A118738 -id:A118738 - cp's OEIS frontend

A011754 Number of ones in the binary expansion of 3^n.

1, 2, 2, 4, 3, 6, 6, 5, 6, 8, 9, 13, 10, 11, 14, 15, 11, 14, 14, 17, 17, 20, 19, 22, 16, 18, 24, 30, 25, 25, 25, 26, 26, 34, 29, 32, 27, 34, 36, 32, 28, 39, 38, 39, 34, 34, 45, 38, 41, 33, 41, 46, 42, 35, 39, 42, 39, 40, 42, 48, 56, 56, 49, 57, 56, 51, 45, 47, 55, 55, 64, 68, 58

Offset: 0

Allan C. Wechsler, Dec 11 1999

Conjecture: a(n)/n tends to log(3)/(2*log(2)) = 0.792481250... (A094148). - Ed Pegg Jr, Dec 05 2002
Senge & Straus prove that for every m, there is some N such that for all n > N, a(n) > m. Dimitrov & Howe make this effective, proving that for n > 25, a(n) > 22. - Charles R Greathouse IV, Aug 23 2021
Ed Pegg's conjecture means that about half of the bits of 3^n are nonzero. It appears that the same is true for 5^n (A000351, cf. A118738) and 7^n (A000420). - M. F. Hasler, Apr 17 2024

S. Wolfram, "A new kind of science", p. 903.

Hugo Pfoertner, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
Vassil S. Dimitrov and Everett W. Howe, Powers of 3 with few nonzero bits and a conjecture of Erdős, arXiv:2105.06440 [math.NT], 2021.
Taylor Dupuy and David E. Weirich, Bits of 3^n in binary, Wieferich primes and a conjecture of Erdős, Journal of Number Theory, Volume 158, January 2016, Pages 268-280.
Hugo Pfoertner, Plot of a(n) - 0.79248*n, +-Pi*sqrt(n), n up to 10^6.
H. G. Senge and E. G. Straus, PV-numbers and sets of multiplicity, Periodica Mathematica Hungarica 3 (1973), pp. 93-100.

Cf. A007088, A000120 (Hamming weight), A000244 (3^n), A004656, A261009, A094148.

Cf. A118738 (same for 5^n).

a011754 = a000120 . a000244  -- Reinhard Zumkeller, Aug 14 2015

[&+Intseq(3^n, 2): n in [0..79]]; // Vincenzo Librandi, Nov 28 2018

f:= n -> convert(convert(3^n,base,2),`+`):
map(f, [$0..100]); # Robert Israel, Apr 17 2024

Table[DigitCount[3^n, 2][[1]], {n, 0, 100}] (* Stefan Steinerberger, Apr 03 2006 *)
DigitCount[3^Range[0,100],2,1] (* Harvey P. Dale, Apr 06 2012 *)

a(n)=hammingweight(3^n) \\ Charles R Greathouse IV, Feb 09 2017

A011754 = lambda n: (3**n).bit_count() # M. F. Hasler, Apr 17 2024

a(n) = A000120(3^n). - Benoit Cloitre, Dec 06 2002

a(n) = A000120(A000244(n)). - Reinhard Zumkeller, Aug 14 2015

More terms from Stefan Steinerberger, Apr 03 2006

A061785 a(n) = m such that 2^m < 5^n < 2^(m+1).

Original entry on oeis.org

2, 4, 6, 9, 11, 13, 16, 18, 20, 23, 25, 27, 30, 32, 34, 37, 39, 41, 44, 46, 48, 51, 53, 55, 58, 60, 62, 65, 67, 69, 71, 74, 76, 78, 81, 83, 85, 88, 90, 92, 95, 97, 99, 102, 104, 106, 109, 111, 113, 116, 118, 120, 123, 125, 127, 130, 132, 134, 136, 139, 141, 143, 146, 148

Offset: 1

Lekraj Beedassy, May 09 2003

The Beatty sequence for log_2(5) (A020858). The asymptotic density of this sequence is log_5(2) (A152675). - Amiram Eldar, Apr 09 2021

One less than the length of 5^n written in binary. Could and should be extended to a(0) = 0 (with definition corrected to "2^m <= ..."). - M. F. Hasler, Apr 17 2024

			a(2) = 4 since 2^4 < 5^2 < 2^(4+1).

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

Cf. A020858, A152675.

