A011754 -id:A011754 - cp's OEIS frontend

A001370 Sum of digits of 2^n.

1, 2, 4, 8, 7, 5, 10, 11, 13, 8, 7, 14, 19, 20, 22, 26, 25, 14, 19, 29, 31, 26, 25, 41, 37, 29, 40, 35, 43, 41, 37, 47, 58, 62, 61, 59, 64, 56, 67, 71, 61, 50, 46, 56, 58, 62, 70, 68, 73, 65, 76, 80, 79, 77, 82, 92, 85, 80, 70, 77

Offset: 0

N. J. A. Sloane, Simon Plouffe

Same digital roots as A065075 (sum of digits of the sum of the preceding numbers) and A004207 (sum of digits of all previous terms); they enter into the cycle {1 2 4 8 7 5}. - Alexandre Wajnberg, Dec 11 2005
It is believed that a(n) ~ n*9*log_10(2)/2, but this is an open problem. - N. J. A. Sloane, Apr 21 2013
The Radcliffe preprint shows that a(n) > log_4(n). - M. F. Hasler, May 18 2017
Sierpiński shows that if n >= A137284(k-1) then a(n) >= k (Problem 209). - David Radcliffe, Dec 26 2022

Archimedeans Problems Drive, Eureka, 26 (1963), 12.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Zak Seidov, Table of n, a(n) for n = 0..10000
David Radcliffe, The growth of digital sums of powers of two. Preprint, 2015.
David G. Radcliffe, The growth of digital sums of powers of two, arXiv:1605.02839 [math.NT], 2016.
W. Sierpiński, 250 Problems in Elementary Number Theory, 1970.
C. L. Stewart, On the representation of an integer in two different bases, Journal für die reine und angewandte Mathematik 319 (1980): 63-72.

Cf. sum of digits of k^n: A004166 (k=3), A065713 (k=4), A066001(k=5), A066002 (k=6), A066003(k=7), A066004 (k=8), A065999 (k=9), A066005 (k=11), A066006 (k=12).

Cf. A007953, A000079, A261009, A011754, A137284.

a001370 = a007953 . a000079  -- Reinhard Zumkeller, Aug 14 2015

seq(convert(convert(2^n,base,10),`+`),n=0..1000); # Robert Israel, Mar 29 2015

Table[Total[IntegerDigits[2^n]], {n, 0, 55}] (* Vincenzo Librandi, Oct 08 2013 *)

a(n)=sumdigits(2^n); \\ Michel Marcus, Nov 01 2013

[sum(map(int, str(2**n))) for n in range(56)] # David Radcliffe, Mar 29 2015

a(n) = A007953(A000079(n)). - Michel Marcus, Nov 01 2013

A118738 Number of ones in binary expansion of 5^n.

Original entry on oeis.org

1, 2, 3, 6, 5, 6, 7, 8, 12, 13, 11, 15, 13, 14, 17, 20, 20, 20, 24, 19, 26, 29, 25, 27, 30, 19, 31, 33, 29, 36, 37, 33, 39, 34, 42, 40, 44, 42, 38, 46, 53, 54, 49, 52, 52, 53, 50, 49, 54, 60, 58, 60, 54, 64, 58, 74, 61, 67, 74, 65, 61, 77, 74, 81, 86, 78, 87, 85, 82, 89, 83, 79

Offset: 0

Zak Seidov, May 22 2006

Also binary weight of 10^n, which is verified easily enough: 10^n = 2^n * 5^n; it is obvious that 2^n in binary is a single 1 followed by n 0's, therefore, in the binary expansion of 2^n * 5^n, the 2^n contributes only the trailing zeros. - Alonso del Arte, Oct 28 2012

Conjecture: a(n)/n -> log_4(5) = 1.160964... as n -> oo. - M. F. Hasler, Apr 17 2024

			a(2) = 3 because 5^2 = 25 is 11001, which has 3 on bits.

Robert Israel, Table of n, a(n) for n = 0..10000
Hugo Pfoertner, Plot of a(n) - 1.160964*n, +-4*sqrt(n), n up to 10^6.

Cf. A000120 (Hamming weight), A000351 (5^n), A061785 (floor(log_2(5^n))), A118737 (number of bits 0 in 5^n).

