). - Reinhard Zumkeller, Oct 01 2007

A212314 Numbers m such that B(m^3) = 3*B(m), where B(m) is the binary weight of m (A000120).

Original entry on oeis.org

0, 5, 9, 10, 17, 18, 20, 33, 34, 36, 39, 40, 49, 65, 66, 68, 69, 72, 78, 80, 98, 105, 129, 130, 132, 135, 136, 138, 144, 156, 160, 169, 196, 199, 209, 210, 229, 257, 258, 260, 263, 264, 270, 272, 276, 277, 288, 291, 297, 312, 313, 320, 338, 359, 365, 392, 395, 398, 418

Offset: 1

Alex Ratushnyak, Oct 24 2013

2*k is a term if and only if k is a term. - Robert Israel, Nov 06 2022

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000120, A077436, A083567, A118655, A212315.

Includes A136318.

select(n -> convert(convert(n^3,base,2),`+`) = 3*convert(convert(n,base,2),`+`), [$0..1000]); # Robert Israel, Nov 06 2022

isok(m) = hammingweight(m^3) == 3*hammingweight(m); \\ Michel Marcus, Nov 06 2022

import math
for n in range(10000):
    c1 = c2 = 0
    t = n
    while t:
        c1+=t&1
        t>>=1
    t = n*n*n
    while t:
        c2+=t&1
        t>>=1
    if c1*3==c2: print(str(n), end=',')

s = lambda n: sum((n^3).digits(2)) - 3*sum(n.digits(2))
[n for n in (0..418) if s(n)==0]  # Peter Luschny, Oct 24 2013

A000120(a(n)^3) = A000120(a(n)) * 3.

A261586 Odd numbers n such that the sum of the binary digits of n equals the sum of the binary digits of n^2.

Original entry on oeis.org

1, 3, 7, 15, 31, 63, 79, 91, 127, 157, 159, 183, 187, 255, 279, 287, 317, 319, 351, 365, 375, 379, 445, 511, 573, 575, 637, 639, 703, 735, 751, 759, 763, 815, 893, 975, 1023, 1071, 1087, 1145, 1149, 1151, 1215, 1255, 1277, 1279, 1407, 1449, 1455, 1463

Offset: 1

Charles R Greathouse IV, Aug 25 2015

A077436 consists of elements of this sequence times powers of 2.
Hare, Laishram, & Stoll show that this sequence is infinite. 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.
Subsequence of A077436.

			15 = 1111_2 and 15^2 = 11100001_2, both of which have a Hamming weight (sum of binary digits) equal to 4.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
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.

Cf. A077436, A254066.

[n: n in [1..1500 by 2] | &+Intseq(n, 2) eq &+Intseq(n^2, 2) ]; // Vincenzo Librandi, Aug 30 2015

Select[Range[1, 1463, 2], Total@ IntegerDigits[#, 2] == Total@ IntegerDigits[#^2, 2] &] (* Michael De Vlieger, Aug 29 2015 *)

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

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

A094695 Numbers having in binary representation fewer ones than in their squares.

Original entry on oeis.org

5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 25, 26, 27, 29, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 93, 97, 98, 99

Offset: 1

Reinhard Zumkeller, May 20 2004

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

Cf. A000120, A001737, A007088, A077436, A094694, A159918.

Select[Range[100],DigitCount[#,2,1]Harvey P. Dale, May 29 2017 *)

A000120(a(n)) < A000120(A000290(a(n))).

A212315 Numbers m such that B(m) = B(triangular(m)), where B(m) is the binary weight of m (A000120).

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 39, 45, 63, 78, 79, 91, 93, 127, 139, 143, 158, 159, 175, 182, 187, 189, 222, 255, 286, 287, 318, 319, 351, 367, 375, 379, 381, 407, 446, 487, 511, 535, 543, 572, 574, 575, 607, 627, 638, 639, 703, 724, 727, 731, 747, 750, 759, 763, 765, 799, 823, 830

Offset: 1

Alex Ratushnyak, Oct 24 2013

Dumitru Damian, Table of n, a(n) for n = 1..10000

Cf. A000120, A077436, A083567, A118655, A212314.

def isa(n): return n.bit_count()==((n*(n+1))>>1).bit_count()
print(list(n for n in range(10**3) if isa(n)))  # Dumitru Damian, Mar 04 2023

A000120(a(n)) = A000120(a(n)*(a(n)+1)/2).

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).

cp's OEIS Frontend

A114391 Positions where A114390 is a square.

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A118655 Numbers k such that Hamming weight of k equals Hamming weight of k^3.

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

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A049343 Numbers m such that 2m and m^2 have same digit sum.

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A212314 Numbers m such that B(m^3) = 3*B(m), where B(m) is the binary weight of m (A000120).

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A261586 Odd numbers n such that the sum of the binary digits of n equals the sum of the binary digits of n^2.

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

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