Cf. A118738 (Hamming weight of 5^n).

Table[Floor[n*Log2[5]], {n, 100}] (* Amiram Eldar, Apr 09 2021 *)

a(n) = floor(n*log(5)/log(2)) \\ Michel Marcus, Jul 27 2013

def A061785(n): return (5**n).bit_length()-1 # Chai Wah Wu, Jul 22 2025

a(n) = floor(n*log_2(5)). - M. F. Hasler, Apr 17 2024

Corrected and extended by John W. Layman, May 09 2003

A118737 Number of zeros in binary expansion of 5^n.

Original entry on oeis.org

0, 1, 2, 1, 5, 6, 7, 9, 7, 8, 13, 11, 15, 17, 16, 15, 18, 20, 18, 26, 21, 20, 27, 27, 26, 40, 30, 30, 37, 32, 33, 39, 36, 43, 37, 42, 40, 44, 51, 45, 40, 42, 49, 48, 51, 52, 57, 61, 58, 54, 59, 59, 67, 60, 68, 54, 70, 66, 61, 72, 79, 65, 70, 66, 63, 73, 67, 71, 76, 72, 80, 86, 78

Offset: 0

Zak Seidov, May 22 2006

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000351, A023416, A061785, A118738.

seq(numboccur(0,convert(5^n,base,2)),n=0..100); # Robert Israel, Dec 24 2017

a[n_] := DigitCount[5^n, 2, 0]; Table[a[n], {n, 0, 72}] (* Ray Chandler, Sep 29 2006 *)
DigitCount[5^Range[0,80],2,0] (* Harvey P. Dale, Aug 30 2021 *)

a(n) = #binary(5^n)-hammingweight(5^n) \\ Felix Fröhlich, Dec 24 2017

a(n) + A118738(n) = A061785(n)+1 for n >= 1. - Robert Israel, Dec 24 2017

a(n) = A023416(A000351(n)). - Felix Fröhlich, Dec 24 2017

A218451 a(n) = 10^n minus its binary weight.

Original entry on oeis.org

0, 8, 97, 994, 9995, 99994, 999993, 9999992, 99999988, 999999987, 9999999989, 99999999985, 999999999987, 9999999999986, 99999999999983, 999999999999980, 9999999999999980, 99999999999999980, 999999999999999976, 9999999999999999981, 99999999999999999974

Offset: 0

Alonso del Arte, Oct 28 2012, based on examples for A011371 from Hieronymus Fischer, Jun 06 2012

Also exponent of highest power of 2 which divides (10^n)!. - Robert G. Wilson v, Jan 10 2025

			a(2) = 97 because 10^2 = 100, which is 1100100 in binary, having 3 on bits, and 100 - 3 = 97.
a(3) = 994 because 10^3 = 1000, which is 1111101000 in binary, having 6 on bits, and 1000 - 6 = 994.

Amiram Eldar, Table of n, a(n) for n = 0..1000

Cf. A000120, A011371, A118738.

Table[10^n - DigitCount[10^n, 2, 1], {n, 0, 19}]

a(n)=10^n-hammingweight(5^n) \\ Charles R Greathouse IV, Oct 28 2012

a(n) = A011371(10^n) = 10^n - A000120(10^n).
a(n) = 10^n - A118738(n).
a(n) = A007814(A000142(A011557(n))). - Robert G. Wilson v, Jan 10 2025

More terms from Amiram Eldar, Jul 16 2023

A369857 Number of nonzero bits (a.k.a. binary or Hamming weight) of 7^n.

Original entry on oeis.org

1, 3, 3, 6, 5, 7, 9, 11, 14, 17, 15, 20, 11, 20, 23, 20, 22, 24, 20, 31, 30, 24, 24, 36, 29, 29, 37, 33, 37, 42, 43, 43, 40, 50, 53, 44, 39, 53, 43, 60, 57, 53, 60, 59, 62, 68, 65, 66, 59, 68, 75, 71, 84, 77, 65, 74, 81, 87, 85, 83, 77, 89, 83, 95, 84, 89, 86, 98, 102, 94, 97, 104

Offset: 0

M. F. Hasler, Apr 17 2024

nonn

Conjecture: a(n)/n -> log_4(7) = 1.403677461..., i.e., about half of the bits of 7^n are nonzero.