Cf. A011754 (analog for 3^n).

[&+Intseq(5^n, 2): n in [0..100]]; // Vincenzo Librandi, Nov 13 2024

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

Table[DigitCount[5^n, 2, 1], {n, 0, 71}] (* Ray Chandler, Sep 29 2006 *)

a(n) = hammingweight(5^n) \\ Iain Fox, Dec 24 2017

A118738 = lambda n: (5**n).bit_count() # For Python 3.10 and later. - M. F. Hasler, Apr 17 2024

a(n) + A118737(n) = A061785(n) + 1 for n >= 1. - Robert Israel, Dec 24 2017 [corrected by Amiram Eldar, Jul 27 2023]

a(n) = A000120(A000351(n)) = Hammingweight(5^n). - M. F. Hasler, Apr 17 2024

A078839 Numbers k such that the binary expansion of 3^k has the same number of 0's and 1's.

Original entry on oeis.org

2, 12, 69, 73, 150, 184, 252, 328, 339, 464, 483, 541, 729, 747, 758, 763, 1014, 1047, 1090, 1094, 1158, 1264, 1359, 1601, 1679, 1693, 1698, 1780, 2368, 2641, 2815, 3292, 3393, 3606, 3682, 3857, 3909, 3919, 3963, 4087, 4111, 4289, 4314, 5017, 5398, 5466

Offset: 1

Benoit Cloitre, Dec 06 2002

Does the limit of a(n)/n^2 as n -> infinity exist?

Hugo Pfoertner, Table of n, a(n) for n = 1..1600 (terms 1..1000 from Amiram Eldar)
Amiram Eldar, Plot of a(n)/n^2 for n = 1..1000
Hugo Pfoertner, Plot of a(n)/n^2 using Plot 2.

Cf. A000244, A004656, A011754, A031443.

balanced[n_] := Module[{d=IntegerDigits[n, 2]}, Plus@@d==Length[d]/2]; Select[Range[0, 5500], balanced[3^# ]&]

is(n)=hammingweight(n=3^n)==hammingweight(bitneg(n,#binary(n))) \\ Charles R Greathouse IV, Mar 29 2013

A261009 Write 2^n in base 3, add up the "digits".

Original entry on oeis.org

1, 2, 2, 4, 4, 4, 4, 6, 4, 8, 8, 10, 10, 8, 10, 16, 12, 14, 12, 16, 14, 18, 16, 12, 10, 12, 14, 20, 20, 22, 24, 26, 24, 22, 22, 22, 18, 20, 26, 28, 28, 28, 26, 30, 30, 30, 26, 26, 26, 32, 38, 40, 38, 38, 28, 34, 40, 42, 38, 40, 46, 40, 38, 42, 48, 44, 42, 40, 42, 48, 48, 44

Offset: 0

N. J. A. Sloane, Aug 14 2015

Comment from Jean-Paul Allouche, Oct 25 2015: As mentioned by Holdum et al. (2015) the following problem, cited in "Concrete Mathematics" by Graham, Knuth, and Patashnik (1994), is still open: prove that for all n > 256, binomial(2n,n) is either divisible by 4 or by 9 (cf. A000984). This can be easily reduced to show that, for all k >= 9, 2*a(k) - a(k+1) >= 4. This has been proved up to huge values of k (Holdum et al. mention k = 10^{13}).

For additional information about the divisibility of binomial(2n,n) by squares see the comments and references in A000984, - N. J. A. Sloane, Oct 29 2015

			2^7 = 128_10 = 11202_3, so a(7) = 1+1+2+0+2 = 6.

Giovanni Resta, Table of n, a(n) for n = 0..10000
Cernenoks J., Iraids J., Opmanis M., Opmanis R., Podnieks K., Integer complexity: experimental and analytical results II, arXiv:1409.0446 [math.NT] (September 2014)
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994
Sebastian Tim Holdum, Frederik Ravn Klausen, Peter Michael Reichstein Rasmussen, Powers in prime bases and a problem on central coefficients, Integers 15 (2015), #A43
K. Podnieks, Digits of pi: limits to the seeming randomness, arXiv:1411.3911 [math.NT], 2014.

Cf. A000079, A000984, A053735, A007089.

Sum of digits of k^n in base b for various pairs (k,b): A001370 (2,10), A011754 (3,2), A261009 (2,3), A261010 (5,3).

a261009 = a053735 . a000079  -- Reinhard Zumkeller, Aug 14 2015

S:=n->add(i,i in convert(2^n,base,3)); [seq(S(n),n=0..100)];

Table[Total@ IntegerDigits[2^n, 3], {n, 0, 100}] (* Giovanni Resta, Aug 14 2015 *)

a(n) = vecsum(digits(2^n, 3)); \\ Michel Marcus, Aug 14 2015

a(n) = A053735(A000079(n)). - Michel Marcus, Aug 14 2015

A372097 Exponents k where A000120(3^k) - A070939(3^k)/2 reaches a new minimum.

Original entry on oeis.org

0, 2, 4, 7, 16, 24, 40, 49, 53, 102, 104, 126, 174, 226, 379, 768, 831, 832, 1439, 1452, 1914, 2291, 2731, 3000, 3363, 3472, 5608, 5883, 6725, 6787, 7438, 8786, 10280, 11948, 12190, 13135, 15170, 15645, 22407, 26232, 27099, 32773, 33085, 40189, 40523, 48068, 51187

Offset: 1

Hugo Pfoertner, Apr 25 2024

nonn

These are the k-values of the lower envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number. The corresponding negated differences are given in A372098.

Hugo Pfoertner, Table of n, a(n) for n = 1..99
Hugo Pfoertner, Illustration of scatter band bounded by lower and upper records, up to exponents k=8*10^6.

Cf. A000120, A000244, A011754, A070939, A078839, A372098, A372099, A372100.

a372097(upto) = {my (dm=-oo); for (k=0, upto, my (p=3^k, h=hammingweight(p), b=#binary(p)/2,d=b-h); if (d>dm, print1(k,", "); dm=d))};
a372097(60000)

A372098 a(n) = A070939(3^k) - 2*A000120(3^k) with k = A372097(n).

Original entry on oeis.org

-1, 0, 1, 2, 4, 7, 8, 12, 15, 18, 25, 26, 30, 51, 75, 78, 84, 129, 133, 148, 170, 180, 183, 189, 209, 265, 279, 285, 287, 336, 369, 388, 406, 412, 445, 469, 496, 581, 711, 737, 741, 742, 873, 939, 994, 1044, 1078, 1111, 1157, 1158, 1492, 1636, 1767, 1914, 1933

Offset: 1

Hugo Pfoertner, Apr 25 2024

sign

a(n)/2 are the negated differences at supporting points of the lower envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number.

Hugo Pfoertner, Table of n, a(n) for n = 1..99

Cf. A000120, A000244, A011754, A070939, A078839, A372097, A372099, A372100.

A372099 Exponents k where A000120(3^k) - A070939(3^k)/2 reaches a new maximum.

Original entry on oeis.org

0, 1, 3, 5, 11, 27, 71, 119, 140, 158, 198, 218, 441, 537, 538, 868, 1092, 2128, 2294, 2343, 2811, 2911, 3849, 4003, 4655, 5079, 5279, 5920, 6269, 6603, 10181, 10574, 12801, 12803, 15563, 15784, 16054, 16253, 17127, 18257, 20187, 21934, 34633, 49209, 76791, 78938

Offset: 1

Hugo Pfoertner, Apr 25 2024

nonn

These are the k-values of the upper envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number. The corresponding differences are given in A372100.

Hugo Pfoertner, Table of n, a(n) for n = 1..96
Hugo Pfoertner, Illustration of scatter band bounded by lower and upper records, up to exponents k=8.5*10^6.

Cf. A000120, A000244, A011754, A070939, A078839, A372097, A372098, A372100.

a372099(upto) = {my(dm=oo); for (k=0, upto, my (p=3^k, h=hammingweight(p), b=#binary(p)/2, d=b-h); if (d

A372100 a(n) = 2*A000120(3^k) - A070939(3^k) with k = A372099(n).