			The first few powers of 7 and their binary representation are as follows:
   n  | 0 |  1  |    3   |     4     |       5      |        6        | ...
  ----+---+-----+--------+-----------+--------------+-----------------+----
  7^n | 1 |  7  |   49   |    343    |     2401     |      16807      | ...
  ----+---+-----+--------+-----------+--------------+-----------------+----
  bin | 1 | 111 | 110001 | 101010111 | 100101100001 | 100000110100111 | ...
  ----+---+-----+--------+-----------+--------------+-----------------+----
  a(n)| 1 |  3  |    3   |     6     |       5      |        7        | ...
  ----+---+-----+--------+-----------+--------------+-----------------+----

Hugo Pfoertner, Plot of a(n) - 1.40368*n, +-4*sqrt(n), n up to 10^6.

Cf. A000120 (Hamming weight), A000420 (7^n).

Cf. A011754, A118738 (analog for 3^n and 5^n).

Maple
```
A369857 := A000120 @ A000420
```

apply( {A369857(n) = hammingweight(7^n)}, [0..99])

A369857 = lambda n: (7**n).bit_count() # for Python >= 3.10

a(n) = A000120(A000420(n)). (Definition of this sequence.)

A371470 Triangle read by rows: for 1 <= k <= n, T(n,k) is the least sum of decimal digits of numbers with n binary digits and binary weight k.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 1, 2, 6, 7, 2, 3, 3, 4, 5, 4, 5, 6, 7, 9, 10, 8, 1, 2, 2, 3, 10, 11, 4, 2, 3, 4, 5, 7, 12, 13, 5, 4, 3, 4, 5, 6, 6, 7, 8, 7, 8, 6, 7, 1, 2, 3, 4, 6, 7, 5, 4, 2, 3, 2, 3, 3, 4, 6, 13, 14, 6, 5, 3, 4, 4, 3, 4, 6, 7, 8, 18, 19, 5, 6, 6, 5, 6, 6, 6, 7, 8, 9, 14, 19, 20, 7, 8, 5

Offset: 1

Robert Israel, May 31 2024

			T(5,3) = 3 because the numbers with 5 binary digits of which 3 are 1 are 19, 21, 22, 25, 26 and 28, and the least sum of decimal digits of these is 3 (for 21).
Triangle starts:
  1;
  2, 3;
  4, 5, 7;
  8, 1, 2, 6;
  7, 2, 3, 3, 4;
  5, 4, 5, 6, 7, 9;

Robert Israel, Table of n, a(n) for n = 1..325 (rows 1 to 25, flattened)

Cf. A000120, A007953, A118738, A123384, A373289.

M:= proc(n,k) local i,R,m,r,v;
   R:= ListTools:-Reverse(map(t -> 2^(n-1)+add(2^(n-1-t[i]),i=1..k-1), combinat:-choose(n-1,k-1 )));
   m:= infinity;
   for r in R do
     v:= convert(convert(r,base,10),`+`);
     if v < m then m:= v fi;
   od;
   m
end proc:
for n from 1 to 12 do
  seq(M(n,k),k=1..n)
od;

a(n) = A007953(A373289(n)).

T(A123384(i),A118738(i)) = 1.

A373289 Triangle read by rows: for 1 <= k <= n, T(n,k) is a number with n binary digits and binary weight k whose sum of decimal digits is least; in case of a tie, choose the least such number.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 10, 11, 15, 16, 20, 21, 30, 31, 32, 40, 41, 51, 61, 63, 64, 80, 100, 101, 110, 111, 127, 128, 130, 200, 201, 211, 221, 223, 255, 256, 320, 400, 300, 301, 311, 501, 510, 511, 512, 520, 521, 600, 601, 1000, 1001, 1011, 1021, 1023, 1024, 1040, 1030, 1100, 1101, 2000, 2001, 2010

Offset: 1

Robert Israel, May 30 2024

			T(5,3) = 21 because 21 = 10101_2 has 5 binary digits of which 3 are 1, and it has a lower sum of decimal digits, 3, than the other numbers (19, 22, 25, 26 and 28) with 5 binary digits of which 3 are 1.
Triangle starts
   1;
   2,  3;
   4,  5,  7;
   8, 10, 11, 15;
  16, 20, 21, 30, 31;
  32, 40, 41, 51, 61, 63;

Robert Israel, Table of n, a(n) for n = 1..325 (rows 1 to 25, flattened)

Cf. A000120, A007953, A118738, A123384, A371470.