Original entry on oeis.org

1, 2, 3, 4, 8, 17, 23, 29, 38, 39, 44, 56, 57, 58, 91, 114, 145, 147, 156, 168, 182, 208, 219, 239, 277, 297, 300, 307, 331, 360, 367, 442, 452, 477, 487, 492, 507, 513, 568, 571, 614, 893, 963, 1275, 1283, 1288, 1440, 1563, 1702, 1957, 2019, 2440, 2471, 2566, 3004

Offset: 1

Hugo Pfoertner, Apr 25 2024

nonn

a(n)/2 are the differences at supporting points of the upper envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number.

Hugo Pfoertner, Table of n, a(n) for n = 1..96

Cf. A000120, A000244, A011754, A070939, A078839, A372097, A372098, A372099.

A261010 Write 5^n in base 3, add up the "digits".

Original entry on oeis.org

1, 3, 5, 7, 7, 9, 9, 13, 15, 13, 13, 17, 19, 21, 21, 27, 25, 25, 25, 23, 27, 33, 31, 39, 35, 45, 37, 57, 45, 47, 45, 45, 53, 47, 55, 51, 57, 59, 67, 67, 69, 65, 67, 65, 71, 79, 71, 65, 67, 75, 65, 71, 73, 83, 69, 79, 81, 85, 79, 89, 87, 95, 89, 85, 97, 99, 93, 101, 107

Offset: 0

N. J. A. Sloane, Aug 14 2015

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Cernenoks J., Iraids J., Opmanis M., Opmanis R., Podnieks K., Integer complexity: experimental and analytical results II, arXiv:1409.0446 [math.NT] (September 2014)
K. Podnieks, Digits of pi: limits to the seeming randomness, arXiv:1411.3911 [math.NT], 2014.

Sum of digits of k^n in base b for various pairs (k,b): A001370 (2,10), A011754 (3,2), A261009 (2,3), A261010 (5,3).

S:=n->add(i,i in convert(5^n,base,3)); [seq(S(n),n=0..100)];

def digits(n, b=10): # digits of n in base 2 <= b <= 62
    x, y = n, ''
    while x >= b:
        x, r = divmod(x,b)
        y += str(r) if r < 10 else (chr(r+87) if r < 36 else chr(r+29))
    y += str(x) if x < 10 else (chr(x+87) if x < 36 else chr(x+29))
    return y[::-1]
def A261010(n):
    return sum([int(d) for d in digits(5**n,3)]) # Chai Wah Wu, Aug 14 2015

A364650 Number of powers of 3 whose binary representation contains exactly n 1's.

Original entry on oeis.org

1, 2, 1, 1, 1, 3, 0, 1, 1, 1, 2, 0, 1, 3, 1, 1, 2, 1, 1, 1, 0, 1

Offset: 1

Pontus von Brömssen, Jul 31 2023

Number of numbers k >= 0 such that A011754(k) = n.
Senge and Straus prove that a(n) is finite for all n.
After a(22), the sequence undoubtedly continues 0, 1, 3, 2, 1, 1, 1, 1, 0, 2, 1, 4, 1, 1, 0, 2, 4, 1, 2, 3, 0, 0, 2, 1, 1, 1, 1, 0, ..., but there seem to be proofs only for the first 22 terms (Dimitrov and Howe).

			There are a(6) = 3 powers of 3 that have exactly 6 binary 1's: 3^5 (11110011 in binary), 3^6 (1011011001), and 3^8 (1100110100001).
There is no power of 3 with exactly 7 binary 1's, so a(7) = 0.

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.
H. G. Senge and E. G. Straus, PV-numbers and sets of multiplicity, Periodica Mathematica Hungarica 3 (1973), 93-100.

Cf. A011754.

cp's OEIS Frontend

A001370 Sum of digits of 2^n.

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A118738 Number of ones in binary expansion of 5^n.

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A078839 Numbers k such that the binary expansion of 3^k has the same number of 0's and 1's.

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A261009 Write 2^n in base 3, add up the "digits".

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A372097 Exponents k where A000120(3^k) - A070939(3^k)/2 reaches a new minimum.

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A372098 a(n) = A070939(3^k) - 2*A000120(3^k) with k = A372097(n).

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A372099 Exponents k where A000120(3^k) - A070939(3^k)/2 reaches a new maximum.

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PARI