M:= proc(n,k) local i,R,m,b,r,v;
   R:= ListTools:-Reverse(map(t -> 2^(n-1)+add(2^(n-1-t[i]),i=1..k-1), combinat:-choose(n-1,k-1 )));
   m:= infinity;
   for r in R do
     v:= convert(convert(r,base,10),`+`);
     if v < m then b:= r; m:= v fi;
   od;
   b
end proc:
for n from 1 to 12 do
  seq(M(n,k),k=1..n)
od;

T(A123384(i),A118738(i)) = 10^i.
T(m,1) = 2^(m-1).
T(m,m) = 2^m - 1.

A379446 a(n) is the number of ones in the binary expansion of 10^(10^n).

Original entry on oeis.org

2, 11, 105, 1163, 11683, 115979, 1161413, 11606847, 116093517, 1160951533, 11609679812, 116096181467, 1160963225086

Offset: 0

Hugo Pfoertner, Dec 26 2024

Cf. A000120, A011557, A118738, A153201, A379448.

PARI
```
a379446(n) = hammingweight(5^(10^n))
```

Conjectured: Limit_{n->oo} a(n)/10^n = log(5)/log(4). (A153201)

a(12) from Markus Sigg, Dec 28 2024

A371971 a(n) is the least exponent m minimizing the percentage of set bits in the binary representation of prime(n)^m.

Original entry on oeis.org

24, 25, 12, 6, 16, 2, 10, 2, 7, 22, 16, 8, 15, 2, 14, 3, 20, 2, 4, 7, 21, 4, 19, 6, 13, 6, 7, 7, 3, 10, 2, 1, 9, 5, 2, 2, 6, 10, 36, 11, 13, 2, 2, 10, 9, 5, 2, 34, 5, 2, 4, 4, 8, 2, 9, 4, 6, 9, 3, 14, 38, 9, 14, 16, 2, 7, 4, 16, 3, 9, 6, 2, 4, 2, 10, 15, 10, 14, 6, 71, 31, 4

Offset: 2

Hugo Pfoertner, Apr 17 2024

Hugo Pfoertner, Logfile of downward searches for {3, 5, 7, 11, 13}^m, m = 10^6 .. 1.
Hugo Pfoertner, Plot of Hammingweight(3^m)/log_2(3^m), m=2..645.

Cf. A000120, A070939, A011754, A118738, A369857.

a371971(n,limit=10000) = {my(p=prime(n), q=1, m=1); for (k=1,limit, my (pp=p^k, h=hammingweight(pp), g=#binary(pp), qq=h/g); if (qq

A320394 Number of ones in binary expansion n^5.

Original entry on oeis.org

0, 1, 1, 6, 1, 6, 6, 7, 1, 9, 6, 10, 6, 11, 7, 12, 1, 10, 9, 9, 6, 11, 10, 14, 6, 11, 11, 15, 7, 15, 12, 15, 1, 10, 10, 13, 9, 11, 9, 13, 6, 16, 11, 15, 10, 21, 14, 18, 6, 15, 11, 13, 11, 13, 15, 19, 7, 17, 15, 14, 12, 14, 15, 18, 1, 10, 10, 14, 10, 18, 13, 18

Offset: 0

Vincenzo Librandi, Nov 28 2018

The binary weight of n^5.

Cf. A000120, A000584, A118738, A192085.

Magma
```
[&+Intseq(n^5, 2): n in [0..79]];
```

Table[DigitCount[n^5, 2][[1]], {n, 0, 100}]

a(n) = hammingweight(n^5); \\ Michel Marcus, Nov 28 2018

a(n) = A000120(A000584(n)).

cp's OEIS Frontend

A011754 Number of ones in the binary expansion of 3^n.

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A061785 a(n) = m such that 2^m < 5^n < 2^(m+1).

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A118737 Number of zeros in binary expansion of 5^n.

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A218451 a(n) = 10^n minus its binary weight.

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A369857 Number of nonzero bits (a.k.a. binary or Hamming weight) of 7^n.

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A371470 Triangle read by rows: for 1 <= k <= n, T(n,k) is the least sum of decimal digits of numbers with n binary digits and binary weight k.

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A373289 Triangle read by rows: for 1 <= k <= n, T(n,k) is a number with n binary digits and binary weight k whose sum of decimal digits is least; in case of a tie, choose the least such number.